text:
User talk:Mike.lifeguard/Archive 14

History
Thanks for the invite, and the merge. The content was worth preserving in some way, I will try to 'bookify' it. I have a couple of other ideas for new books, but will find my around way here by adding to current content. See you around, Cygnis insignis 05:52, 2 March 2008 (UTC)
 * I looked around for a model to fix the transwikied page, I tried to fix the links in this one A Wikimanual of Gardening/Digitalis. It appears that something, maybe the circular ref in the Pest and diseases section, is breaking the code further down the page. Let me know if you see what the problem is. Cygnis insignis 06:56, 2 March 2008 (UTC)
 * Looks like the problem was a transcluded redirect, which I fixed. – Mike.lifeguard  | talk 07:06, 2 March 2008 (UTC)
 * Cheers. I made a bit of start on Cephalotus, I will read some more of the help pages and have another go later. I have noted that there is a project for this book, a useful quick start. Regards, Cygnis insignis 15:00, 2 March 2008 (UTC)

Naming Conventions
Hello Mike. The Naming Convention list is looking good! Was going to try and clear it down some more. What is the plan with the programming books? Were we going to rename them without a : in the title? --AdRiley (talk) 12:28, 3 March 2008 (UTC)
 * Yeah, I've been doing Programming:Language to Language Programming, and nobody's yelled at me yet :) – Mike.lifeguard  | talk 14:21, 3 March 2008 (UTC)

GMAX and Trainz book
GMAX is free software used to create 3d Models. Widely used in Trainz, a computer simulator for trains and also Microsoft Flight Simulator. Trainz has a tradition of user created content, roughly 100,000 free items at the last count.

The Trainz book has a number of objectives the part I'm concentrating on is tutorials for creating content. Since ideas evolve as does the software Wikibooks is ideal since corrections can be added to tutorials at a later date.

I should update the book to add the source for GMAX and the exporters. All my images are screen shots created by myself and hopefully I have tagged them correctly. I understand that screen shot images are more expensive to store and download than text but in this case there is very little alternative.

Cheerio John —The preceding unsigned comment was added by Johnwhelan (talk • contribs) 22:15, March 3, 2008.
 * There's no problem with the screenshots except they have no rationale. You can probably get away with copy-pasting a few over all the images since they're all used in basically the same way. You'll need one for the Windows screenshots, one for the GMAX screenshots, and one for the rest. If you do one image of each with an appropriate rationale, I'll fix the rest for you. – Mike.lifeguard  | talk 02:38, 4 March 2008 (UTC)

I've just uploaded resize4.jpg and think I've put in a correct rationale. Could you verify this for me. I'm more than happy to edit the others but at the moment although I can see the description in the file list I'm unable to see how to edit it. Cheerio John
 * I fixed it up a bit. The rationale had the wrong module and was missing |not free= and I added . If you want to do them all yourself, go for it. Again, you can probably copy-paste what's on this one onto the rest of the screenshots with minimal modification (ie the module its used in might have to change, but that's it).
 * As well, you have to edit the image description pages - don't reupload the images, as that won't work. Just hit the edit tab to edit the page, copy in the code from Image:Resize4.jpg, modify it as necessary and hit save. Hope this helps! – Mike.lifeguard  | talk 17:41, 4 March 2008 (UTC)

If I go special pages, list files I can see the files with their description, I don't see an edit tab to edit the description. If I look at the book I can edit the page on which the screen shots appear but that doesn't allow me to edit the description as far as I can see. Since I wasn't certain where you are located and I'm used to an active forum where questions get answered by people in different time zones I posted in the technical help bit as well. I'm learning about wikibooks slowly. Could you give me a link where I can see my files and edit the descriptions.

Thanks John
 * If you click the image on any page where it appears, you'll go to the description page. Also, you can find links to the image description page in Special:Contributions/Johnwhelan or your upload log.
 * Please sign your posts with four tildes ( ~ ), which will leave a signature including a timestamp; this makes discussions easier to follow. Thanks! – Mike.lifeguard  | talk 21:49, 4 March 2008 (UTC)

Think I've caught them all, is there any way you can check to see if I missed any?

Thanks John Johnwhelan (talk) 23:54, 4 March 2008 (UTC)
 * Missed a few: Image:Edit in explorer.jpg, Image:Folder image.jpg, Image:Msedit2.jpg, Image:Upload.jpg. The rest look good though. – Mike.lifeguard  | talk 19:03, 5 March 2008 (UTC)

I'm afraid I haven't worked out how to delete unwanted images so since you've nominated them for deletion I think we'll just let the process delete them. Thanks John

Johnwhelan (talk) 01:00, 8 March 2008 (UTC)
 * Only admins can delete things. If you're going to use them, you can just remove the delete template and fix up the rationale with whatever module they're used in. If you're sure you don't need them, I'll go ahead and delete them. – Mike.lifeguard  | talk 01:06, 8 March 2008 (UTC)

NowCommons images
G'day Mike, it's generally a good idea to leave images that have been moved to Commons for a few days, to make sure that they've "taken", as Commons sometimes rejects the images. I know I was the last uploader for some of those images, but Commons might yet create a fuss about me recreating those images based on unlicensed images. (I'd push back, of course :-) Don't do anything with the ones you've deleted, but please consider giving them a few days before deleting the local copy next time. cheers, Webaware talk 09:16, 5 March 2008 (UTC)
 * I did check at commons to make sure licensing was ok, and kept the ones where it wasn't, so all is well. Thanks for the reminder though. – Mike.lifeguard  | talk 18:59, 5 March 2008 (UTC)

picture
I wanted to use a picture. I used a family photo. I am not sure what I am missing. I'm not very familiar with Wiki. Can you help me? Christa Portlock (ck 209) —The preceding unsigned comment was added by Ck209 (talk • contribs) 19:52, March 5, 2008.
 * If you didn't take the picture, you can't release it under any license since you don't own the copyright. You can find other images like this which are free to reuse at Wikimedia Commons. If you did take it, please add the code below, filling in the requested information.

 * As well, if you can re-upload the image at higher quality, it'd be nice. Thanks! – Mike.lifeguard  | talk 23:59, 5 March 2008 (UTC)

Userpage
Hey Mike, I borrowed your userpage navbox idea. Hope you don't mind. Νεοπτόλεμος ( talk | email | contribs ) 03:06, 6 March 2008 (UTC)
 * Not at all :) – Mike.lifeguard  | talk 04:37, 6 March 2008 (UTC)

OpenOffice.Org/Prelink "blanking"
Thanks, that's what I'm searching for (the /delete|reason/ tag). I'm not used to these wikis tags. The Kid (talk) 19:04, 7 March 2008 (UTC)
 * [[Image:Symbol delete vote.svg|15px]] Deleted – Mike.lifeguard  | talk 19:13, 7 March 2008 (UTC)

Getting in contributors for the book
I have been contributing and will be contributing to the inorganic chemistry book here on Wikibooks. The problem that faces me is that I cannot submit Polished content in a single edit! I have been browsing around wikibooks for quite a time and have been really discouraged to see the lack of enthusiasm in science textbooks. I suspect this is due to the sparse contributions.

Let's just hope that if i keep contributing, collaborators will join in. It is here that I have a little query - how do i announce that this book is undergoing continuos editing? We are not talking about announcing it as a new book - because it has almost always been there. Maybe more like, sending out a sort of "Volunteers wanted" advert!

Here is the book in question (module actually) : http://en.wikibooks.org/wiki/Inorganic_Chemistry/Qualitative_Analysis/Tests_for_anions

Infact, you must have already een it once given that you messaged me about not blanking out pages.

Thanks! Hope to get a reply soon! —The preceding unsigned comment was added by Thewinster (talk • contribs) 16:32, March 7, 2008.
 * Yes, contributors are the #1 thing we need. You can try to spread the word at various WikiProjects at Wikipedia. Besides that, you can add, though that's not really accurate since you're working on it. Most books have no contributors; most books that are being edited usually have only one contributor, so you're far from the only one working on a book alone. Sorry I can't help more. – Mike.lifeguard  | talk 21:09, 7 March 2008 (UTC)
 * That was all I needed! And when it comes to science textbooks, I guess the only reason of inactivity is that nobody would want to start from scratch. I'd let this work develop into a few lectures, so that anyone who refers them might voluntarily organize it. That is what wiki is all about! Thanks again.

--Thewinster (talk) 16:55, 8 March 2008 (UTC)

Mike
I created this wikibook last August and am now just breaking it up into chapters as it appears it will be easier to read and download. I am not very up on all of the terms and such. It is my original material. Let me know if I should be doing something. rclarkenRod (talk) 00:34, 9 March 2008 (UTC)
 * I figured as much. All I wanted was for you to use the edit summary when removing large amounts of text. Something like "moving to subpage" is fine. Thanks. – Mike.lifeguard  | talk 00:36, 9 March 2008 (UTC)

Bloom clock update
Hi Mike. Time for another bot run on the clock.

For all pages in v:Category:Bcptop version 7.10.24, replace lines in the top template as follows:

 * Grab:
 * version tracking for top template (for robotic use) = 7.10.24

 * And replace with:
 * Included Subclasses =
 * Global temperate seasons =
 * version tracking for top template (for robotic use) = 8.3.9

Thanks as always! I'll need a second run to update the bottom template sometime next week too, I'll let you know when the fields are ready.-- SB_Johnny | PA! 13:54, 9 March 2008 (UTC)

 * Oh, oops. Minor issue there: you'll only find the template in pages in the BCP/ pseudonamespace. The top template transcludes categories to the keys, so you'll have to ignore pages in the Bloom Clock/ pseudonamespace (they're just transclusion lists of BCP pages). -- SB_Johnny | PA! 13:59, 9 March 2008 (UTC)

 * Also, there are a small number of pages using template version 8.3.3 (v:Category:Bcptop version 8.3.3, but I'll manually update those.
 * [[Image:Symbol wait.svg|15px]] In progress – Mike.lifeguard  | talk 16:32, 9 March 2008 (UTC)
 * [[Image:Yes check.svg|15px]] Done and all looks good. – Mike.lifeguard  | talk 17:28, 9 March 2008 (UTC)

LaleenaWiki
Hello - just thought you'd want to know that I lost my password to Laleena, so I'm now LaleenaWiki. Thanks, LaleenaWiki (talk) 18:42, 9 March 2008 (UTC)
 * Yes, I saw that. Can you not recover the password via email? – Mike.lifeguard  | talk 18:43, 9 March 2008 (UTC)
 * No, I can't. I don't put up an e-mail address, for fear of a hacker giving me a virus through the e-mail. LaleenaWiki (talk) 18:48, 9 March 2008 (UTC)

glossary edit
you left comment, I had tried to crudely draft a glossary list as I saw it didn' seem to be there - I added the "stub" marker ...just trying to help--Billymac00 (talk) 02:45, 10 March 2008 (UTC)
 * Yup, that's fine. – Mike.lifeguard  | talk 18:43, 11 March 2008 (UTC)

Thank You!:-D
Thank you for the compliment! You made my day! Hhhmmm, I usually specialize in layout, I don't do much in aethetics, I usually leave that to someone else, and I'm not a deletionist, so that's a bit problematic for me. Thanks though, we'll keep in touch.68.148.164.166 (talk) 02:48, 10 March 2008 (UTC)

 * Hey, saw your userpage on Wikipedia, and said that your more active here. ASL?68.148.164.166 (talk) 02:48, 10 March 2008 (UTC)

Question About NPOV
You left a comment (related to NPOV) on my book "Sign for Selecting...". I left a message in return, but I am confused as to how to resolve this issue. I am also concerned with upcoming postings, because this book will offer a certain point of view, though it is authentic to the Bahai religion. This is stated in the introduction and About Author section. Could you please email me at rclarken@nmu.edu as I check email more frequently. Thank you. Rod (talk) 18:56, 11 March 2008 (UTC)
 * emailed. – Mike.lifeguard  | talk 20:33, 11 March 2008 (UTC)

Dragons
Hello again, you helped me a little while ago with a plant how-to. I have found a couple more candidates, but want to do things the right way round this time ;) I don't suppose I can move page histories from wikipedia, but let me know what else I can do when I find these things. Shall I just ask you or is there a list or page to add them to? The article I want to move is Bearded Dragon - another 'how-to' mostly. I haven't checked it for copyvio, but it seems to have been added incrementally. The book that might contain this page is Animal Care.

