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The Pink Panther Strikes Again

The Pink Panther Strikes Again is a 1976 comedy film. The fifth film in The Pink Panther series, its plot begins three years after the conclusion of The Return of the Pink Panther. Unused footage from the film was later included in Trail of the Pink Panther (1982), released after Peter Sellers's death.

Plot
After three years in a psychiatric hospital, former Chief Inspector of the Sûreté Charles Dreyfus has recovered his sanity and no longer is obsessed with killing Inspector Jacques Clouseau. Dreyfus is informed by his psychiatrist that he is to appear before the sanity board that afternoon pending release. Clouseau, who has since replaced Dreyfus as Chief Inspector, arrives unannounced to speak on behalf of his former boss and within five minutes manages to drive Dreyfus insane again. Dreyfus later escapes from the hospital and again tries to kill Clouseau by planting a bomb while the chief inspector (by periodic arrangement) duels with his manservant Cato. The bomb destroys Clouseau's apartment and injures Cato. However, Clouseau is unharmed, as he is lifted from the room by an inflatable hunchback disguise. Deciding that a more elaborate plan is needed to eliminate Clouseau, Dreyfus enlists career criminals and abducts professor Hugo Fassbender, a renowned nuclear physicist and the professor's daughter Margo. Dreyfus forces the professor to build a "doomsday weapon" in exchange for their freedom.

Clouseau travels to England to investigate the kidnapping and wreaks havoc in the Fassbender home while ineptly interrogating the domestic staff, including Jarvis, Fassbender's cross-dressing butler. Although Jarvis is later killed by the kidnappers, to whom he had become a dangerous witness, Clouseau discovers a clue that leads him to the Oktoberfest in Munich, West Germany. Meanwhile, Dreyfus, using Fassbender's invention, disintegrates the United Nations Building in New York City and blackmails the leaders of the world, including the president of the United States and his secretary of state (based on Gerald Ford and Henry Kissinger respectively), into assassinating Clouseau. Many nations instruct their operatives to kill Clouseau to gain Dreyfus's favor and possibly the doomsday machine. As a result of their orders and Clouseau's obliviousness, all of the assassins kill each another until only the agents of the Soviet Union and Egypt remain.

One of Dreyfus's henchmen, disguised as Clouseau, is killed by the Egyptian assassin after mistaking him for Clouseau. The Egyptian is seduced by Russian operative Olga Bariosova, who makes the same mistake and falls in love with him. After the Egyptian departs, the real Clouseau arrives in his hotel room. He is surprised to find Olga in his bed and is perplexed by her affections. From her information, Clouseau ascertains Dreyfus's location at a castle in Bavaria. Dreyfus is elated at the erroneous report of Clouseau's demise but suffers from a toothache and sends for a dentist. Arriving in Bavaria, Clouseau learns that a dentist is needed at the castle. He disguises himself as a German dentist and finally gains entry to the castle (his earlier attempts at sneaking into the castle had been foiled by his general ineptitude and the castle's drawbridge). Unrecognized by Dreyfus, Clouseau intoxicates both of them with nitrous oxide. While both are laughing uncontrollably, Clouseau mistakenly pulls the wrong tooth, and Dreyfus realizes that the dentist is actually Clouseau in disguise. Clouseau escapes, and a vengeful Dreyfus prepares to use the machine to destroy England. Clouseau, eluding Dreyfus's henchmen, unwittingly foils Dreyfus's plans when a medieval catapult outside the castle launches him on top of the doomsday machine, causing it to malfunction and fire on Dreyfus and the castle. As the remaining henchmen, Fassbender, his daughter and Clouseau escape the dissolving castle, Dreyfus plays "Tiptoe Through the Tulips" on the castle's pipe organ while he disintegrates, until he and the castle vanish into thin air.

Returning to Paris, Clouseau finds Olga waiting for him in his bed. However, their tryst is interrupted first by Clouseau's apparent inability to remove his clothes, and then by Cato's latest surprise attack, which causes all three to be hurled into the river Seine when the reclining bed snaps back upright and crashes through the wall.

Cast notes

 * ABC News anchorman and journalist Howard K. Smith appeared in the film as himself but his scenes were cut, though his name remains in the film's credits.
 * Julie Andrews, the wife of director Blake Edwards, provided the singing voice for the female impersonator Ainsley Jarvis. The scene in the nightclub when Jarvis sings is in many ways similar to scenes in Edwards's later film Victor Victoria (1982), in which Andrews plays a woman pretending to be a man who is a female impersonator.
 * Graham Stark, a longtime friend of Sellers, appears in a small role as the desk clerk of a small German hotel. Since his role as Hercule LaJoy in A Shot in the Dark, he had appeared in small roles in every Pink Panther sequel except Inspector Clouseau, in which Sellers did not play Clouseau.
 * Scenes featuring Harvey Korman as professor Auguste Balls and Marne Maitland as deputy commissioner Lasorde were deleted from the film, but were later seen in full in Trail of the Pink Panther in 1982. Stark would assume the role of Professor Balls in the next film, Revenge of the Pink Panther (1978).
 * Omar Sharif appears uncredited as the Egyptian assassin.
 * Tom Jones sings the Oscar-nominated song "Come to Me".
 * The role of Olga Bariosova was originally played by Maud Adams, who was replaced after filming a few scenes. Edwards then intended to cast Nicola Pagett after seeing her in Upstairs, Downstairs but instead cast Pagett's castmate Lesley-Anne Down.
 * Though the character of the President of the United States (portrayed by Dick Crockett) is unnamed in the film, he is obviously based on then-current president Gerald Ford. Crockett bore a marked physical resemblance to Ford, whose exaggerated reputation for clumsiness as depicted in the film was a national joke at the time. The president's unnamed somber Secretary of State (portrayed by Byron Kane) is obviously based on secretary Henry Kissinger.

Production
The Pink Panther Strikes Again was rushed into production by United Artists following the success of The Return of the Pink Panther. Blake Edwards had adapted one of two scripts that he and Frank Waldman had written for a proposed Pink Panther television series as the basis for that film, and he adapted the other as the starting point for The Pink Panther Strikes Again. As a result, it is the only Pink Panther sequel that has a storyline (Dreyfus in the insane asylum) that directly follows that of its previous film. The plot does not concern the famous Pink Panther diamond of previous films, but is played more as a parody of James Bond films.

The film was in production from December 1975 to September 1976, with principal photography taking place between February and June 1976. The strained relationship between Sellers and Edwards had further deteriorated by the time that production of The Pink Panther Strikes Again was under way. Sellers was ailing both mentally and physically, and Edwards later commented on the actor's mental state during production of the film: "If you went to an asylum and you described the first inmate you saw, that's what Peter had become. He was certifiable."

The original cut of the film ran for about 120 minutes but was trimmed to 103 minutes for theatrical release. Edwards originally conceived The Pink Panther Strikes Again as an even longer 180 minute epic, zany chase film, in a similar vein to his earlier comedy The Great Race, but UA vetoed the long version and the film was kept to a more conventional length. The excised footage was later used in Trail of the Pink Panther. There have been rumors that a 180 minute cut did exist as well, but given that only around 17 minutes of unused material was used in Trail, this does make it unlikely since it would mean an entire extra hour's worth of footage has gone unused and unreleased. No photograph, cast or crew comments or script evidence of any additional scenes not included in either Trail or Strikes Again has ever emerged either, leading to the conclusion that there likely never actually was a 3 hour cut and rumors of its existence may have simply come from Edwards's conception instead of from the actual filming. The Pink Panther Strikes Again was marketed with the tagline "Why are the world's chief assassins after Inspector Clouseau? Why not? Everybody else is."