It is a popular page (stats), but I won't be when I gut it! However, the traffic will come here, dragon owners will find what they are looking for, and I am free to fix the article. Please also let me know how I might further improve A Wikimanual of Gardening/Cephalotus. Thanks, Cygnis insignis 19:32, 12 March 2008 (UTC)
 * You can list stuff that needs to be moved here at WB:RFI, and an admin will do it. I've imported it, and it's now here. As for WMOG, you should ask User:SB Johnny. – Mike.lifeguard  | talk 19:44, 12 March 2008 (UTC)
 * Cheers, I will start fixing it up soon. And I will introduce myself to that user. Regards, Cygnis insignis 19:58, 12 March 2008 (UTC)

Categorization
What category would my wikibook fall into? Category:Writing? Category:Fiction? Category:Literature? I don't see a list of the categories we have. Cilantrohead (talk) 06:24, 14 March 2008 (UTC)
 * Adding should be fine.  – Mike.lifeguard  | talk 06:41, 14 March 2008 (UTC)

United States Government/Domestic Issues
A stub is just a stub, if the edit is on topic marking it as a doesn't seem productive to me, only if you have the intention in removing (as in speed delete) all stubs, if so I have a problem with that... (no need to reply to this if you had not a particular reason to tag it, just be careful that this can be destructive) --Panic (talk) 23:41, 14 March 2008 (UTC)
 * No, I just wondered what was meant there as it's a series of headings (or was at the time). As the template says, remove it if the module is legit. – Mike.lifeguard  | talk 23:46, 14 March 2008 (UTC)

Image:ReverseEngineeringPush.JPG
Hi, saw your message on my Talk page just now. I was trying to replace the above Jpeg with one I made, but after uploading all I saw was the old image resized to match the new one's shape. Also, a thumbnail with the correct image appeared on the original author's 'Images' page. Should I upload it by itself and replace all references? Gannalech (talk) 09:42, 15 March 2008 (UTC)
 * That's very strange. I'm going to try reverting to the version you made. If that works, just update the information on the image page. If not, then you might as well upload a separate image and use that one. I'll let you know what happens. – Mike.lifeguard  | talk 14:32, 15 March 2008 (UTC)
 * Looks like it worked; please confirm. – Mike.lifeguard  | talk 14:36, 15 March 2008 (UTC)

WB:PATROL
Thanks for your offer to become a patroller. After considerable thought, I am not sure I can really commit the time at the moment. I feel that this is a significant responsibility (e.g. verifying copyright violations etc), and to do it appropriately could use up a lot of effort. Given that my time contributing is limited by RL, I am concerned this might/would take away time from my main project at Radiation_Oncology. What I plan to do over next couple of weeks is to take a look at new pages, and see how it goes. Also, I know there is an issue over at wikipedia about the "inclusionists" vs. the "deletionists", and the potentially negative impact the deletionists are having on the dropping rate of contributions. What's the overall wikibooks stance thus far? Thanks! Tdvorak (talk) 23:35, 16 March 2008 (UTC)
 * I can't speak for Wikipedia, since I barely contribute there. I think the Wikibooks community is more stable, in part because we're smaller, but also because we're a "spinoff" of Wikipedia, so we had to figure out what should be included ans what shouldn't fairly early on. Of course, there has been back-and-forth on certain issues like game guides, but overall I don't think we have the same strain between the two camps that Wikipedia has. Nevertheless, what happens at Wikipedia doesn't tend to affect us much here since we are a different project. The only way I can see the inclusion/exclusion thing affecting us is by how much influx of content we have, since they like to dump their unwanted stuff here more often than perhaps we'd like.
 * As for being a patroller, no problem there. If you feel like you want to help out I'd be willing to add your name to WB:RFP.
 * By the way, you've done wonders with Radiation Oncology, and I hope you'll continue for some time. Keep it up! – Mike.lifeguard  | talk 00:30, 17 March 2008 (UTC)

Bureaucratification
Wanted to make sure that you knew my vote opposing your bureaucratship was not in malice. The rights really aren't much and are used quite rarely around here. Keep rocking the 'books. -within focus 00:25, 17 March 2008 (UTC)
 * I certainly don't think it was. It did solidify a few thoughts I had, hence the withdrawal. See you 'round – Mike.lifeguard  | talk 00:32, 17 March 2008 (UTC)

How to rename a Book
I need to rename a book "How to Repar Dishwashers", but have not been able to use this procedure:

http://en.wikibooks.org/wiki/Help:Editing_FAQ

Q: How can I easily change the title of a book? I keep browsing around in the help section but I can't seem to find a way to change the title of my book. Can anyone help?

A: Use the MOVE button on top of the page to change the title of your book

Because I do not see a "MOVE" button in either the display or edit mode for the root page. Can you tell me how to rename this book? I have tried inserting "Rename|How To Repair Dishwashers" (with the brackets), but nothing has happened. Thank you DIYprojects (talk) 01:57, 17 March 2008 (UTC)
 * There should be a move tab at the top of the page, beside where you find the edit tab. If it's not there, then it means you're not allowed to move the page (your account must be 4 days old to move pages). – Mike.lifeguard  | talk 02:44, 17 March 2008 (UTC)

Thank you- I have 3 days to go. DIYprojects (talk) 04:28, 17 March 2008 (UTC)
 * I think I ended up moving everything for you. Let me know if I missed something. – Mike.lifeguard  | talk 20:11, 17 March 2008 (UTC)

title: User talk:Mike.lifeguard/Archive 14


text: User talk:165.139.138.1

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User talk:165.139.138.1


text: User talk:216.11.243.60

title: User talk:216.11.243.60


text:
User talk:87.122.93.147

Welcome!
Welcome to Wikibooks, a collection of open-content textbooks! You don't have to log in to read or edit articles on Wikibooks, but creating an account is quick, free and non-intrusive, requires no personal information, and gives you many benefits, including:
 * The use of a username of your choice
 * The ability to view all your contributions via a "My contributions" link
 * Your own user page
 * Your own talk page which, if you choose, also allows users to send you messages without knowing your e-mail address
 * The use of your own personal watchlist to which you can add articles that interest you
 * The ability to rename pages.
 * The ability to upload images
 * The ability to customize the appearance and behavior of the website
 * The eligibility to become an administrator
 * The right to be heard in votes and elections
 * Your IP address will no longer be visible to other users.

We hope that you choose to become a Wikibookian and create an account. Feel free to ask me any questions you may have on my talk page. By the way, make sure to sign your posts and comments with four tildes (~~~~), which will let others know who left it.

Νεοπτόλεμος ( talk | email | contribs ) 19:42, 11 March 2008 (UTC)

title: User talk:87.122.93.147


text: User talk:128.97.205.225

title: User talk:128.97.205.225


text: User talk:216.48.136.60

title: User talk:216.48.136.60


text: User talk:163.6.107.10

title: User talk:163.6.107.10


text:
User talk:89.176.28.156

Join Us!
--Webaware talk 22:04, 11 March 2008 (UTC)

Turns out I already had an account and didn't even realise it. -- Dooglus (talk) 23:15, 11 March 2008 (UTC)

title: User talk:89.176.28.156


text:
Talk:US History/Secession

NPOV
This section needs a serious and thorough rewrite. Article is written with large pro-South bias. 140.247.40.201 06:47, 17 April 2006 (UTC)

i agree there needs to be re edititd by a Nutral party like a proffeser

...who kin spel gude.

I teach American history but will make no contribution, since my contributions to the Kant and Adorno essays were taken from me, incorporated in the current text which will be sold as soon as Wales thinks he can sell it, and then my user status was destroyed by AmerindianArts et al. because I don't take crap from convenience store clerks.

This facility is beyond hope and Wikipedia/media needs to be put of business, on the basis of tax law alone: it is an .org that means to be a .com and a conspiracy of insiders to make money by stealing the unpaid intellectual production of unwitting volunteers, and destroying their reputations if they question the insiders.

Jimbo Wales is a redneck with a redneck dream of making his fortune by way of slavery. That the slavery of "open content" is virtual time-sliced makes no difference. This is why this odious "textbook" downplays the cause of the civil war: slavery.

See you in prison, Wales.

Edward G. Nilges

Southern Bias
The whole section seems to have a Southern bias, especially under Jefferson Davis. I'll try and fix it up after the AP exams are over. I've also never really edited anything, but there's always a start. Actually, I don't see the real purpose to the section. It doesn't fit at all with the rest.Sameerkale 00:25, 5 May 2006 (UTC)

 * In reading through this, I would admit that there is a bit of an NPOV problem here, but at the same time this is an issue that has not really been effectively dealt with in American history and is an interesting point to bring up: under what sort of circumstances can a state be capable of leaving.  IMHO this has some modern implications, particularly with the vote for statehood by Puerto Rico where they would like the option of being able to leave the federal union if the future relationship between them and the rest of the United States becomes politically difficult.  The "one way" trip into statehood is certainly a detractor to joining America as a state. And I say this not as a southerner, but rather somebody whose ancestors either fought for the North or ditched the whole fight altogether by heading to the western frontier. --Rob Horning 20:42, 6 March 2007 (UTC)

Every state has the natural right of secession: cf. Hobbes, Leviathan. It has this right in precisely the same way a citizen has the natural right to rob a convenience store: cf. Hobbes, Leviathan. It has this right in precisely the same way John Law has the right to detain a convenience store armed robber, using deadly force if needed: cf. Hobbes, Leviathan. It has this right in precisely the same way Abraham Lincoln, the greatest President the United States ever had, had the right to unleash Ulysses S. Grant on you boys: cf. Hobbes, Leviathan.

But no state has more than this natural right. The United States is a sovereign nation with rights under the charter of the United Nations to enforce its sovereignity using deadly force as needed.

You white trash think you're cute. You might be interested to know, especially if you think you're American patriots, that a KGB agent is predicting the break-up of the United States by 2010 owing to economic and moral decline, and you rednecks are an example of the latter. A world without the United States is a world of gangsters such as your filthy, slave-driving ancestors, who have always been the worst element in the USA.

Edward G. Nilges

Simple logic dictates that sovereignity attaches either to the states or the federal government. It can't be shared: a house divided against itself cannot stand. Bullshit artists like to evade this question: thus the bullshit artists of the European Union cannot answer the question of secession, but a real man in Brussells would raise troops if France tried to leave if he wanted the EU to be taken seriously thereafter: and Lincoln, unlike you BOYS was a MAN.

I'm serious. He'd levy Dutchmen, fantastical Spaniards, and Germans, and take Paris IF he wanted the EU to be a functioning entity, just as Lincoln raised troops.

This is NOT "tyranny". Lincoln knew that secession gets people killed and destroys economies.

You people seem to me to have serious problems with your fathers, and the Father's demands which you seem to have failed to meet. Thus you seriously open your yaps and repeat the Northern coward's "erring sisters depart in peace" policy as if Lincoln or Seward could do so and remain in office, much less end slavery, which is what they did (what part of better angels of our nature don't you understand?)

The fact is that there won't be world peace until we have a world government with an armed and dangerous United Nations, and this horrifies you because it's what Slavoj Zizek, the Slovenian philosopher calls the big other. You are Mama's boys who are wasting your lives doing slave work for Jimmy Wales because you are in permanent rebellion, as a bunch of wastrels and fools, against the Father.

Edward G. Nilges

title: Talk:US History/Secession


text:
US History/Secession

Secession and the Southern Confederacy
With the demise of the Whig Party and the split of the Northern and Southern branches of the Democratic Party, the opportunity afforded itself for the recently organized Republican Party to increase its political power in both chambers of Congress and to successfully elect Abraham Lincoln to the Presidency. Wendell Phillips acknowledged that the Republican Party was "a sectional party, organized against the South." Several other leading Republicans even went so far as to advocate civil war in order to keep the Southern States in a condition of subordination to a Northern majority.

Tariffs
Southern leaders, such as John C. Calhoun, had warned that if the North ever gained control of the federal Government the rights of the Southern people would be lost. In the Republicans' pledge to confine slavery within the existing States and to prevent its spread into the common Territories was perceived an intent to destroy the rights of the Southern people wholesale. Many Republicans, such as the former Whig and Henry Clay admirer, Abraham Lincoln, also openly advocated a high tariff and internal improvement system (which Clay had named, "The American System"). Historically, high tariffs benefited Northern industry and had adverse effects on the price of exported Southern cotton.

Why They Fought
Consequently, the conflict between the North and the South had much more to do with differing views on the relation of the States to the federal Government, the extent of State power, and economics rather than the issues of slavery or African-American rights. In fact, some of the Northern people deplored Abolitionism and were opposed to African-American equality. Even Lincoln openly declared himself in opposition to African-American citizenship. Most of the Northern States had various anti-African American laws on the books and Lincoln's own State of Illinois altered its constitution in 1862 to prohibit the immigration of free African Americans entirely.

Secession
Upon receiving news of Lincoln's election, the South Carolina Convention voted for secession on December 20, 1860. In the next few months, Mississippi, Florida, Alabama, Georgia, Louisiana, and Texas had all seceded and joined South Carolina in forming the Confederate States of America. The other four Southern States - Virginia, North Carolina, Tennessee, and Arkansas - originally voted against secession, but later joined the Southern Confederacy when Lincoln's call for 75,000 militia was issued on April 15, 1861.

Definition
Secession was generally accepted as a revolutionary, if not a constitutional right, by both North and South prior to the actual secession of the seven Gulf States. In fact, secession was first threatened in the early years of the Union by the State of Massachusetts, and the threat was repeated several times over the decades preceding the War Between the States. A Northern Confederacy of the New England States was proposed and nearly formed in protest of the War of 1812. Of course, Southern leaders such as Jefferson Davis believed that since the original thirteen States had voluntarily acceded to the Union, they could also rescind that accession and lawfully secede. This act of secession was to be voted upon and declared to the world by the same sovereign power which had brought the State into the Union - that of the people assembled in convention. According to this logic, those States which were admitted to the Union after 1789 also retained this right of secession, since the main ground of their admission was that they would stand "on equal footing" with the other States. The Tenth Amendment to the Constitution was also appealed to with the claim that the several States never surrendered their sovereignty to the federal Government, and could therefore recall their delegated powers from their common agent by withdrawing from the Union.

Lincoln's View
Abraham Lincoln, on the other hand, insisted that the relation of the States to the Federal Government was akin to that of counties to States. He believed that the Union preceded the States, rather than vice versa, and that State sovereignty was a myth. Consequently, secession was treason and could only result in anarchy. For these views, he relied upon Daniel Webster's speeches in the Senate in the early 1830s.

Jefferson Davis
Jefferson Davis (June 3, 1808 – December 6, 1889) was an American soldier and politician, most famous for serving as the first and only President of the Confederate States, leading the Confederate States of America during the American Civil War. Before the Civil War, Davis served in the Mississippi Legislature, the U.S. House and U.S. Senate. He fought in the Mexican-American War as a colonel of a volunteer regiment. Later he became Secretary of War in the cabinet of U.S. President Franklin Pierce.

Histoire des États-Unis d'Amérique/La guerre de sécession

title: US History/Secession


text:
User:Dooglus

Currently working through Blender_3D:_Noob_to_Pro and trying to improve it as I go.

title: User:Dooglus


text:
Linear Algebra/Definition of Vector Space

We shall study structures with two operations, an addition and a scalar multiplication, that are subject to some simple conditions. We will reflect more on the conditions later, but on first reading notice how reasonable they are. For instance, surely any operation that can be called an addition (e.g., column vector addition, row vector addition, or real number addition) will satisfy all the conditions in Definition 1.1 below.


text:
User:Dev789

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title: User:Dev789


text:
Linear Algebra/Linear Independence

The prior section shows that a vector space can be understood as an unrestricted linear combination of some of its elements— that is, as a span. For example, the space of linear polynomials $$\{a+bx\,\big|\, a,b\in\mathbb{R}\}$$ is spanned by the set $$\{1,x\}$$. The prior section also showed that a space can have many sets that span it. The space of linear polynomials is also spanned by $$\{1,2x\}$$ and $$\{1,x,2x\}$$.

At the end of that section we described some spanning sets as "minimal", but we never precisely defined that word. We could take "minimal" to mean one of two things. We could mean that a spanning set is minimal if it contains the smallest number of members of any set with the same span. With this meaning $$\{1,x,2x\}$$ is not minimal because it has one member more than the other two. Or we could mean that a spanning set is minimal when it has no elements that can be removed without changing the span. Under this meaning $$\{1,x,2x\}$$ is not minimal because removing the $$ 2x $$ and getting $$ \{1,x\} $$ leaves the span unchanged.

The first sense of minimality appears to be a global requirement, in that to check if a spanning set is minimal we seemingly must look at all the spanning sets of a subspace and find one with the least number of elements. The second sense of minimality is local in that we need to look only at the set under discussion and consider the span with and without various elements. For instance, using the second sense, we could compare the span of $$\{1,x,2x\}$$ with the span of $$\{1,x\}$$ and note that the $$2x$$ is a "repeat" in that its removal doesn't shrink the span.

In this section we will use the second sense of "minimal spanning set" because of this technical convenience. However, the most important result of this book is that the two senses coincide; we will prove that in the section after this one.

Álgebra linear/Dependência linear

title: Linear Algebra/Linear Independence


text:
Linear Algebra/Basis and Dimension

The prior section ends with the statement that a spanning set is minimal when it is linearly independent and a linearly independent set is maximal when it spans the space. So the notions of minimal spanning set and maximal independent set coincide. In this section we will name this idea and study its properties.

title: Linear Algebra/Basis and Dimension


text:
Linear Algebra/Dimension

In the prior subsection we defined the basis of a vector space, and we saw that a space can have many different bases. For example, following the definition of a basis, we saw three different bases for $$\mathbb{R}^2$$. So we cannot talk about "the" basis for a vector space. True, some vector spaces have bases that strike us as more natural than others, for instance, $$\mathbb{R}^2$$'s basis $$\mathcal{E}_2$$ or $$\mathbb{R}^3$$'s basis $$\mathcal{E}_3$$ or $$\mathcal{P}_2$$'s basis $$\langle 1,x,x^2 \rangle $$. But, for example in the space $$\{a_2x^2+a_1x+a_0\,\big|\, 2a_2-a_0=a_1\}$$, no particular basis leaps out at us as the most natural one. We cannot, in general, associate with a space any single basis that best describes that space.

We can, however, find something about the bases that is uniquely associated with the space. This subsection shows that any two bases for a space have the same number of elements. So, with each space we can associate a number, the number of vectors in any of its bases.

This brings us back to when we considered the two things that could be meant by the term "minimal spanning set". At that point we defined "minimal" as linearly independent, but we noted that another reasonable interpretation of the term is that a spanning set is "minimal" when it has the fewest number of elements of any set with the same span. At the end of this subsection, after we have shown that all bases have the same number of elements, then we will have shown that the two senses of "minimal" are equivalent.

Before we start, we first limit our attention to spaces where at least one basis has only finitely many members.

(One reason for sticking to finite-dimensional spaces is so that the representation of a vector with respect to a basis is a finitely-tall vector, and so can be easily written.) From now on we study only finite-dimensional vector spaces. We shall take the term "vector space" to mean "finite-dimensional vector space". Other spaces are interesting and important, but they lie outside of our scope.

To prove the main theorem we shall use a technical result.

Again, although we sometimes say "finite-dimensional" as a reminder, in the rest of this book all vector spaces are assumed to be finite-dimensional. An instance of this is that in the next result the word "space" should be taken to mean "finite-dimensional vector space".

The main result of this subsection, that all of the bases in a finite-dimensional vector space have the same number of elements, is the single most important result in this book because, as Example 2.9 shows, it describes what vector spaces and subspaces there can be. We will see more in the next chapter.

Exercises
Assume that all spaces are finite-dimensional unless otherwise stated.

/Solutions/

title: Linear Algebra/Dimension


text:
Linear Algebra/Vector Spaces and Linear Systems

We will now reconsider linear systems and Gauss' method, aided by the tools and terms of this chapter. We will make three points.

For the first point, recall the first chapter's Linear Combination Lemma and its corollary: if two matrices are related by row operations $$A\longrightarrow\cdots\longrightarrow B$$ then each row of $$B$$ is a linear combination of the rows of $$A$$. That is, Gauss' method works by taking linear combinations of rows. Therefore, the right setting in which to study row operations in general, and Gauss' method in particular, is the following vector space.

Thus, row operations leave the row space unchanged. But of course, Gauss' method performs the row operations systematically, with a specific goal in mind, echelon form.

Thus, in the language of this chapter, Gaussian reduction works by eliminating linear dependencies among rows, leaving the span unchanged, until no  nontrivial linear relationships remain (among the nonzero rows). That is, Gauss' method produces a basis for the row space.