During the film's title sequence, there are references to television's Alfred Hitchcock Presents and Batman as well as the films King Kong, The Sound of Music (starring Edwards's wife Julie Andrews), Dracula, Singin' in the Rain, Steamboat Bill, Jr. and Sweet Charity, placing the Pink Panther character and the animated persona of Inspector Clouseau into recognizable events from the films. There is also a reference to Jaws in the ending credits sequence. The scene in which Clouseau impersonates a dentist who uses laughing gas and extracts the wrong tooth is inspired by Bob Hope's role in The Paleface (1948).

Richard Williams (later of Who Framed Roger Rabbit fame) supervised the animation of the opening and closing sequences for the second and final time; original animators DePatie-Freleng Enterprises would return on the next film with animation influenced by Williams's style. Sellers was unhappy with the final cut of the film and publicly criticized Edwards for misusing his talents. Their tense relationship is noted in Revenge of the Pink Panther 's opening credits that list it as a "Sellers-Edwards" production. French comic-book writer René Goscinny, the original writer of the Asterix series, was reportedly trying to sue Edwards for plagiarism in 1977 after noticing strong similarities to Goscinny's script titled Le maître du monde (The Master of the World), which he had sent to Sellers in 1975.

Reception
On review aggregator Rotten Tomatoes, the film has an approval rating of 75% based on 24 reviews, with an average score of 7.30/10. The film earned theatrical rentals of $19.5 million in the United States and Canada from a gross of $33.8 million. Internationally, it earned rentals of $10.5 million for a worldwide total of $30 million. By March 1978, the film had grossed $75 million worldwide and was hoping to earn another $8 million by the end of the year.

Roger Ebert of the Chicago Sun-Times awarded the film two and a half stars out of four and wrote, "If I'm less than totally enthusiastic about The Pink Panther Strikes Again, maybe it was because I've been over this ground with Clouseau many times before," stating that a time would have to come "when inspiration gives way to habit, and I think the Pink Panther series is just about at that point. That's not to say this film isn't funny—it has moments as good as anything Sellers and Edwards have ever done—but that it's time for them to move on. They worked together once on the funniest movie either one has ever done, The Party. Now it's time to try something new again."

Vincent Canby of The New York Times wrote that the characters of Clouseau and Dreyfus "were made for each other," and further stated, "I'm not sure why Mr. Sellers and Mr. Lom are such a hilarious team, though it may be because each is a fine comic actor with a special talent for portraying the sort of all-consuming, epic self-absorption that makes slapstick farce initially acceptable—instead of alarming—and finally so funny." Canby also enjoyed Clouseau's French accent, and wrote, "Both Mr. Sellers and Mr. Edwards delight in old gags, and part of the joy of The Pink Panther Strikes Again is watching the way they spin out what is essentially a single routine".

Awards and nominations

 * American Film Institute Lists
 * AFI's 100 Years...100 Laughs – Nominated
 * AFI's 100 Years...100 Movie Quotes:
 * "Does your dog bite?" – Nominated

Play adaptation
Around 1981, the film was adapted into a play by William Gleason, mostly for high-school or community-theatre productions. The storyline bears similarities to that of the film, although some locations are changed, and women dressed as pink panthers also perform scene changes.

title: The Pink Panther Strikes Again


text:
Declaration of Boulogne

The Declaration on the Essence of Esperantism (Deklaracio pri la esenco de Esperantismo), commonly referred to as the Declaration of Boulogne (Bulonja Deklaracio), is a historic document that establishes several important premises for the Esperanto movement. The Declaration was written by L. L. Zamenhof and ratified in 1905 by the attendees of the first World Esperanto Congress, held in Boulogne-sur-Mer, France.

Content

The Declaration of Boulogne consists of an introduction and five points.

In the introduction, Zamenhof clarifies that the five points of the Declaration are necessary to establish because many people misunderstand the nature of the Esperanto movement. The five points are largely in response to these widely held misconceptions.

 * 1) Esperantism is a movement that supports the introduction of an international auxiliary language (IAL). No further meaning can be attached to it. It is politically, religiously, and morally neutral, and it does not seek to replace any existing languages, only to supplement them.
 * 2) It recognizes that Esperanto is the most realistic IAL that exists, and they work to further it based on this goal.
 * 3) Esperanto belongs to no one. Anyone can use it for any reason they like.
 * 4) Fundamento de Esperanto is the single, perpetual obligatory authority over Esperanto, and it cannot be modified. Otherwise, Esperanto depends on no legal authority, neither a governing body nor an individual, including Zamenhof himself. If a linguistic matter is not covered in Fundamento, it is up to the individual on how to handle the matter.
 * 5) An Esperantist is a fluent Esperanto speaker. Involvement in the Esperanto community is encouraged but not required.

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

title: Declaration of Boulogne


text:
Talk:Peter J. Schoomaker

Dear Sir,

My husband was in the army back in 1971 when he was 18 years old, just get out of high school. He was shot two times once in the knee and once in the stomach. He has a lot of pain in his knee and has had problems with his stomach from day one. He is 100% disabled with Social Security and he has filed 2 are 3 time for 100% with the VA and hasn't gotten it from when he filed back in Oct.2004 and doesn't know anything. He came home from the army to get well and the sgt. said the paper work would follow along with his purple heart paper he didn't get any of them. He just get a ribbon to wear home. He was service connected at the time he came home but he didn't know it. The reason he didn't follow up with a doctors at the VA hospitl after he came home from the service is the he didn't know how. Some of his bowls were removed in the surgery. He has diarrhea 9 to 10 times a day because of it. He has been checked for agent orange and it was positive. He has had two strokes. He has diabetes. He is legally blind in one eye and the other eye is 20/30. He has had open heart surgery from congestive heart failure. He is very tierd all of the time and weak. He has recently been diagnosed with COPD, a lung diesease. He uses inhalors just so he can breath. He has to have hearing aids for both ears. He served our country and our county as a sherriff's deputy for 19 and 1/2 years. How many of our soilders don't know what kind of help they can get from the VA because the paperwork don't follow. And when they do find out about it no one wants to help them. He is 60% disabled with the VA, what is 40% more? I think he would qualify for more than that with his disablities. And no one can answer the questions of what all agent orange can do to you. I would appreciate any help you can give us in this matter. I stay at home and take care of him 24/7.

Thank You, Regina Pollard. kaynckay@charter.net reginapollard@yahoo.com 828-851-3418

title: Talk:Peter J. Schoomaker


text:
Shoot-and-scoot

Shoot-and-scoot (alternatively, fire-and-displace or fire-and-move) is an artillery tactic of firing at a target and then immediately moving away from the location from where the shots were fired to avoid counter-battery fire, e.g., from enemy artillery.

Caucasian War
The first recorded use of this tactic came from the Caucasian War where the Chechen Naib Talkhig of Shali became famous for his shoot-and-scoot tactics, termed nomadic artillery by Russians, in the 1830-1850s. According to the Russian historian and professor Nikolay Smirnov, he was one of the first commanders to use this tactic.

World War II
The need for such tactics in World War II became obvious from the noticeable smoke signature produced by the use of anti-tank infantry weapons such as the German Panzerfaust anti-tank grenade launcher, the American M1 bazooka and its German Panzerschreck derivative anti-tank rocket launchers, and also by the various models of Nebelwerfer and Wurfrahmen 40 German barrage rocket systems. In modern times, moving after firing is important, as there are a multitude of electronic systems, such as counter-battery radar that can automatically detect artillery fire in near real-time and direct counter-battery fire from friendly artillery.