Using this technique, we can also find bases for spans not directly involving row vectors.

Our interest in column spaces stems from our study of linear systems. An example is that this system

\begin{array}{*{3}{rc}r} c_1 &+  &3c_2  &+  &7c_3  &=  &d_1  \\ 2c_1 &+  &3c_2  &+  &8c_3  &=  &d_2  \\ &  &c_2   &+  &2c_3  &=  &d_3  \\ 4c_1 &   &      &+  &4c_3  &=  &d_4 \end{array} $$

has a solution if and only if the vector of $$ d $$'s is a linear combination of the other column vectors,

c_1\begin{pmatrix} 1 \\ 2 \\ 0 \\ 4 \end{pmatrix} +c_2\begin{pmatrix} 3 \\ 3 \\ 1 \\ 0 \end{pmatrix} +c_3\begin{pmatrix} 7 \\ 8 \\ 2 \\ 4 \end{pmatrix} =\begin{pmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \end{pmatrix} $$

meaning that the vector of $$ d $$'s is in the column space of the matrix of coefficients.

So the instructions for the prior example are "transpose, reduce, and transpose back".

We can even, at the price of tolerating the as-yet-vague idea of vector spaces being "the same", use Gauss' method to find bases for spans in other types of vector spaces.

Thus, our first point in this subsection is that the tools of this chapter give us a more conceptual understanding of Gaussian reduction.

For the second point of this subsection, consider the effect on the column space of this row reduction.

 * $$\begin{array}{rcl}

\begin{pmatrix} 1 &2  \\ 2  &4 \end{pmatrix} &\xrightarrow[]{-2\rho_1+\rho_2} &\begin{pmatrix} 1 &2  \\ 0  &0 \end{pmatrix} \end{array} $$

The column space of the left-hand matrix contains vectors with a second component that is nonzero. But the column space of the right-hand matrix is different because it contains only vectors whose second component is zero. It is this knowledge that row operations can change the column space that makes next result surprising.

Another way, besides the prior result, to state that Gauss' method has something to say about the column space as well as about  the row space is to consider again Gauss-Jordan reduction. Recall that it ends with the reduced echelon form of a matrix, as here.

 * $$\begin{array}{rcl}

\begin{pmatrix} 1 &3  &1  &6  \\ 2  &6  &3  &16 \\ 1  &3  &1  &6 \end{pmatrix} &\xrightarrow[]{}\;\cdots\;\xrightarrow[]{} &\begin{pmatrix} 1 &3  &0  &2  \\ 0  &0  &1  &4  \\ 0  &0  &0  &0 \end{pmatrix} \end{array} $$

Consider the row space and the column space of this result. Our first point made above says that a basis for the row space is easy to get: simply collect together all of the rows with leading entries. However, because this is a reduced echelon form matrix, a basis for the column space is just as easy: take the columns containing the leading entries, that is,  $$ \langle \vec{e}_1,\vec{e}_2 \rangle  $$. (Linear independence is obvious. The other columns are in the span of this set, since they all have a third component of zero.) Thus, for a reduced echelon form matrix, bases for the row and column spaces can be found in essentially the same way— by taking the parts of the  matrix, the rows or columns, containing the leading entries.

So our second point in this subsection is that the column space and row space of a matrix have the same dimension. Our third and final point is that the concepts that we've seen arising naturally in the study of vector spaces are exactly the ones that we have studied with linear systems.

So if the system has at least one particular solution then for the set of solutions, the number of parameters equals $$n-r$$, the number of variables minus the rank of the matrix of coefficients.

Exercises
/Solutions/


text:
Linear Algebra/Combining Subspaces

''This subsection is optional. It is required only for the last sections of Chapter Three and Chapter Five and for occasional exercises, and can be passed over without loss of continuity.''

This chapter opened with the definition of a vector space, and the middle consisted of a first analysis of the idea. This subsection closes the chapter by finishing the analysis, in the sense that "analysis" means "method of determining the ... essential features of something by separating it into parts".

A common way to understand things is to see how they can be built from component parts. For instance, we think of $$ \mathbb{R}^3 $$ as put together, in some way, from the $$ x $$-axis, the $$ y $$-axis, and $$ z $$-axis. In this subsection we will make this precise;we will describe how to decompose a vector space into a combination of some of its subspaces. In developing this idea of subspace combination, we will keep the $$\mathbb{R}^3$$ example in mind as a benchmark model.

Subspaces are subsets and sets combine via union. But taking the combination operation for subspaces to be the simple union operation isn't what we want. For one thing, the union of the $$ x $$-axis, the $$ y $$-axis, and $$ z $$-axis is not all of $$\mathbb{R}^3$$, so the benchmark model would be left out. Besides, union is all wrong for this reason: a union of subspaces need not be a subspace (it need not be closed; for instance, this $$\mathbb{R}^3$$ vector

\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} $$

is in none of the three axes and hence is not in the union). In addition to the members of the subspaces, we must at least also include all of the linear combinations.

(The notation, writing the "$$ + $$" between sets in addition to using it between vectors, fits with the practice of using this symbol for any natural accumulation operation.)

The above definition gives one way in which a space can be thought of as a combination of some of its parts. However, the prior example shows that there is at least one interesting property of our benchmark model that is not captured by the definition of the sum of subspaces. In the familiar decomposition of $$\mathbb{R}^3$$, we often speak of a vector's "$$x$$part" or "$$y$$part" or "$$z$$part". That is, in this model, each vector has a unique decomposition into parts that come from the parts making up the whole space. But in the decomposition used in Example 4.4, we cannot refer to the "$$xy$$part" of a vector— these three sums

\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} =\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} +\begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} =\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} +\begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix} =\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} +\begin{pmatrix} 0 \\ 1 \\ 3 \end{pmatrix} $$

all describe the vector as comprised of something from the first plane plus something from the second plane, but the "$$xy$$part" is different in each.

That is, when we consider how $$\mathbb{R}^3$$ is put together from the three axes "in some way", we might mean "in such a way that every vector has at least one decomposition", and that leads to the definition above. But if we take it to mean "in such a way that every vector has one and only one decomposition" then we need another condition on combinations. To see what this condition is, recall that vectors are uniquely represented in terms of a basis. We can use this to break a space into a sum of subspaces such that any vector in the space breaks uniquely into a sum of members of those subspaces.

These examples illustrate a natural way to decompose a space into a sum of subspaces in such a way that each vector decomposes uniquely into a sum of vectors from the parts. The next result says that this way is the only way.

The special case of two subspaces is worth mentioning separately.

In this subsection we have seen two ways to regard a space as built up from component parts. Both are useful; in particular, in this book the direct sum definition is needed to do the Jordan Form construction in the fifth chapter.

Exercises
/Solutions/

title: Linear Algebra/Combining Subspaces


text:
Linear Algebra/Topic: Fields

Linear combinations involving only fractions or only integers are much easier for computations than combinations involving real numbers, because computing with irrational numbers is awkward. Could other number systems, like the rationals or the integers, work in the place of $$ \mathbb{R} $$ in the definition of a vector space?

Yes and no. If we take "work" to mean that the results of this chapter remain true then an analysis of which properties of the reals we have used in this chapter gives the following list of conditions an algebraic system needs in order to "work" in the place of $$\mathbb{R}$$.

The number system consisting of the set of real numbers along with the usual addition and multiplication operation is a field, naturally. Another field is the set of rational numbers with its usual addition and multiplication operations. An example of an algebraic structure that is not a field is the integer number system — it fails the final condition.

Some examples are surprising. The set $$ \{0,1\} $$ under these operations: $$ \begin{array}{c|cc} + & 0  & 1  \\ \hline 0 & 0  & 1  \\ 1  & 1  & 0 \end{array} \qquad \begin{array}{c|cc} \cdot & 0  & 1  \\ \hline 0  & 0  & 0   \\ 1   & 0  & 1 \end{array} $$ is a field (see Problem 4).

We could develop Linear Algebra as the theory of vector spaces with scalars from an arbitrary field, instead of sticking to taking the scalars only from $$ \mathbb{R} $$. In that case, almost all of the statements in this book would carry over by replacing "$$ \mathbb{R} $$" with "$$ \mathcal{F} $$", and thus by taking coefficients, vector entries, and matrix entries to be elements of $$ \mathcal{F} $$ ("almost" because statements involving distances or angles are exceptions). Here are some examples; each applies to a vector space $$ V $$ over a field $$ \mathcal{F} $$.

  For any $$ \vec{v}\in V $$ and $$ a\in\mathcal{F} $$, $$ 0\cdot\vec{v}=\vec{0} $$, and  $$ -1\cdot\vec{v}+\vec{v}=\vec{0} $$, and $$ a\cdot\vec{0}=\vec{0} $$.   The span (the set of linear combinations) of a subset of $$ V $$ is a subspace of $$ V $$.

 Any subset of a linearly independent set is also linearly independent.

 In a finite-dimensional vector space, any two bases have the same number of elements.

(Even statements that don't explicitly mention $$ \mathcal{F} $$ use field properties in their proof.)

We won't develop vector spaces in this more general setting because the additional abstraction can be a distraction. The ideas we want to bring out already appear when we stick to the reals.

The only exception is in Chapter Five. In that chapter we must factor polynomials, so we will switch to considering vector spaces over the field of complex numbers. We will discuss this more, including a brief review of complex arithmetic, when we get there.

Exercises
/Solutions/

title: Linear Algebra/Topic: Fields


text:
Linear Algebra/Topic: Crystals

Everyone has noticed that table salt comes in little cubes. Remarkably, the explanation for the cubical external shape is the simplest one possible: the internal shape, the way the atoms lie, is also cubical. The internal structure is pictured below. Salt is sodium chloride, and the small spheres shown are sodium while the big ones are chloride. (To simplify the view, only the sodiums and chlorides on the front, top, and right are shown.) The specks of salt that we see when we spread a little out on the table consist of many repetitions of this fundamental unit. That is, these cubes of atoms stack up to make the larger cubical structure that we see. A solid, such as table salt, with a regular internal structure is a crystal.

We can restrict our attention to the front face. There, we have this pattern repeated many times. The distance between the corners of this cell is about 3.34 Ångstroms (an Ångstrom is $$10^{-10}$$ meters). Obviously that unit is unwieldly for describing points in the crystal lattice. Instead, the thing to do is to take as a unit the length of each side of the square. That is, we naturally adopt this basis.

\langle \begin{pmatrix} 3.34 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 3.34 \end{pmatrix} \rangle $$

Then we can describe, say, the corner in the upper right of the picture above as $$3\vec{\beta}_1+2\vec{\beta}_2$$.

Another crystal from everyday experience is pencil lead. It is graphite, formed from carbon atoms arranged in this shape. This is a single plane of graphite. A piece of graphite consists of many of these planes layered in a stack. (The chemical bonds between the planes are much weaker than the bonds inside the planes, which explains why graphite writes---it can be sheared so that the planes slide off and are left on the paper.) A convenient unit of length can be made by decomposing the hexagonal ring into three regions that are rotations of this unit cell. A natural basis then would consist of the vectors that form the sides of that unit cell. The distance along the bottom and slant is $$1.42$$ ngstroms, so this

\langle \begin{pmatrix} 1.42 \\ 0 \end{pmatrix}, \begin{pmatrix} 1.23 \\ .71 \end{pmatrix} \rangle $$

is a good basis.