Rocket launcher
Shoot and scoot was originally made possible and used by Soviet Katyusha rocket artillery, whose rudimentary construction of rails on a truck chassis made it comparatively light and mobile, while its quick 7-10 second full salvo, slow reload, and complete lack of protection made switching positions its main tactic and best defence.

NATO
Shoot and scoot tactics were first adopted by NATO in the early 1960s by its nuclear artillery units using towed 8-inch howitzers and truck-mounted MGR-1 Honest John rockets. These operated as single guns or launchers, typically deployed to a hide, came into action when given a nuclear fire mission, fired and immediately came out of action and moved to another hide adjacent to another firing position. These tactics became normal for missile units worldwide and were successfully used by Iraqi missile units in the 1990–1991 Gulf War.

M270
Similar tactics were adopted by M270 Multiple Launch Rocket System (MLRS) platoons when it entered service, although in this case the launchers usually went first to a reload site after firing. It may be used by Indian Pinaka MBRL or Russian Grad, Uragan, and Smerch, although originally systems like Grad used more traditional tactics.

Self-propelled gun batteries
The tactic has also been adapted for more general use with self-propelled gun batteries (such as the US M109 Paladin or German PzH2000).

AS-90
A refined version are gun manoeuvre areas developed by UK for use by AS-90 batteries when the counter-battery threat is high. These areas cover several square kilometres and the guns move around in them in pairs. The difficult decision is deciding how long guns can remain in one place before moving. This requires judgement about the enemy's counter-battery responsiveness. In these areas, the battery's command elements remain in one place and there are various options for replenishing the guns' on-board ammunition. The tactic is affected by the rate of fire of the manoeuvring guns and the time it takes them to come out of action. However, it is also very wearing on the gun crews and for sustained use relief crews are essential.

G6 Howitzer
South African Defence Force G6 Howitzers used the technique very effectively in Angola against Cuban forces in the Battle of Cuito Cuanavale.

title: Shoot-and-scoot


text:
Wikipedia:Articles for deletion/Quietude/archive1

 This page is an archive of the discussion about the proposed deletion of the article below. This page is no longer live. Further comments should be made on the article's talk page rather than here so that this page is preserved as an historic record. The result of the debate was transwiki. Rossami (talk) 06:58, 13 Mar 2005 (UTC)

quietude
It's a New Age -thing, but I can't see it being more than a dicdef. Squidwina 16:01, 24 Feb 2005 (UTC)
 * Yes, it's a dicdef. Transwiki as such. Also, considering the context, a redirect to something like meditation may be useful. Radiant! 21:21, Feb 24, 2005 (UTC)
 * Transwiki. Megan1967 23:49, 24 Feb 2005 (UTC)
 * Transwiki, but no redirect. The word, as I understand it, is a bit too widely defined for just meditation to apply: repose, dormancy, hibernation, etc. HyperZonktalk 17:35, Feb 25, 2005 (UTC)

This page is now preserved as an archive of the debate and, like some other VfD subpages, is no longer 'live'. Subsequent comments on the issue, the deletion, or the decision-making process should be placed on the relevant 'live' pages. Please do not edit this page.


text: Category:2005 in Canada

title: Category:2005 in Canada


text:
Perron–Frobenius theorem

In matrix theory, the Perron–Frobenius theorem, proved by and, asserts that a real square matrix with positive entries has a unique eigenvalue of largest magnitude and that eigenvalue is real. The corresponding eigenvector can be chosen to have strictly positive components, and also asserts a similar statement for certain classes of nonnegative matrices. This theorem has important applications to probability theory (ergodicity of Markov chains); to the theory of dynamical systems (subshifts of finite type); to economics (Okishio's theorem, Hawkins–Simon condition ); to demography (Leslie population age distribution model); to social networks (DeGroot learning process); to Internet search engines (PageRank); and even to ranking of American football teams. The first to discuss the ordering of players within tournaments using Perron–Frobenius eigenvectors is Edmund Landau.

Statement
Let positive and non-negative respectively describe matrices with exclusively positive real numbers as elements and matrices with exclusively non-negative real numbers as elements. The eigenvalues of a real square matrix A are complex numbers that make up the spectrum of the matrix. The exponential growth rate of the matrix powers Ak as k → ∞ is controlled by the eigenvalue of A with the largest absolute value (modulus). The Perron–Frobenius theorem describes the properties of the leading eigenvalue and of the corresponding eigenvectors when A is a non-negative real square matrix. Early results were due to and concerned positive matrices. Later, found their extension to certain classes of non-negative matrices.

Positive matrices
Let $$A = (a_{ij}) $$ be an $$ n \times n $$ positive matrix: $$ a_{ij} > 0 $$ for $$ 1 \le i,j \le n $$. Then the following statements hold.

 * 1) There is a positive real number r, called the Perron root or the Perron–Frobenius eigenvalue (also called the leading eigenvalue, principal eigenvalue or dominant eigenvalue), such that r is an eigenvalue of A and any other eigenvalue λ (possibly complex) in absolute value is strictly smaller than r, |λ| < r. Thus, the spectral radius $$\rho(A) $$ is equal to r. If the matrix coefficients are algebraic, this implies that the eigenvalue is a Perron number.
 * 2) The Perron–Frobenius eigenvalue is simple: r is a simple root of the characteristic polynomial of A. Consequently, the eigenspace associated to r is one-dimensional. (The same is true for the left eigenspace, i.e., the eigenspace for AT, the transpose of A.)
 * 3) There exists an eigenvector v = (v1,...,vn)T of A with eigenvalue r such that all components of v are positive: A v = r v, vi > 0 for 1 ≤ i ≤ n. (Respectively, there exists a positive left eigenvector w : wT A = wT r, wi > 0.) It is known in the literature under many variations as the Perron vector, Perron eigenvector, Perron-Frobenius eigenvector, leading eigenvector, principal eigenvector or dominant eigenvector.
 * 4) There are no other positive (moreover non-negative) eigenvectors except positive multiples of v (respectively, left eigenvectors except ww'w''), i.e., all other eigenvectors must have at least one negative or non-real component.
 * 5) $$ \lim_{k \rightarrow \infty} A^k/r^k = v w^T$$, where the left and right eigenvectors for A are normalized so that wTv = 1. Moreover, the matrix vwT is the projection onto the eigenspace corresponding to r. This projection is called the Perron projection.
 * 6) Collatz–Wielandt formula: for all non-negative non-zero vectors x, let f(x) be the minimum value of [Ax]i / xi taken over all those i such that xi ≠ 0. Then f is a real valued function whose maximum over all non-negative non-zero vectors x is the Perron–Frobenius eigenvalue.
 * 7) A "Min-max" Collatz–Wielandt formula takes a form similar to the one above: for all strictly positive vectors x, let g(x) be the maximum value of [Ax]i / xi taken over i. Then g is a real valued function whose minimum over all strictly positive vectors x is the Perron–Frobenius eigenvalue.
 * 8) Birkhoff–Varga formula: Let x and y be strictly positive vectors.  Then, $$r = \sup_{x>0} \inf_{y>0} \frac{y^\top A x}{y^\top x} = \inf_{x>0} \sup_{y>0} \frac{y^\top A x}{y^\top x} = \inf_{x>0} \sup_{y>0} \sum_{i,j=1}^n y_i a_{ij} x_j/\sum_{i=1}^n y_i x_i.$$
 * 9) Donsker–Varadhan–Friedland formula: Let p be a probability vector and x a strictly positive vector.  Then,  $$r = \sup_p \inf_{x>0} \sum_{i=1}^n p_i[Ax]_i/x_i.$$
 * 10) Fiedler formula: $$r = \sup_{z > 0} \ \inf_{x>0, \ y>0,\ x \circ y = z} \frac{y^\top A x}{y^\top x} = \sup_{z > 0} \ \inf_{x>0, \ y>0,\ x \circ y = z}\sum_{i,j=1}^n y_i a_{ij} x_j/\sum_{i=1}^n y_i x_i.$$
 * 11) The Perron–Frobenius eigenvalue satisfies the inequalities $$\min_i \sum_{j} a_{ij} \le r \le \max_i \sum_{j} a_{ij}.$$

All of these properties extend beyond strictly positive matrices to primitive matrices (see below). Facts 1–7 can be found in Meyer chapter 8 claims 8.2.11–15 page 667 and exercises 8.2.5,7,9 pages 668–669.