The selection of convienent bases extends to three dimensions. Another familiar crystal formed from carbon is diamond. Like table salt, it is built from cubes, but the structure inside each cube is more complicated than salt's. In addition to carbons at each corner, there are carbons in the middle of each face. (To show the added face carbons clearly, the corner carbons have been reduced to dots.) There are also four more carbons inside the cube, two that are a quarter of the way up from the bottom and two that are a quarter of the way down from the top. (As before, carbons shown earlier have been reduced here to dots.) The distance along any edge of the cube is $$2.18$$ Ångstroms. Thus, a natural basis for describing the locations of the carbons, and the bonds between them, is this.

\langle \begin{pmatrix} 2.18 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 2.18 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 2.18 \end{pmatrix} \rangle $$

Even the few examples given here show that the structures of crystals is complicated enough that some organized system to give the locations of the atoms, and how they are chemically bound, is needed. One tool for that organization is a convienent basis. This application of bases is simple, but it shows a context where the idea arises naturally. The work in this chapter just takes this simple idea and develops it.

Exercises
/Solutions/

title: Linear Algebra/Topic: Crystals


text:
Linear Algebra/Topic: Dimensional Analysis

"You can't add apples and oranges," the old saying goes. It reflects our experience that in applications the quantities have units and keeping track of those units is worthwhile. Everyone has done calculations such as this one that use the units as a check.

60\,\frac{\text{sec}}{\text{min}} \cdot 60\,\frac{\text{min}}{\text{hr}} \cdot 24\,\frac{\text{hr}}{\text{day}} \cdot 365\,\frac{\text{day}}{\text{year}} = 31\,536\,000\,\frac{\text{sec}}{\text{year}} $$

However, the idea of including the units can be taken beyond bookkeeping. It can be used to draw conclusions about what relationships are possible among the physical quantities.

To start, consider the physics equation: $$\text{distance}=16\cdot(\text{time})^2$$. If the distance is in feet and the time is in seconds then this is a true statement about falling bodies. However it is not correct in other unit systems; for instance, it is not correct in the meter-second system. We can fix that by making the $$16$$ a dimensional constant.

\text{dist}=16\,\frac{\text{ft}}{\text{sec}^2}\cdot (\text{time})^2 $$

For instance, the above equation holds in the yard-second system.

\text{distance in yards}=16\,\frac{(1/3)\,\text{yd}}{\text{sec}^2} \cdot (\text{time in sec})^2 =\frac{16}{3}\,\frac{\text{yd}}{\text{sec}^2} \cdot (\text{time in sec})^2 $$

So our first point is that by "including the units" we mean that we are restricting our attention to equations that use dimensional constants.

By using dimensional constants, we can be vague about units and say only that all quantities are measured in combinations of some units of length $$L$$, mass $$M$$, and time $$T$$. We shall refer to these three as dimensions (these are the only three dimensions that we shall need in this Topic). For instance, velocity could be measured in $$\text{feet}/\text{second}$$ or $$\text{fathoms}/\text{hour}$$, but in all events it involves some unit of length divided by some unit of time so the dimensional formula of velocity is $$L/T$$. Similarly, the dimensional formula of density is $$M/L^3$$. We shall prefer using negative exponents over the fraction bars and we shall include the dimensions with a zero exponent, that is, we shall write the dimensional formula of velocity as $$L^1M^0T^{-1}$$ and that of density as $$L^{-3}M^1T^0$$.

In this context, "You can't add apples to oranges" becomes the advice to check that all of an equation's terms have the same dimensional formula. An example is this version of the falling body equation: $$d-gt^2=0$$. The dimensional formula of the $$d$$ term is $$L^1M^0T^0$$. For the other term, the dimensional formula of $$g$$ is $$L^1M^0T^{-2}$$ ($$g$$ is the dimensional constant given above as $$16\,\text{ft}/\text{sec}^2$$) and the dimensional formula of $$t$$ is $$L^0M^0T^1$$, so that of the entire $$gt^2$$ term is $$L^1M^0T^{-2}(L^0M^0T^1)^2=L^1M^0T^0$$. Thus the two terms have the same dimensional formula. An equation with this property is dimensionally homogeneous.

Quantities with dimensional formula $$L^0M^0T^0$$ are dimensionless. For example, we measure an angle by taking the ratio of the subtended arc to the radius which is the ratio of a length to a length $$L^1M^0T^0/L^1M^0T^0$$ and thus angles have the dimensional formula $$L^0M^0T^0$$.

The classic example of using the units for more than bookkeeping, using them to draw conclusions, considers the formula for the period of a pendulum.

p=\text{--some expression involving the length of the string, etc.--} $$

The period is in units of time $$L^0M^0T^1$$. So the quantities on the other side of the equation must have dimensional formulas that combine in such a way that their $$L$$'s and $$M$$'s cancel and only a single $$T$$ remains. The table on below has the quantities that an experienced investigator would consider possibly relevant. The only dimensional formulas involving $$L$$ are for the length of the string and the acceleration due to gravity. For the $$L$$'s of these two to cancel, when they appear in the equation they must be in ratio, e.g., as $$(\ell/g)^2$$, or as $$\cos(\ell/g)$$, or as $$(\ell/g)^{-1}$$. Therefore the period is a function of $$\ell/g$$.

This is a remarkable result: with a pencil and paper analysis, before we ever took out the pendulum and made measurements, we have determined something about the relationship among the quantities.

To do dimensional analysis systematically, we need to know two things (arguments for these are in, Chapter II and IV). The first is that each equation relating physical quantities that we shall see involves a sum of terms, where each term has the form

m_1^{p_1}m_2^{p_2}\cdots m_k^{p_k} $$

for numbers $$m_1$$, ..., $$m_k$$ that measure the quantities.

For the second, observe that an easy way to construct a dimensionally homogeneous expression is by taking a product of dimensionless quantities or by adding such dimensionless terms. Buckingham's Theorem states that any complete relationship among quantities with dimensional formulas can be algebraically manipulated into a form where there is some function $$f$$ such that

f(\Pi_1,\ldots,\Pi_n)=0 $$

for a complete set $$\{\Pi_1,\ldots,\Pi_n\}$$ of dimensionless products. (The first example below describes what makes a set of dimensionless products "complete".) We usually want to express one of the quantities, $$m_1$$ for instance, in terms of the others, and for that we will assume that the above equality can be rewritten

m_1=m_2^{-p_2}\cdots m_k^{-p_k}\cdot \hat{f}(\Pi_2,\ldots,\Pi_n) $$

where $$\Pi_1=m_1m_2^{p_2}\cdots m_k^{p_k}$$ is dimensionless and the products $$\Pi_2$$, ..., $$\Pi_n$$ don't involve $$m_1$$ (as with $$f$$, here $$\hat{f}$$ is just some function, this time of $$n-1$$ arguments). Thus, to do dimensional analysis we should find which dimensionless products are possible.

For example, consider again the formula for a pendulum's period.  By the first fact cited above, we expect the formula to have (possibly sums of terms of) the form $$p^{p_1}\ell^{p_2}m^{p_3}g^{p_4}\theta^{p_5}$$. To use the second fact, to find which combinations of the powers $$p_1$$, ..., $$p_5$$ yield dimensionless products, consider this equation.

(L^0M^0T^1)^{p_1}(L^1M^0T^0)^{p_2}(L^0M^1T^0)^{p_3} (L^1M^0T^{-2})^{p_4}(L^0M^0T^0)^{p_5} =L^0M^0T^0 $$

It gives three conditions on the powers.

\begin{array}{*{5}{rc}r} &\  &p_2  &\   &    &+  &p_4   &  &  &=  &0  \\ &   &     &    &p_3 &   &      &  &  &=  &0  \\ p_1&   &     &    &    &-  &2p_4  &  &  &=  &0 \end{array} $$

Note that $$p_3$$ is $$0$$ and so the mass of the bob does not affect the period. Gaussian reduction and parametrization of that system gives this

\{\begin{pmatrix} p_1 \\ p_2 \\ p_3 \\ p_4 \\ p_5 \end{pmatrix}= \begin{pmatrix} 1 \\ -1/2  \\  0  \\  1/2  \\  0 \end{pmatrix}p_1+ \begin{pmatrix} 0  \\  0   \\  0  \\  0  \\  1 \end{pmatrix}p_5 \,\big|\, p_1,p_5\in\mathbb{R}\} $$

(we've taken $$p_1$$ as one of the parameters in order to express the period in terms of the other quantities).

Here is the linear algebra. The set of dimensionless products contains all terms $$p^{p_1}\ell^{p_2}m^{p_3}a^{p_4}\theta^{p_5}$$ subject to the conditions above. This set forms a vector space under the "$$+$$" operation of multiplying two such products and the "$$\cdot$$" operation of raising such a product to the power of the scalar (see Problem 5). The term "complete set of dimensionless products" in Buckingham's Theorem means a basis for this vector space.

We can get a basis by first taking $$p_1=1$$, $$p_5=0$$ and then $$p_1=0$$, $$p_5=1$$. The associated dimensionless products are $$\Pi_1=p\ell^{-1/2}g^{1/2}$$ and $$\Pi_2=\theta$$. Because the set $$\{\Pi_1,\Pi_2\}$$ is complete, Buckingham's Theorem says that

p = \ell^{1/2}g^{-1/2}\cdot\hat{f}(\theta) = \sqrt{\ell/g}\cdot\hat{f}(\theta) $$

where $$\hat{f}$$ is a function that we cannot determine from this analysis (a first year physics text will show by other means that for small angles it is approximately the constant function $$\hat{f}(\theta)=2\pi$$).

Thus, analysis of the relationships that are possible between the quantities with the given dimensional formulas has produced a fair amount of information: a pendulum's period does not depend on the mass of the bob, and it rises with the square root of the length of the string.

For the next example we try to determine the period of revolution of two bodies in space orbiting each other under mutual gravitational attraction. An experienced investigator could expect that these are the relevant quantities.

To get the complete set of dimensionless products we consider the equation

(L^0M^0T^1)^{p_1}(L^1M^0T^0)^{p_2}(L^0M^1T^0)^{p_3}(L^0M^1T^0)^{p_4} (L^3M^{-1}T^{-2})^{p_5}=L^0M^0T^0 $$

which results in a system

\begin{array}{*{5}{rc}r} & &p_2  &   &     &   &      &+  &3p_5 &=  &0  \\ & &     &   &p_3  &+  &p_4   &-  &p_5  &=  &0  \\ p_1 & &     &   &     &   &      &-  &2p_5 &=  &0 \end{array} $$

with this solution.

\{\begin{pmatrix} 1 \\ -3/2 \\ 1/2 \\ 0 \\ 1/2 \end{pmatrix}p_1 +\begin{pmatrix} 0 \\ 0   \\  -1 \\ 1 \\ 0 \end{pmatrix}p_4 \,\big|\, p_1,p_4\in\mathbb{R}\} $$

As earlier, the linear algebra here is that the set of dimensionless products of these quantities forms a vector space, and we want to produce a basis for that space, a "complete" set of dimensionless products. One such set, gotten from setting $$p_1=1$$ and $$p_4=0$$, and also setting $$p_1=0$$ and $$p_4=1$$ is $$\{\Pi_1=pr^{-3/2}m_1^{1/2}G^{1/2},\,\Pi_2=m_1^{-1}m_2\}$$. With that, Buckingham's Theorem says that any complete relationship among these quantities is stateable this form.

p = r^{3/2}m_1^{-1/2}G^{-1/2}\cdot\hat{f}(m_1^{-1}m_2) = \frac{r^{3/2}}{\sqrt{Gm_1}}\cdot\hat{f}(m_2/m_1) $$

Remark. An important application of the prior formula is when $$m_1$$ is the mass of the sun and $$m_2$$ is the mass of a planet. Because $$m_1$$ is very much greater than $$m_2$$, the argument to $$\hat{f}$$ is approximately $$0$$, and we can wonder whether this part of the formula remains approximately constant as $$m_2$$ varies. One way to see that it does is this. The sun is so much larger than the planet that the mutual rotation is approximately about the sun's center. If we vary the planet's mass $$m_2$$ by a factor of $$x$$ (e.g., Venus's mass is $$x=0.815$$ times Earth's mass), then the force of attraction is multiplied by $$x$$, and $$x$$ times the force acting on $$x$$ times the mass gives, since $$F=ma$$, the same acceleration, about the same center (approximately). Hence, the orbit will be the same and so its period will be the same, and thus the right side of the above equation also remains unchanged (approximately). Therefore, $$\hat{f}(m_2/m_1)$$ is approximately constant as $$m_2$$ varies. This is Kepler's Third Law: the square of the period of a planet is proportional to the cube of the mean radius of its orbit about the sun.

The final example was one of the first explicit applications of dimensional analysis. Lord Raleigh considered the speed of a wave in deep water and suggested these as the relevant quantities. The equation

(L^1M^0T^{-1})^{p_1}(L^{-3}M^1T^0)^{p_2} (L^1M^0T^{-2})^{p_3}(L^1M^0T^0)^{p_4}=L^0M^0T^0 $$

gives this system

\begin{array}{*{4}{rc}r} p_1 &-  &3p_2 &+  &p_3  &+  &p_4  &=  &0  \\ &  &p_2  &   &     &   &     &=  &0  \\ -p_1 &  &     &-  &2p_3 &   &     &=  &0 \end{array} $$

with this solution space

\{\begin{pmatrix} 1 \\ 0 \\ -1/2 \\ -1/2 \end{pmatrix}p_1 \,\big|\, p_1\in \mathbb{R}\} $$

(as in the pendulum example, one of the quantities $$d$$ turns out not to be involved in the relationship). There is one dimensionless product, $$\Pi_1=vg^{-1/2}\lambda^{-1/2}$$, and so $$v$$ is $$\sqrt{\lambda g}$$ times a constant ($$\hat{f}$$ is constant since it is a function of no arguments).

As the three examples above show, dimensional analysis can bring us far toward expressing the relationship among the quantities. For further reading, the classic reference is —this brief book is delightful. Another source is .. A description of dimensional analysis's place in modeling is in ..

Exercises
/Solutions/


text:
Linear Algebra/Isomorphisms

In the examples following the definition of a vector space we developed the intuition that some spaces are "the same" as others. For instance, the space of two-tall column vectors and the space of two-wide row vectors are not equal because their elements—column vectors and row vectors—are not equal, but we have the idea that these spaces differ only in how their elements appear. We will now make this idea precise.

This section illustrates a common aspect of a mathematical investigation. With the help of some examples, we've gotten an idea. We will next give a formal definition, and then we will produce some results backing our contention that the definition captures the idea. We've seen this happen already, for instance, in the first section of the Vector Space chapter. There, the study of linear systems led us to consider collections closed under linear combinations. We defined such a collection as a vector space, and we followed it with some supporting results.

Of course, that definition wasn't an end point, instead it led to new insights such as the idea of a basis. Here too, after producing a definition, and supporting it, we will get two surprises (pleasant ones). First, we will find that the definition applies to some unforeseen, and interesting, cases. Second, the study of the definition will lead to new ideas. In this way, our investigation will build a momentum.

title: Linear Algebra/Isomorphisms


text:
Linear Algebra/Dimension Characterizes Isomorphism

In the prior subsection, after stating the definition of an isomorphism, we gave some results supporting the intuition that such a map describes spaces as "the same". Here we will formalize this intuition. While two spaces that are isomorphic are not equal, we think of them as almost equal— as equivalent. In this subsection we shall show that the relationship "is isomorphic to" is an equivalence relation.

As a consequence of that result, we know that the universe of vector spaces is partitioned into classes: every space is in one and only one isomorphism class.   $$V\cong W$$

This follows from the next two lemmas.

That ends the proof of Theorem 2.2. We say that the isomorphism classes are characterized by dimension because we can describe each class simply by giving the number that is the dimension of all of the spaces in that class.

This subsection's results give us a collection of representatives of the isomorphism classes.

The proofs above pack many ideas into a small space. Through the rest of this chapter we'll consider these ideas again, and fill them out. For a taste of this, we will expand here on the proof of Lemma 2.4.

We will close this section with a summary.

Recall that in the first chapter we defined two matrices as row equivalent if they can be derived from each other by elementary row operations (this was the meaning of same-ness that was of interest there). We showed that is an equivalence relation and so the collection of matrices is partitioned into classes, where all the matrices that are row equivalent fall together into a single class. Then, for insight into which matrices are in each class, we gave representatives for the classes, the reduced echelon form matrices.

In this section, except that the appropriate notion of same-ness here is vector space isomorphism, we have followed much the same outline. First we defined isomorphism, saw some examples, and established some properties. Then we showed that it is an equivalence relation, and now we have a set of class representatives, the real vector spaces $$\mathbb{R}^1$$, $$\mathbb{R}^2$$, etc.    As before, the list of representatives helps us to understand the partition. It is simply a classification of spaces by dimension.

In the second chapter, with the definition of vector spaces, we seemed to have opened up our studies to many examples of new structures besides the familiar $$\mathbb{R}^n$$'s. We now know that isn't the case. Any finite-dimensional vector space is actually "the same" as a real space. We are thus considering exactly the structures that we need to consider.

The rest of the chapter fills out the work in this section. In particular, in the next section we will consider maps that preserve structure, but are not necessarily correspondences.


text:
Linear Algebra/Homomorphisms

The definition of isomorphism has two conditions. In this section we will consider the second one, that the map must preserve the algebraic structure of the space. We will focus on this condition by studying maps that are required only to preserve structure; that is, maps that are not required to be correspondences.

Experience shows that this kind of map is tremendously useful in the study of vector spaces. For one thing, as we shall see in the second subsection below, while isomorphisms describe how spaces are the same, these maps describe how spaces can be thought of as alike.

title: Linear Algebra/Homomorphisms


text:
Linear Algebra/Definition of Homomorphism

Obviously, any isomorphism is a homomorphism— an isomorphism is a homomorphism that is also a correspondence. So, one way to think of the "homomorphism" idea is that it is a generalization of "isomorphism", motivated by the observation that many of the properties of isomorphisms have only to do with the map's structure preservation property and not to do with it being a correspondence. As examples, these two results from the prior section do not use one-to-one-ness or onto-ness in their proof, and therefore apply to any homomorphism.

Part 1 is often used to check that a function is linear.

However, some of the results that we have seen for isomorphisms fail to hold for homomorphisms in general. Consider the theorem that an isomorphism between spaces gives a correspondence between their bases. Homomorphisms do not give any such correspondence; Example 1.2 shows that there is no such correspondence, and another example is the zero map between any two nontrivial spaces. Instead, for homomorphisms a weaker but still very useful result holds.

Just as the isomorphisms of a space with itself are useful and interesting, so too are the homomorphisms of a space with itself.

We finish this subsection about maps by recalling that we can linearly combine maps. For instance, for these maps from $$ \mathbb{R}^2 $$ to itself

\begin{pmatrix} x \\ y \end{pmatrix} \stackrel{f}{\longmapsto} \begin{pmatrix} 2x \\ 3x-2y \end{pmatrix} \quad\text{and}\quad \begin{pmatrix} x \\ y \end{pmatrix} \stackrel{g}{\longmapsto} \begin{pmatrix} 0 \\ 5x \end{pmatrix} $$

the linear combination $$ 5f-2g $$ is also a map from $$R^2$$ to itself.

\begin{pmatrix} x \\ y \end{pmatrix} \stackrel{5f-2g}{\longmapsto} \begin{pmatrix} 10x \\ 5x-10y \end{pmatrix} $$

We started this section by isolating the structure preservation property of isomorphisms. That is, we defined homomorphisms as a generalization of isomorphisms. Some of the properties that we studied for isomorphisms carried over unchanged, while others were adapted to this more general setting.

It would be a mistake, though, to view this new notion of homomorphism as derived from, or somehow secondary to, that of isomorphism. In the rest of this chapter we shall work mostly with homomorphisms, partly because any statement made about homomorphisms is automatically true about isomorphisms, but more because, while the isomorphism concept is perhaps more natural, experience shows that the homomorphism concept is actually more fruitful and more central to further progress.

Exercises
Álgebra linear/Transformações lineares


text:
Linear Algebra/Rangespace and Nullspace

Isomorphisms and homomorphisms both preserve structure. The difference is that homomorphisms needn't be onto and needn't be one-to-one. This means that homomorphisms are a more general kind of map, subject to fewer restrictions than isomorphisms. We will examine what can happen with homomorphisms that is prevented by the extra restrictions satisfied by isomorphisms.

We first consider the effect of dropping the onto requirement, of not requiring as part of the definition that a homomorphism be onto its codomain. For instance, the injection map $$ \iota:\mathbb{R}^2\to \mathbb{R}^3 $$

\begin{pmatrix} x \\ y \end{pmatrix} \mapsto \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} $$

is not an isomorphism because it is not onto. Of course, being a function, a homomorphism is onto some set, namely its range; the map $$\iota$$ is onto the $$xy$$-plane subset of $$ \mathbb{R}^3$$.

(We shall soon see the connection between the rank of a map and the rank of a matrix.)

The prior result shows that, in passing from the definition of isomorphism to the more general definition of homomorphism, omitting the "onto" requirement doesn't make an essential difference. Any homomorphism is onto its rangespace.

However, omitting the "one-to-one" condition does make a difference. A homomorphism may have many elements of the domain that map to one element of the codomain. Below is a "bean " sketch of a many-to-one map between sets. It shows three elements of the codomain that are each the image of many members of the domain. Recall that for any function $$h:V\to W$$, the set of elements of $$V$$ that are mapped to $$ \vec{w}\in W $$ is the inverse image $$h^{-1}(\vec{w})=\{\vec{v}\in V\,\big|\, h(\vec{v})=\vec{w}\}$$. Above, the three sets of many elements on the left are inverse images.