The left and right eigenvectors w and v are sometimes normalized so that the sum of their components is equal to 1; in this case, they are sometimes called stochastic eigenvectors. Often they are normalized so that the right eigenvector v sums to one, while $$w^T v=1$$.

Non-negative matrices
There is an extension to matrices with non-negative entries. Since any non-negative matrix can be obtained as a limit of positive matrices, one obtains the existence of an eigenvector with non-negative components; the corresponding eigenvalue will be non-negative and greater than or equal, in absolute value, to all other eigenvalues. However, for the example $$A = \left(\begin{smallmatrix}0 & 1\\ 1 & 0\end{smallmatrix}\right)$$, the maximum eigenvalue r = 1 has the same absolute value as the other eigenvalue −1; while for $$A = \left(\begin{smallmatrix}0 & 1\\ 0 & 0\end{smallmatrix}\right)$$, the maximum eigenvalue is r = 0, which is not a simple root of the characteristic polynomial, and the corresponding eigenvector (1, 0) is not strictly positive.

However, Frobenius found a special subclass of non-negative matrices — irreducible matrices — for which a non-trivial generalization is possible. For such a matrix, although the eigenvalues attaining the maximal absolute value might not be unique, their structure is under control: they have the form $$\omega r$$, where $$r$$ is a real strictly positive eigenvalue, and $$\omega$$ ranges over the complex h' th roots of 1 for some positive integer h called the period of the matrix. The eigenvector corresponding to $$r$$ has strictly positive components (in contrast with the general case of non-negative matrices, where components are only non-negative). Also all such eigenvalues are simple roots of the characteristic polynomial. Further properties are described below.

Classification of matrices
Let A be a n × n square matrix over field F. The matrix A is irreducible if any of the following equivalent properties holds.

Definition 1 : A does not have non-trivial invariant coordinate subspaces. Here a non-trivial coordinate subspace means a linear subspace spanned by any proper subset of standard basis vectors of Fn. More explicitly, for any linear subspace spanned by standard basis vectors ei 1, ..., ei k, 0 < k < n its image under the action of A is not contained in the same subspace.

Definition 2: A cannot be conjugated into block upper triangular form by a permutation matrix P:
 * $$PAP^{-1} \ne

\begin{pmatrix} E & F \\ O & G \end{pmatrix},$$ where E and G are non-trivial (i.e. of size greater than zero) square matrices.

Definition 3: One can associate with a matrix A a certain directed graph GA. It has n vertices labeled 1,...,n, and there is an edge from vertex i to vertex j precisely when aij ≠ 0. Then the matrix A is irreducible if and only if its associated graph GA is strongly connected.

If F is the field of real or complex numbers, then we also have the following condition.

Definition 4: The group representation of $$(\mathbb R, +)$$ on $$\mathbb{R}^n$$ or $$(\mathbb C, +)$$ on $$\mathbb{C}^n$$ given by $$t \mapsto\exp(tA)$$ has no non-trivial invariant coordinate subspaces. (By comparison, this would be an irreducible representation if there were no non-trivial invariant subspaces at all, not only considering coordinate subspaces.)

A matrix is reducible if it is not irreducible.

A real matrix A is primitive if it is non-negative and its mth power is positive for some natural number m (i.e. all entries of Am are positive).

Let A be real and non-negative. Fix an index i and define the period of index i  to be the greatest common divisor of all natural numbers m such that (Am)ii > 0. When A is irreducible, the period of every index is the same and is called the 'period of A''. ' In fact, when A is irreducible, the period can be defined as the greatest common divisor of the lengths of the closed directed paths in G''A (see Kitchens page 16). The period is also called the index of imprimitivity (Meyer page 674) or the order of cyclicity. If the period is 1, A is aperiodic. It can be proved that primitive matrices are the same as irreducible aperiodic non-negative matrices.

All statements of the Perron–Frobenius theorem for positive matrices remain true for primitive matrices. The same statements also hold for a non-negative irreducible matrix, except that it may possess several eigenvalues whose absolute value is equal to its spectral radius, so the statements need to be correspondingly modified. In fact the number of such eigenvalues is equal to the period.

Results for non-negative matrices were first obtained by Frobenius in 1912.

Perron–Frobenius theorem for irreducible non-negative matrices
Let $$A$$ be an irreducible non-negative $$N\times N$$ matrix with period $$h$$ and spectral radius $$\rho(A) = r$$. Then the following statements hold.

 * The number $$r\in\mathbb{R}^+$$ is a positive real number and it is an eigenvalue of the matrix $$A$$. It is called Perron–Frobenius eigenvalue.
 * The Perron–Frobenius eigenvalue $$r$$ is simple. Both right and left eigenspaces associated with $$r$$ are one-dimensional.
 * $$A$$ has both a right and a left eigenvectors, respectively $$\mathbf v$$ and $$\mathbf w$$, with eigenvalue $$r$$ and whose components are all positive. Moreover these are the only eigenvectors whose components are all positive are those associated with the eigenvalue $$r$$.
 * The matrix $$A$$ has exactly $$h$$ (where $$h$$ is the period) complex eigenvalues with absolute value $$r$$. Each of them is a simple root of the characteristic polynomial and is the product of $$r$$ with an $$h$$th root of unity.
 * Let $$\omega = 2\pi/h$$. Then the matrix $$A$$ is similar to $$e^{i\omega}A$$, consequently the spectrum of $$A$$ is invariant under multiplication by $$e^{i\omega}$$ (i.e. to rotations of the complex plane by the angle $$\omega$$).
 * If $$h>1$$ then there exists a permutation matrix $$P$$ such that
 * $$PAP^{-1}=

\begin{pmatrix} O & A_1 & O & O & \ldots & O \\ O & O & A_2 & O & \ldots & O \\ \vdots & \vdots &\vdots & \vdots & & \vdots \\ O & O & O & O & \ldots & A_{h-1} \\ A_h & O & O & O & \ldots & O \end{pmatrix}, $$

where $$O$$ denotes a zero matrix and the blocks along the main diagonal are square matrices.

$$ let $$f(\mathbf x) $$ be the minimum value of $$[A\mathbf x]_i/x_i $$'' taken over all those $$i $$ such that $$x_i\neq0 $$. Then $$f $$ is a real valued function whose maximum is the Perron–Frobenius eigenvalue.
 * Collatz–Wielandt formula: for all non-negative non-zero vectors ''$$\mathbf x

 * The Perron–Frobenius eigenvalue satisfies the inequalities
 * $$\min_i \sum_{j} a_{ij} \le r \le \max_i \sum_{j} a_{ij}.$$

The example $$A =\left(\begin{smallmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 0 \end{smallmatrix}\right)$$ shows that the (square) zero-matrices along the diagonal may be of different sizes, the blocks Aj need not be square, and h need not divide n.

Further properties
Let A be an irreducible non-negative matrix, then: P A^q P^{-1}= \begin{pmatrix} A_1 & O & O & \dots & O \\ O & A_2 & O & \dots & O \\ \vdots & \vdots & \vdots & & \vdots \\ O & O & O & \dots & A_d \\ \end{pmatrix} $$, where Ai are irreducible matrices having the same maximal eigenvalue. The number of these matrices d is the greatest common divisor of q and h, where h is period of A.
 * 1) (I+A)n−1 is a positive matrix. (Meyer claim 8.3.5 p. 672). For a non-negative A, this is also a sufficient condition.
 * 2) Wielandt's theorem.  If |B|<A, then ρ(B)≤ρ(A). If equality holds (i.e. if μ=ρ(A)eiφ is eigenvalue for B), then B = eiφ D AD−1 for some diagonal unitary matrix D (i.e. diagonal elements of D equals to eiΘl, non-diagonal are zero).
 * 3) If some power Aq is reducible, then it is completely reducible, i.e. for some permutation matrix P, it is true that: $$
 * 1) If c(x) = xn + ck 1 xn-k1 + ck 2 xn-k2 + ... + ck s xn-ks is the characteristic polynomial of A in which only the non-zero terms are listed, then the period of A equals the greatest common divisor of k1, k2, ..., ks.
 * 2) Cesàro averages: $$ \lim_{k \rightarrow \infty} 1/k\sum_{i=0,...,k} A^i/r^i = ( v w^T),$$ where the left and right eigenvectors for A are normalized so that wTv = 1. Moreover, the matrix v wT is the spectral projection corresponding to r, the Perron projection.
 * 3) Let r be the Perron–Frobenius eigenvalue, then the adjoint matrix for (r-A) is positive.
 * 4) If A has at least one non-zero diagonal element, then A is primitive.
 * 5) If 0 ≤ A < B, then rA ≤ rB. Moreover, if B is irreducible, then the inequality is strict: rA < rB.

A matrix A is primitive provided it is non-negative and Am is positive for some m, and hence Ak is positive for all k ≥ m. To check primitivity, one needs a bound on how large the minimal such m can be, depending on the size of A:
 * If A is a non-negative primitive matrix of size n, then An 2 − 2n + 2 is positive. Moreover, this is the best possible result, since for the matrix M below, the power Mk is not positive for every k < n2 − 2n + 2, since (Mn 2 − 2n+1 )1,1 = 0.
 * $$M=

\left(\begin{smallmatrix} 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 \\ 1 & 1 & 0 & 0 & \cdots & 0 \end{smallmatrix}\right) $$

Applications
Numerous books have been written on the subject of non-negative matrices, and Perron–Frobenius theory is invariably a central feature. The following examples given below only scratch the surface of its vast application domain.

Non-negative matrices
The Perron–Frobenius theorem does not apply directly to non-negative matrices. Nevertheless, any reducible square matrix A may be written in upper-triangular block form (known as the normal form of a reducible matrix)
 * PAP−1 = $$ \left( \begin{smallmatrix}

B_1 & * & * & \cdots & * \\ 0 & B_2 & * & \cdots & * \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & * \\ 0 & 0 & 0 & \cdots & B_h \end{smallmatrix} \right)$$

where P is a permutation matrix and each Bi is a square matrix that is either irreducible or zero. Now if A is non-negative then so too is each block of PAP−1, moreover the spectrum of A is just the union of the spectra of the Bi.

The invertibility of A can also be studied. The inverse of PAP−1 (if it exists) must have diagonal blocks of the form Bi−1 so if any Bi isn't invertible then neither is PAP−1 or A. Conversely let D be the block-diagonal matrix corresponding to PAP−1, in other words PAP−1 with the asterisks zeroised. If each Bi is invertible then so is D and D−1(PAP−1) is equal to the identity plus a nilpotent matrix. But such a matrix is always invertible (if Nk = 0 the inverse of 1 − N is 1 + N + N2 + ... + Nk−1) so PAP−1 and A are both invertible.

Therefore, many of the spectral properties of A may be deduced by applying the theorem to the irreducible Bi. For example, the Perron root is the maximum of the ρ(Bi). While there will still be eigenvectors with non-negative components it is quite possible that none of these will be positive.

Stochastic matrices
A row (column) stochastic matrix is a square matrix each of whose rows (columns) consists of non-negative real numbers whose sum is unity. The theorem cannot be applied directly to such matrices because they need not be irreducible.

If A is row-stochastic then the column vector with each entry 1 is an eigenvector corresponding to the eigenvalue 1, which is also ρ(A) by the remark above. It might not be the only eigenvalue on the unit circle: and the associated eigenspace can be multi-dimensional. If A is row-stochastic and irreducible then the Perron projection is also row-stochastic and all its rows are equal.

Algebraic graph theory
The theorem has particular use in algebraic graph theory. The "underlying graph" of a nonnegative n-square matrix is the graph with vertices numbered 1, ..., n and arc ij if and only if Aij ≠ 0. If the underlying graph of such a matrix is strongly connected, then the matrix is irreducible, and thus the theorem applies. In particular, the adjacency matrix of a strongly connected graph is irreducible.

Finite Markov chains
The theorem has a natural interpretation in the theory of finite Markov chains (where it is the matrix-theoretic equivalent of the convergence of an irreducible finite Markov chain to its stationary distribution, formulated in terms of the transition matrix of the chain; see, for example, the article on the subshift of finite type).

Compact operators
More generally, it can be extended to the case of non-negative compact operators, which, in many ways, resemble finite-dimensional matrices. These are commonly studied in physics, under the name of transfer operators, or sometimes Ruelle–Perron–Frobenius operators (after David Ruelle). In this case, the leading eigenvalue corresponds to the thermodynamic equilibrium of a dynamical system, and the lesser eigenvalues to the decay modes of a system that is not in equilibrium. Thus, the theory offers a way of discovering the arrow of time in what would otherwise appear to be reversible, deterministic dynamical processes, when examined from the point of view of point-set topology.

Proof methods
A common thread in many proofs is the Brouwer fixed point theorem. Another popular method is that of Wielandt (1950). He used the Collatz–Wielandt formula described above to extend and clarify Frobenius's work. Another proof is based on the spectral theory from which part of the arguments are borrowed.

Perron root is strictly maximal eigenvalue for positive (and primitive) matrices
If A is a positive (or more generally primitive) matrix, then there exists a real positive eigenvalue r (Perron–Frobenius eigenvalue or Perron root), which is strictly greater in absolute value than all other eigenvalues, hence r is the spectral radius of A.

This statement does not hold for general non-negative irreducible matrices, which have h eigenvalues with the same absolute eigenvalue as r, where h is the period of A.

Proof for positive matrices
Let A be a positive matrix, assume that its spectral radius ρ(A) = 1 (otherwise consider A/ρ(A)). Hence, there exists an eigenvalue λ on the unit circle, and all the other eigenvalues are less or equal 1 in absolute value. Suppose that another eigenvalue λ ≠ 1 also falls on the unit circle. Then there exists a positive integer m such that Am is a positive matrix and the real part of λm is negative. Let ε be half the smallest diagonal entry of Am and set T = Am − εI which is yet another positive matrix. Moreover, if Ax = λx then Amx = λmx thus λm − ε is an eigenvalue of T. Because of the choice of m this point lies outside the unit disk consequently ρ(T) > 1. On the other hand, all the entries in T are positive and less than or equal to those in Am so by Gelfand's formula ρ(T) ≤ ρ(Am) ≤ ρ(A)m = 1. This contradiction means that λ=1 and there can be no other eigenvalues on the unit circle.