The above examples have only to do with the fact that we are considering functions, specifically, many-to-one functions. They show the inverse images as sets of vectors that are related to the image vector $$\vec{w}$$. But these are more than just arbitrary functions, they are homomorphisms; what do the two preservation conditions say about the relationships?

In generalizing from isomorphisms to homomorphisms by dropping the one-to-one condition, we lose the property that we've stated intuitively as: the domain is "the same as" the range. That is, we lose that the domain corresponds perfectly to the range in a one-vector-by-one-vector way.

What we shall keep, as the examples below illustrate, is that a homomorphism describes a way in which the domain is "like", or "analogous to", the range.

We won't describe how every homomorphism that we will use is an analogy because the formal sense that we make of "alike in that ..." is "a homomorphism exists such that ...". Nonetheless, the idea that a homomorphism between two spaces expresses how the domain's vectors fall into classes that act like the range's vectors is a good way to view homomorphisms.

Another reason that we won't treat all of the homomorphisms that we see as above is that many vector spaces are hard to draw (e.g., a space of polynomials). However, there is nothing bad about gaining insights from those spaces that we are able to draw, especially when those insights extend to all vector spaces. We derive two such insights from the three examples 2.7, 2.8, and 2.9.

First, in all three examples, the inverse images are lines or planes, that is, linear surfaces. In particular, the inverse image of the range's zero vector is a line or plane through the origin— a subspace of the domain.

Now for the second insight from the above pictures. In Example 2.7, each of the vertical lines is squashed down to a single point— $$\pi$$, in passing from the domain to the range, takes all of these one-dimensional vertical lines and "zeroes them out", leaving the range one dimension smaller than the domain. Similarly, in Example 2.8, the two-dimensional domain is mapped to a one-dimensional range by breaking the domain into lines (here, they are diagonal lines), and compressing each of those lines to a single member of the range. Finally, in Example 2.9, the domain breaks into planes which get "zeroed out", and so the map starts with a three-dimensional domain but ends with a one-dimensional range— this map "subtracts" two from the dimension. (Notice that, in this third example, the codomain is two-dimensional but the range of the map is only one-dimensional, and it is the dimension of the range that is of interest.)

We know that an isomorphism exists between two spaces if and only if their dimensions are equal. Here we see that for a homomorphism to exist, the dimension of the range must be less than or equal to the dimension of the domain. For instance, there is no homomorphism from $$ \mathbb{R}^2 $$ onto $$ \mathbb{R}^3 $$. There are many homomorphisms from $$ \mathbb{R}^2 $$ into $$ \mathbb{R}^3 $$, but none is onto all of three-space.

The rangespace of a linear map can be of dimension strictly less than the dimension of the domain (Example 2.3's derivative transformation on $$\mathcal{P}_3$$ has a domain of dimension four but a range of dimension three). Thus, under a homomorphism, linearly independent sets in the domain may map to linearly dependent sets in the range (for instance, the derivative sends $$\{1,x,x^2,x^3\}$$ to $$\{0,1,2x,3x^2\}$$). That is, under a homomorphism, independence may be lost. In contrast, dependence stays.

When is independence not lost? One obvious sufficient condition is when the homomorphism is an isomorphism. This condition is also necessary; see Problem 14. We will finish this subsection comparing homomorphisms with isomorphisms by observing that a one-to-one homomorphism is an isomorphism from its domain onto its range.

(In the next section we will see the connection between this use of "nonsingular" for maps and its familiar use for matrices.)

The prior observation allows us to adapt some results about isomorphisms to this setting.

We've now seen that a linear map shows how the structure of the domain is like that of the range. Such a map can be thought to organize the domain space into inverse images of points in the range. In the special case that the map is one-to-one, each inverse image is a single point and the map is an isomorphism between the domain and the range.

Exercises
/Solutions/

title: Linear Algebra/Rangespace and Nullspace


text:
Linear Algebra/Computing Linear Maps

The prior section shows that a linear map is determined by its action on a basis. In fact, the equation

h(\vec{v}) =h(c_1\cdot\vec{\beta}_1+\dots+c_n\cdot\vec{\beta}_n) =c_1\cdot h(\vec{\beta}_1)+\dots +c_n\cdot h(\vec{\beta}_n) $$

shows that, if we know the value of the map on the vectors in a basis, then we can compute the value of the map on any vector $$\vec{v}$$ at all. We just need to find the $$c$$'s to express $$\vec{v}$$ with respect to the basis.

This section gives the scheme that computes, from the representation of a vector in the domain $${\rm Rep}_{B}(\vec{v})$$, the representation of that vector's image in the codomain $${\rm Rep}_{D}(h(\vec{v}))$$, using the representations of $$ h(\vec{\beta}_1) $$, ..., $$ h(\vec{\beta}_n) $$.

title: Linear Algebra/Computing Linear Maps


text:
Linear Algebra/Representing Linear Maps with Matrices

We will express computations like the one above with a matrix notation.

\begin{pmatrix} 0            &1  \\ -1/2          &-1  \\ 1             &0 \end{pmatrix}_{B,D} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}_B = \begin{pmatrix} 0c_1+1c_2 \\ (-1/2)c_1-1c_2 \\ 1c_1+0c_2 \end{pmatrix}_D $$

In the middle is the argument $$\vec{v}$$ to the map, represented with respect to the domain's basis $$B$$ by a column vector with components $$c_1$$ and $$c_2$$. On the right is the value $$h(\vec{v})$$ of the map on that argument, represented with respect to the codomain's basis $$D$$ by a column vector with components $$0c_1+1c_2$$, etc. The matrix on the left is the new thing. It consists of the coefficients from the vector on the right, $$0$$ and $$1$$ from the first row, $$-1/2$$ and $$-1$$ from the second row, and $$1$$ and $$0$$ from the third row.

This notation simply breaks the parts from the right, the coefficients and the $$c$$'s, out separately on the left, into a vector that represents the map's argument and a matrix that we will take to represent the map itself.

Briefly, the vectors representing the $$h(\vec{\beta})$$'s are adjoined to make the matrix representing the map.

{\rm Rep}_{B,D}(h)= \left(\begin{array}{c|c|c} \vdots                        &       &\vdots    \\ {\rm Rep}_{D}(\,h(\vec{\beta}_1)\,)  &\cdots &{\rm Rep}_{D}(\,h(\vec{\beta}_n)\,)  \\ \vdots                         &       &\vdots \end{array}\right) $$

Observe that the number of columns $$n$$ of the matrix is the dimension of the domain of the map, and the number of rows $$m$$ is the dimension of the codomain.

We will use lower case letters for a map, upper case for the matrix, and lower case again for the entries of the matrix. Thus for the map $$ h $$, the matrix representing it is $$ H $$, with entries $$ h_{i,j} $$.

We will think of the matrix $${\rm Rep}_{B,D}(h)$$ and the vector $${\rm Rep}_{B}(\vec{v})$$ as combining to make the vector $${\rm Rep}_{D}(h(\vec{v}))$$.

The point of Definition 1.2 is to generalize Example 1.1, that is, the point of the definition is Theorem 1.4, that the matrix describes how to get from the representation of a domain vector with respect to the domain's basis to the representation of its image in the codomain with respect to the codomain's basis. With Definition 1.5, we can restate this as: application of a linear map is represented by the matrix-vector product of the map's representative and the vector's representative.

We now have two ways to compute the effect of projection, the straightforward formula that drops each three-tall vector's third component to make a two-tall vector, and the above formula that uses representations and matrix-vector multiplication. Compared to the first way, the second way might seem complicated. However, it has advantages. The next example shows that giving a formula for some maps is simplified by this new scheme.

We have already seen the addition and scalar multiplication operations of matrices and the dot product operation of vectors. Matrix-vector multiplication is a new operation in the arithmetic of vectors and matrices. Nothing in Definition 1.5 requires us to view it in terms of representations. We can get some insight into this operation by turning away from what is being represented, and instead focusing on how the entries combine.

A good way to view a matrix-vector product is as the dot products of the rows of the matrix with the column vector.

\begin{pmatrix} &\vdots                        \\ a_{i,1} &a_{i,2}  &\ldots   &a_{i,n}    \\ &\vdots \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} = \begin{pmatrix} \vdots \\ a_{i,1}c_1+a_{i,2}c_2+\ldots+a_{i,n}c_n \\ \vdots \end{pmatrix} $$

Looked at in this row-by-row way, this new operation generalizes dot product.

Matrix-vector product can also be viewed column-by-column.

 * $$\begin{array}{rl}

\left( \begin{array}{cccc} h_{1,1} &h_{1,2}  &\ldots  &h_{1,n}\\ h_{2,1}  &h_{2,2}  &\ldots  &h_{2,n}  \\ &\vdots\\ h_{m,1} &h_{m,2} &\ldots  &h_{m,n} \end{array} \right) \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} &=\begin{pmatrix} h_{1,1}c_1+h_{1,2}c_2+\dots+h_{1,n}c_n \\ h_{2,1}c_1+h_{2,2}c_2+\dots+h_{2,n}c_n \\ \vdots \\ h_{m,1}c_1+h_{m,2}c_2+\dots+h_{m,n}c_n \end{pmatrix}   \\ &=c_1\begin{pmatrix} h_{1,1} \\ h_{2,1} \\ \vdots \\ h_{m,1} \end{pmatrix}

+\dots +c_n\begin{pmatrix} h_{1,n} \\ h_{2,n} \\ \vdots \\ h_{m,n} \end{pmatrix} \end{array}$$

The result has the columns of the matrix weighted by the entries of the vector. This way of looking at it brings us back to the objective stated at the start of this section, to compute $$ h(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n) $$ as $$ c_1h(\vec{\beta}_1)+\dots+c_nh(\vec{\beta}_n) $$.

We began this section by noting that the equality of these two enables us to compute the action of $$h$$ on any argument knowing only $$h(\vec{\beta}_1)$$, ..., $$h(\vec{\beta}_n)$$. We have developed this into a scheme to compute the action of the map by taking the matrix-vector product of the matrix representing the map and the vector representing the argument. In this way, any linear map is represented with respect to some bases by a matrix. In the next subsection, we will show the converse, that any matrix represents a linear map.

Exercises
/Solutions/


text:
Linear Algebra/Any Matrix Represents a Linear Map

The prior subsection shows that the action of a linear map $$h$$ is described by a matrix $$H$$, with respect to appropriate bases, in this way.

\vec{v}=\begin{pmatrix} v_1 \\ \vdots \\ v_n \end{pmatrix}_B \;\overset{h}{\underset{H}{\longmapsto}}\; \begin{pmatrix} h_{1,1}v_1+\dots+h_{1,n}v_n \\ \vdots                     \\ h_{m,1}v_1+\dots+h_{m,n}v_n \end{pmatrix}_D =h(\vec{v}) $$

In this subsection, we will show the converse, that each matrix represents a linear map.