Absolutely the same arguments can be applied to the case of primitive matrices; we just need to mention the following simple lemma, which clarifies the properties of primitive matrices.

Lemma
Given a non-negative A, assume there exists m, such that Am is positive, then Am+1, Am+2, Am+3,... are all positive.

Am+1 = AAm, so it can have zero element only if some row of A is entirely zero, but in this case the same row of Am will be zero.

Applying the same arguments as above for primitive matrices, prove the main claim.

Power method and the positive eigenpair
For a positive (or more generally irreducible non-negative) matrix A the dominant eigenvector is real and strictly positive (for non-negative A respectively non-negative.)

This can be established using the power method, which states that for a sufficiently generic (in the sense below) matrix A the sequence of vectors bk+1 = Abk / | Abk | converges to the eigenvector with the maximum eigenvalue. (The initial vector b0 can be chosen arbitrarily except for some measure zero set). Starting with a non-negative vector b0 produces the sequence of non-negative vectors bk. Hence the limiting vector is also non-negative. By the power method this limiting vector is the dominant eigenvector for A, proving the assertion. The corresponding eigenvalue is non-negative.

The proof requires two additional arguments. First, the power method converges for matrices which do not have several eigenvalues of the same absolute value as the maximal one. The previous section's argument guarantees this.

Second, to ensure strict positivity of all of the components of the eigenvector for the case of irreducible matrices. This follows from the following fact, which is of independent interest:

 * Lemma: given a positive (or more generally irreducible non-negative) matrix A and v as any non-negative eigenvector for A, then it is necessarily strictly positive and the corresponding eigenvalue is also strictly positive.

Proof. One of the definitions of irreducibility for non-negative matrices is that for all indexes i,j there exists m, such that (Am)ij is strictly positive. Given a non-negative eigenvector v, and that at least one of its components say j-th is strictly positive, the corresponding eigenvalue is strictly positive, indeed, given n such that (An)ii >0, hence: rnvi = Anvi ≥ (An)iivi >0. Hence r is strictly positive. The eigenvector is strict positivity. Then given m, such that (Am)ij >0, hence: rmvj = (Amv)j ≥ (Am)ijvi >0, hence vj is strictly positive, i.e., the eigenvector is strictly positive.

Multiplicity one
This section proves that the Perron–Frobenius eigenvalue is a simple root of the characteristic polynomial of the matrix. Hence the eigenspace associated to Perron–Frobenius eigenvalue r is one-dimensional. The arguments here are close to those in Meyer.

Given a strictly positive eigenvector v corresponding to r and another eigenvector w with the same eigenvalue. (The vectors v and w can be chosen to be real, because A and r are both real, so the null space of A-r has a basis consisting of real vectors.) Assuming at least one of the components of w is positive (otherwise multiply w by −1). Given maximal possible α such that u=v- α w is non-negative, then one of the components of u is zero, otherwise α is not maximum. Vector u is an eigenvector. It is non-negative, hence by the lemma described in the previous section non-negativity implies strict positivity for any eigenvector. On the other hand, as above at least one component of u is zero. The contradiction implies that w does not exist.

Case: There are no Jordan cells corresponding to the Perron–Frobenius eigenvalue r and all other eigenvalues which have the same absolute value.

If there is a Jordan cell, then the infinity norm (A/r)k∞ tends to infinity for k → ∞ , but that contradicts the existence of the positive eigenvector.

Given r = 1, or A/r. Letting v be a Perron–Frobenius strictly positive eigenvector, so Av=v, then:

$$ \|v\|_{\infty}= \|A^k v\|_{\infty} \ge \|A^k\|_{\infty} \min_i (v_i), \Rightarrow \|A^k\|_{\infty} \le \|v\|/\min_i (v_i) $$ So Ak∞ is bounded for all k. This gives another proof that there are no eigenvalues which have greater absolute value than Perron–Frobenius one. It also contradicts the existence of the Jordan cell for any eigenvalue which has absolute value equal to 1 (in particular for the Perron–Frobenius one), because existence of the Jordan cell implies that Ak∞ is unbounded. For a two by two matrix:

J^k= \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix} ^k = \begin{pmatrix} \lambda^k & k\lambda^{k-1} \\ 0 & \lambda^k \end{pmatrix}, $$ hence Jk∞ = |k + λ| (for |λ| = 1), so it tends to infinity when k does so. Since Jk = C−1 AkC, then Ak ≥ Jk/ (C−1 C ), so it also tends to infinity. The resulting contradiction implies that there are no Jordan cells for the corresponding eigenvalues.

Combining the two claims above reveals that the Perron–Frobenius eigenvalue r is simple root of the characteristic polynomial. In the case of nonprimitive matrices, there exist other eigenvalues which have the same absolute value as r. The same claim is true for them, but requires more work.

No other non-negative eigenvectors
Given positive (or more generally irreducible non-negative matrix) A, the Perron–Frobenius eigenvector is the only (up to multiplication by constant) non-negative eigenvector for A.

Other eigenvectors must contain negative or complex components since eigenvectors for different eigenvalues are orthogonal in some sense, but two positive eigenvectors cannot be orthogonal, so they must correspond to the same eigenvalue, but the eigenspace for the Perron–Frobenius is one-dimensional.

Assuming there exists an eigenpair (λ, y) for A, such that vector y is positive, and given (r, x), where x – is the left Perron–Frobenius eigenvector for A (i.e. eigenvector for AT), then rxTy = (xT A) y = xT (Ay) = λxTy, also xT y > 0, so one has: r = λ. Since the eigenspace for the Perron–Frobenius eigenvalue r is one-dimensional, non-negative eigenvector y is a multiple of the Perron–Frobenius one.

Collatz–Wielandt formula
Given a positive (or more generally irreducible non-negative matrix) A, one defines the function f on the set of all non-negative non-zero vectors x such that f(x) is the minimum value of [Ax]i / xi taken over all those i such that xi ≠ 0. Then f is a real-valued function, whose maximum is the Perron–Frobenius eigenvalue r.

For the proof we denote the maximum of f by the value R. The proof requires to show  R = r. Inserting the Perron-Frobenius eigenvector v into f, we obtain f(v) = r and conclude r ≤ R. For the opposite inequality, we consider an arbitrary nonnegative vector x and let ξ=f(x). The definition of f gives 0 ≤ ξx ≤ Ax (componentwise). Now, we use the positive right eigenvector w for A for the Perron-Frobenius eigenvalue r, then  ξ wT x = wT ξx ≤ wT (Ax) = (wT A)x = r wT x . Hence f(x) = ξ ≤ r, which implies R ≤ r.

Perron projection as a limit: Ak/rk
Let A be a positive (or more generally, primitive) matrix, and let r be its Perron–Frobenius eigenvalue.
 * 1) There exists a limit Ak/rk for k → ∞, denote it by P.
 * 2) P is a projection operator: P2 = P, which commutes with A: AP = PA.
 * 3) The image of P is one-dimensional and spanned by the Perron–Frobenius eigenvector v (respectively for PT—by the Perron–Frobenius eigenvector w for AT).
 * 4) P = vwT, where v,w are normalized such that wT v = 1.
 * 5) Hence P is a positive operator.

Hence P is a spectral projection for the Perron–Frobenius eigenvalue r, and is called the Perron projection. The above assertion is not true for general non-negative irreducible matrices.

Actually the claims above (except claim 5) are valid for any matrix M such that there exists an eigenvalue r which is strictly greater than the other eigenvalues in absolute value and is the simple root of the characteristic polynomial. (These requirements hold for primitive matrices as above).

Given that M is diagonalizable, M is conjugate to a diagonal matrix with eigenvalues r1, ..., rn on the diagonal (denote r1 = r). The matrix Mk/rk will be conjugate (1, (r2/r)k, ..., (rn/r)k), which tends to (1,0,0,...,0), for k → ∞, so the limit exists. The same method works for general M (without assuming that M is diagonalizable).

The projection and commutativity properties are elementary corollaries of the definition: MMk/rk = Mk/rk M ; P2 = lim M2k/r2k = P. The third fact is also elementary: M(Pu) = M lim Mk/rk u = lim rMk+1/rk+1u, so taking the limit yields M(Pu) = r(Pu), so image of P lies in the r-eigenspace for M, which is one-dimensional by the assumptions.

Denoting by v, r-eigenvector for M (by w for MT). Columns of P are multiples of v, because the image of P is spanned by it. Respectively, rows of w. So P takes a form (a v wT), for some a. Hence its trace equals to (a wT v). Trace of projector equals the dimension of its image. It was proved before that it is not more than one-dimensional. From the definition one sees that P acts identically on the r-eigenvector for M. So it is one-dimensional. So choosing (wTv) = 1, implies P = vwT.

Inequalities for Perron–Frobenius eigenvalue
For any non-negative matrix A its Perron–Frobenius eigenvalue r satisfies the inequality:

 * $$ r \; \le \; \max_i \sum_j a_{ij}.$$

This is not specific to non-negative matrices: for any matrix A with an eigenvalue $$\scriptstyle\lambda$$ it is true that $$\scriptstyle |\lambda| \; \le \; \max_i \sum_j |a_{ij}|$$. This is an immediate corollary of the Gershgorin circle theorem. However another proof is more direct:

Any matrix induced norm satisfies the inequality $$\scriptstyle\|A\| \ge |\lambda|$$ for any eigenvalue $$\scriptstyle\lambda$$ because, if $$\scriptstyle x$$ is a corresponding eigenvector, $$\scriptstyle\|A\| \ge |Ax|/|x| = |\lambda x|/|x| = |\lambda|$$. The infinity norm of a matrix is the maximum of row sums: $$\scriptstyle \left \| A \right \| _\infty = \max \limits _{1 \leq i \leq m} \sum _{j=1} ^n | a_{ij} |. $$ Hence the desired inequality is exactly $$\scriptstyle\|A\|_\infty \ge |\lambda|$$ applied to the non-negative matrix A.

Another inequality is:

 * $$\min_i \sum_j a_{ij} \; \le \; r .$$

This fact is specific to non-negative matrices; for general matrices there is nothing similar. Given that A is positive (not just non-negative), then there exists a positive eigenvector w such that Aw = rw and the smallest component of w (say wi) is 1. Then r = (Aw)i ≥ the sum of the numbers in row i of A. Thus the minimum row sum gives a lower bound for r and this observation can be extended to all non-negative matrices by continuity.

Another way to argue it is via the Collatz-Wielandt formula. One takes the vector x = (1, 1, ..., 1) and immediately obtains the inequality.

Perron projection
The proof now proceeds using spectral decomposition. The trick here is to split the Perron root from the other eigenvalues. The spectral projection associated with the Perron root is called the Perron projection and it enjoys the following property:

The Perron projection of an irreducible non-negative square matrix is a positive matrix.

Perron's findings and also (1)–(5) of the theorem are corollaries of this result. The key point is that a positive projection always has rank one. This means that if A is an irreducible non-negative square matrix then the algebraic and geometric multiplicities of its Perron root are both one. Also if P is its Perron projection then AP = PA = ρ(A)P so every column of P is a positive right eigenvector of A and every row is a positive left eigenvector. Moreover, if Ax = λx then PAx = λPx = ρ(A)Px which means Px = 0 if λ ≠ ρ(A). Thus the only positive eigenvectors are those associated with ρ(A). If A is a primitive matrix with ρ(A) = 1 then it can be decomposed as P ⊕ (1 − P)A so that An = P + (1 − P)An. As n increases the second of these terms decays to zero leaving P as the limit of An as n → ∞.

The power method is a convenient way to compute the Perron projection of a primitive matrix. If v and w are the positive row and column vectors that it generates then the Perron projection is just wv/vw. The spectral projections aren't neatly blocked as in the Jordan form. Here they are overlaid and each generally has complex entries extending to all four corners of the square matrix. Nevertheless, they retain their mutual orthogonality which is what facilitates the decomposition.

Peripheral projection
The analysis when A is irreducible and non-negative is broadly similar. The Perron projection is still positive but there may now be other eigenvalues of modulus ρ(A) that negate use of the power method and prevent the powers of (1 − P)A decaying as in the primitive case whenever ρ(A) = 1. So we consider the peripheral projection, which is the spectral projection of A corresponding to all the eigenvalues that have modulus ρ(A). It may then be shown that the peripheral projection of an irreducible non-negative square matrix is a non-negative matrix with a positive diagonal.

Cyclicity
Suppose in addition that ρ(A) = 1 and A has h eigenvalues on the unit circle. If P is the peripheral projection then the matrix R = AP = PA is non-negative and irreducible, Rh = P, and the cyclic group P, R, R2, ...., Rh−1 represents the harmonics of A. The spectral projection of A at the eigenvalue λ on the unit circle is given by the formula $$\scriptstyle h^{-1}\sum^h_1\lambda^{-k}R^k$$. All of these projections (including the Perron projection) have the same positive diagonal, moreover choosing any one of them and then taking the modulus of every entry invariably yields the Perron projection. Some donkey work is still needed in order to establish the cyclic properties (6)–(8) but it's essentially just a matter of turning the handle. The spectral decomposition of A is given by A = R ⊕ (1 − P)A so the difference between An and Rn is An − Rn = (1 − P)An representing the transients of An which eventually decay to zero. P may be computed as the limit of Anh as n → ∞.

Counterexamples
The matrices L = $$\left( \begin{smallmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{smallmatrix} \right)$$, P = $$\left( \begin{smallmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ \!\!\!-1 & 1 & 1 \end{smallmatrix} \right)$$, T = $$\left( \begin{smallmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{smallmatrix} \right)$$, M = $$\left( \begin{smallmatrix} 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 \end{smallmatrix} \right)$$ provide simple examples of what can go wrong if the necessary conditions are not met. It is easily seen that the Perron and peripheral projections of L are both equal to P, thus when the original matrix is reducible the projections may lose non-negativity and there is no chance of expressing them as limits of its powers. The matrix T is an example of a primitive matrix with zero diagonal. If the diagonal of an irreducible non-negative square matrix is non-zero then the matrix must be primitive but this example demonstrates that the converse is false. M is an example of a matrix with several missing spectral teeth. If ω = eiπ/3 then ω6 = 1 and the eigenvalues of M are {1,ω2,ω3=-1,ω4} with a dimension 2 eigenspace for +1 so ω and ω5 are both absent. More precisely, since M is block-diagonal cyclic, then the eigenvalues are {1,-1} for the first block, and {1,ω2,ω4} for the lower one

Terminology
A problem that causes confusion is a lack of standardisation in the definitions. For example, some authors use the terms strictly positive and positive to mean > 0 and ≥ 0 respectively. In this article positive means > 0 and non-negative means ≥ 0. Another vexed area concerns decomposability and reducibility: irreducible is an overloaded term. For avoidance of doubt a non-zero non-negative square matrix A such that 1 + A is primitive is sometimes said to be connected. Then irreducible non-negative square matrices and connected matrices are synonymous.

The nonnegative eigenvector is often normalized so that the sum of its components is equal to unity; in this case, the eigenvector is the vector of a probability distribution and is sometimes called a stochastic eigenvector.