Recall that, in the definition of the matrix representation of a linear map, the number of columns of the matrix is the dimension of the map's domain and the number of rows of the matrix is the dimension of the map's codomain. Thus, for instance, a $$ 2 \! \times \! 3 $$ matrix cannot represent a map from $$ \mathbb{R}^5 $$ to $$ \mathbb{R}^4 $$. The next result says that, beyond this restriction on the dimensions, there are no other limitations: the $$ 2 \! \times \! 3 $$ matrix represents a map from any three-dimensional space to any two-dimensional space.

So not only is any linear map described by a matrix but any matrix describes a linear map. This means that we can, when convenient, handle linear maps entirely as matrices, simply doing the computations, without have to worry that a matrix of interest does not represent a linear map on some pair of spaces of interest. (In practice, when we are working with a matrix but no spaces or bases have been specified, we will often take the domain and codomain to be $$\mathbb{R}^n$$ and $$\mathbb{R}^m$$ and use the standard bases. In this case, because the representation is transparent— the representation with respect to the standard basis of $$\vec{v}$$ is $$\vec{v}$$— the column space of the matrix equals the range of the map. Consequently, the column space of $$ H $$ is often denoted by $$ \mathcal{R}(H) $$.)

With the theorem, we have characterized linear maps as those maps that act in this matrix way. Each linear map is described by a matrix and each matrix describes a linear map. We finish this section by illustrating how a matrix can be used to tell things about its maps.

The above results end any confusion caused by our use of the word "rank" to mean apparently different things when applied to matrices and when applied to maps. We can also justify the dual use of "nonsingular". We've defined a matrix to be nonsingular if it is square and is the matrix of coefficients of a linear system with a unique solution, and we've defined a linear map to be nonsingular if it is one-to-one.

We've now seen that the relationship between maps and matrices goes both ways: fixing bases, any linear map is represented by a matrix and any matrix describes a linear map. That is, by fixing spaces and bases we get a correspondence between maps and matrices. In the rest of this chapter we will explore this correspondence. For instance, we've defined for linear maps the operations of addition and scalar multiplication and we shall see what the corresponding matrix operations are. We shall also see the matrix operation that represent the map operation of composition. And, we shall see how to find the matrix that represents a map's inverse.

Exercises
/Solutions/


text:
Linear Algebra/Matrix Operations

The prior section shows how matrices represent linear maps. A good strategy, on seeing a new idea, is to explore how it interacts with some already-established ideas. In the first subsection we will ask how the representation of the sum of two maps $$f+g$$ is related to the representations of the two maps, and how the representation of a scalar product $$r\cdot h$$ of a map is related to the representation of that map. In later subsections we will see how to represent map composition and map inverse.

title: Linear Algebra/Matrix Operations


text:
Linear Algebra/Change of Basis

Representations, whether of vectors or of maps, vary with the bases. For instance, with respect to the two bases $$\mathcal{E}_2$$ and

B=\langle \begin{pmatrix} 1 \\ 1 \end{pmatrix},\begin{pmatrix} 1 \\ -1 \end{pmatrix} \rangle $$

for $$\mathbb{R}^2$$, the vector $$\vec{e}_1$$ has two different representations.

{\rm Rep}_{\mathcal{E}_2}(\vec{e}_1)=\begin{pmatrix} 1 \\ 0 \end{pmatrix} \qquad {\rm Rep}_{B}(\vec{e}_1)=\begin{pmatrix} 1/2 \\ 1/2 \end{pmatrix} $$

Similarly, with respect to $$\mathcal{E}_2,\mathcal{E}_2$$ and $$\mathcal{E}_2,B$$, the identity map has two different representations.

{\rm Rep}_{\mathcal{E}_2,\mathcal{E}_2}(\text{id})= \begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix} \qquad {\rm Rep}_{\mathcal{E}_2,B}(\text{id})= \begin{pmatrix} 1/2 &1/2 \\ 1/2 &-1/2 \end{pmatrix} $$

With our point of view that the objects of our studies are vectors and maps, in fixing bases we are adopting a scheme of tags or names for these objects, that are convienent for computation. We will now see how to translate among these names— we will see exactly how representations vary as the bases vary.

title: Linear Algebra/Change of Basis


text:
Linear Algebra/Sums and Scalar Products

Recall that for two maps $$f$$ and $$g$$ with the same domain and codomain, the map sum $$f+g$$ has this definition.

\vec{v} \;\stackrel{f+g}{\longmapsto}\; f(\vec{v})+g(\vec{v}) $$

The easiest way to see how the representations of the maps combine to represent the map sum is with an example.

Representing a scalar multiple of a map works the same way.

A notable special case of scalar multiplication is multiplication by zero. For any map $$0\cdot h$$ is the zero homomorphism and for any matrix $$0\cdot H$$ is the zero matrix.

Exercises
/Solutions/

title: Linear Algebra/Sums and Scalar Products


text:
Linear Algebra/Matrix Multiplication

After representing addition and scalar multiplication of linear maps in the prior subsection, the natural next map operation to consider is composition.

To see how the representation of the composite arises out of the representations of the two compositors, consider an example.

The theorem is an example of a result that supports a definition. We can picture what the definition and theorem together say with this arrow diagram ("wrt" abbreviates "with respect to").

Above the arrows, the maps show that the two ways of going from $$ V $$ to $$ X $$, straight over via the composition or else by way of $$ W $$, have the same effect

\vec{v}\stackrel{g\circ h}{\longmapsto}g(h(\vec{v})) \qquad \vec{v}\stackrel{h}{\longmapsto}h(\vec{v})\stackrel{g}{\longmapsto}g(h(\vec{v})) $$

(this is just the definition of composition). Below the arrows, the matrices indicate that the product does the same thing— multiplying $$GH$$ into the column vector $${\rm Rep}_{B}(\vec{v})$$ has the same effect as multiplying the column first by $$H$$ and then multiplying the result by $$G$$.

{\rm Rep}_{B,D}(g\circ h) =GH= {\rm Rep}_{C,D}(g)\,{\rm Rep}_{B,C}(h) $$

The definition of the matrix-matrix product operation does not restrict us to view it as a representation of a linear map composition. We can get insight into this operation by studying it as a mechanical procedure. The striking thing is the way that rows and columns combine.

One aspect of that combination is that the sizes of the matrices involved is significant. Briefly, $$m \! \times \! r\text{ times }r \! \times \! n\text{ equals }m \! \times \! n$$.

In terms of the underlying maps, the fact that the sizes must match up reflects the fact that matrix multiplication is defined only when a corresponding function composition

\text{dimension } n \text{ space} \;\stackrel{h}{\longrightarrow}\; \text{dimension } r \text{ space} \;\stackrel{g}{\longrightarrow}\; \text{dimension } m \text{ space} $$

is possible.

Another aspect of the way that rows and columns combine in the matrix product operation is that in the definition of the $$i,j$$ entry

p_{i,j} = g_{i,{\color{red}1}}h_{{\color{red}1},j} +g_{i,{\color{red}2}}h_{{\color{red}2},j} +\dots+g_{i,{\color{red}r}}h_{{\color{red}r},j} $$

the red subscripts on the $$g$$'s are column indicators while those on the $$h$$'s indicate rows. That is, summation takes place over the columns of $$G$$ but over the rows of $$H$$; left is treated differently than right, so $$ GH $$ may be unequal to $$ HG $$. Matrix multiplication is not commutative.

Except for the lack of commutativity, matrix multiplication is algebraically well-behaved. Below are some nice properties and more are in Problem 10 and Problem 11.

We have now seen how the representation of the composition of two linear maps is derived from the representations of the two maps. We have called the combination the product of the two matrices. This operation is extremely important. Before we go on to study how to represent the inverse of a linear map, we will explore it some more in the next subsection.

Exercises
/Solutions/

title: Linear Algebra/Matrix Multiplication


text:
Linear Algebra/Mechanics of Matrix Multiplication

In this subsection we consider matrix multiplication as a mechanical process, putting aside for the moment any implications about the underlying maps. As described earlier, the striking thing about matrix multiplication is the way rows and columns combine. The $$ i,j $$ entry of the matrix product is the dot product of row $$i$$ of the left matrix with column $$j$$ of the right one. For instance, here a second row and a third column combine to make a $$2,3$$ entry.

\begin{pmatrix} 1 & 1 \\ {\color{red} 0} & {\color{red} 1} \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 4 & 6 & {\color{red}8 } & 2\\ 5 & 7 & {\color{red}9 } & 3 \end{pmatrix} = \begin{pmatrix} 9 &13   &17                      &5  \\ 5  &7    &{\color{red}9}          &3  \\ 4 &6    &8                       &2 \end{pmatrix} $$

We can view this as the left matrix acting by multiplying its rows, one at a time, into the columns of the right matrix. Of course, another perspective is that the right matrix uses its columns to act on the left matrix's rows. Below, we will examine actions from the left and from the right for some simple matrices.

The first case, the action of a zero matrix, is very easy.

After zero matrices, the matrices whose actions are easiest to understand are the ones with a single nonzero entry.

Next in complication are matrices with two nonzero entries. There are two cases. If a left-multiplier has entries in different rows then their actions don't interact.

But if the left-multiplier's nonzero entries are in the same row then that row of the result is a combination.

Right-multiplication acts in the same way, with columns.

These observations about matrices that are mostly zeroes extend to arbitrary matrices.

An application of those observations is that there is a matrix that just copies out the rows and columns.

In short, an identity matrix is the identity element of the set of $$n \! \times \! n$$ matrices with respect to the operation of matrix multiplication.

We next see two ways to generalize the identity matrix.

The first is that if the ones are relaxed to arbitrary reals, the resulting matrix will rescale whole rows or columns.

The second generalization of identity matrices is that we can put a single one in each row and column in ways other than putting them down the diagonal.

We finish this subsection by applying these observations to get matrices that perform Gauss' method and Gauss-Jordan reduction.

To see how to perform a pivot, we observe something about those two examples. The matrix that rescales the second row by a factor of three arises in this way from the identity.

\begin{pmatrix} 1 &0  &0  \\ 0  &1  &0  \\ 0  &0  &1 \end{pmatrix} \xrightarrow[]{3\rho_2} \begin{pmatrix} 1 &0  &0  \\ 0  &3  &0  \\ 0  &0  &1 \end{pmatrix} $$

Similarly, the matrix that swaps first and third rows arises in this way.

\begin{pmatrix} 1 &0  &0  \\ 0  &1  &0  \\ 0  &0  &1 \end{pmatrix} \xrightarrow[]{\rho_1\leftrightarrow\rho_3} \begin{pmatrix} 0 &0  &1  \\ 0  &1  &0  \\ 1  &0  &0 \end{pmatrix} $$

We have observed the following result, which we shall use in the next subsection.

Until now we have taken the point of view that our primary objects of study are vector spaces and the maps between them, and have adopted matrices only for computational convenience. This subsection show that this point of view isn't the whole story. Matrix theory is a fascinating and fruitful area.

In the rest of this book we shall continue to focus on maps as the primary objects, but we will be pragmatic— if the matrix point of view gives some clearer idea then we shall use it.

Exercises
/Solutions/