Perron–Frobenius eigenvalue and dominant eigenvalue are alternative names for the Perron root. Spectral projections are also known as spectral projectors and spectral idempotents. The period is sometimes referred to as the index of imprimitivity or the order of cyclicity.

title: Perron–Frobenius theorem


text:
Talk:Cumulative distribution function

Logistic?
Thatsme314 asserted in an edit https://en.wikipedia.org/w/index.php?title=Cumulative_distribution_function&oldid=904343161...

"In the case of a continuous distribution, it gives the area under the probability density function from minus infinity to $$x$$, and is a logistic distribution. Cumulative distribution functions are also used to specify the distribution of multivariate random variables."

Am I misunderstanding what is meant here? The obvious reading is clearly wrong, but perhaps there is a less obvious reading that I am missing. Best clase, it still needed clarification so I deleted it.

--Livingthingdan (talk) 07:36, 7 August 2019 (UTC)

Serious Error
There is written that if the cumulative distribution function is continuos then X is absolutely continuos. This is just false, you need F continuos and with continuous derivative!

cadlag
while the distribution is required to be cadlag? a discussion section on this will be valuable. Jackzhp (talk) 18:46, 15 August 2009 (UTC)

 * Moreover there is a tradition here (I suppose because of Kolmogorov's original notation, but I'm not sure) that the CDF should be left continous... Drkazmer [[Image:Crystal 128 penguin.png|17px]] Just tell me... 23:01, 2 January 2012 (UTC)

EDIT: My sincere apologies, but I don't know where else to report this. Unlike other Wikipedia pages, when this page is googled, its title shows up with the first letter of the first word uncapitalized. Try it. Cheers. -- Anonymous user

Complementary Comulative Distribution function
I assume there is an error after "Proof: Assuming X has density function f, we have for any c > 0", regarding integration limits for E(X) ? —Preceding unsigned comment added by Amir bike (talk • contribs) 05:54, 19 May 2011 (UTC)

It is said that Markov's inequality states that: $$\bar F(x) \leq \frac{\mathbb E(X)}{x} $$ However it is only correct in continuous case, as in discrete case $$P(X \geq x) = \bar F(X) + P(X=x) $$ Although the Inequality still holds, the current version is weaker than the proper Markov's inequality — Preceding unsigned comment added by Colinfang (talk • contribs) 18:59, 4 March 2012 (UTC)

 * The current version is the standard statement of Markov's inequality found in reference books. If there is a stronger result, it could be stated with a citation. If the stronger resuly is still generally known as Markov's inequality, then the Markov's inequality article could be updated as well. But the version in the article (now) states valid conditions under which the results hold. Melcombe (talk) 16:58, 15 April 2012 (UTC)

Utility
It would be helpful if in the entry there was a discussion of the utility performing a CDF plot. This would include when to perform one, and what information is learned from performing the CDF plot. What real world applications would this include? Maybe an example would be helpful — Preceding unsigned comment added by 209.252.149.162 (talk) 14:10, 1 August 2011 (UTC)

Table of cdfs
I have moved the recently added table of cdfs to here for discussion/revision. The version added was ...

There are several problems here, particularly with inconsistent notations. But there are structural problems in defining the cdfs of the discrete distributions, as the formulae given are only valid at the integer points (within the range of the distribution) and would give incorrect values of the cdf at non-integer values. Also several of the functions involved require definitions/wikilinks. So, if the table is to be included, thought needs to be given to possibly dividing it into discrete/continuous tables and/or adding extra columns. Melcombe (talk) 09:25, 21 October 2011 (UTC)
 * I certainly hope we won't see tables that say "cumulative density function" instead of "cumulative distribution function". Michael Hardy (talk) 18:40, 23 February 2019 (UTC)

citation needed.... really?
I think it is a little bit ridiculous to expect a citation that a CDF is càdlàg. It is an almost trivial observation that follows directly from the probability space axioms and the definition of a càdlàg function. Surely this is a routine calculation. --217.84.60.220 (talk) 11:32, 3 November 2012 (UTC)

Please, for didactic, show area relation
EXAMPLE

http://beyondbitsandatomsblog.stanford.edu/spring2010/files/2010/04/CdfAndPdf.gif — Preceding unsigned comment added by 187.66.187.183 (talk) 07:25, 3 February 2013 (UTC)

CDF is definitely LEFT-continuous.
CDF must be left-continuous, not right as stated on the wiki page. Source: current ongoing University studies, 3 separate professors, books from 4 different authors. — Preceding unsigned comment added by 213.181.200.159 (talk) 07:46, 6 March 2013 (UTC)

 * This article follows the convention reached via the link right-continuous. This is "continuous from the right". Perhaps you are thinking of "continuous to the left". 81.98.35.149 (talk) 11:34, 6 March 2013 (UTC)

I'd like to mention that some textbooks define the CDF as the probability that the random variable X at a given point, F_X(a), that X is strictly less the constant, rather than less than or equal to a. This is where the confusion of this talk section is coming from. --128.42.66.81 (talk) 20:23, 25 August 2016 (UTC)

Not that redundant?
The passage that I deleted but which was restored said

 * Point probability
 * The "point probability" that X is exactly b can be found as

 * $$\operatorname{P}(X=b) = F(b) - \lim_{x \to b^{-}} F(x).$$

 * This equals zero if F is continuous at x.

However, at the end of the section "Definition" it says

 * In the case of a random variable X which has distribution having a discrete component at a value x0,
 * $$ \operatorname{P}(X=x_0) =F(x_0)-F(x_0-) ,$$
 * where F(x0-) denotes the limit from the left of F at x0: i.e. lim F(y) as y increases towards x0.

What I deleted looks identical to that, except that it doesn't include the sentence This equals zero if F is continuous at x.

I propose that we re-delete it but put the last-mentioned sentence into the existing section.

 * Okay, no problem. Nijdam (talk) 07:08, 19 April 2013 (UTC)

cdf notation
I went through and changed the notation $$F_X(x)$$ to $$F(x)$$ everywhere in the definition section to try to obtain notational consistency through the article, but the change was reverted by Nijdam with edit summary "Difference between cdf of X and just a cdf". But that conflicts with much notation in the article that uses F(x) for the cdf of X. In the Properties section:

 * the CDF of X will be discontinuous at the points xi and constant in between:

 * $$F(x) = \operatorname{P}(X\leq x) = \sum_{x_i \leq x} \operatorname{P}(X = x_i) = \sum_{x_i \leq x} p(x_i).$$

 * If the CDF F of X is continuous, then X is a continuous random variable; if furthermore F is absolutely continuous, then there exists a Lebesgue-integrable function f(x) such that

 * $$F(b)-F(a) = \operatorname{P}(a< X\leq b) = \int_a^b f(x)\,dx$$

 * for all real numbers a and b. The function f is equal to the derivative of F almost everywhere, and it is called the probability density function of the distribution of X.

In the Examples section:

 * As an example, suppose X is uniformly distributed on the unit interval [0, 1]. Then the CDF of X is given by

 * $$F(x) = \begin{cases}

0 &:\ x < 0\\ x &:\ 0 \le x < 1\\ 1 &:\ 1 \le x. \end{cases}$$

In the Derived functions section:

 * $$\bar F(x) = \operatorname{P}(X > x) = 1 - F(x).$$

In the multivariate case section:

 * for a pair of random variables X,Y, the joint CDF $$F$$ is given by

 * $$F(x,y) = \operatorname{P}(X\leq x,Y\leq y),$$

So we need to establish consistency of notation -- either use FX every time we mention a cdf "of X", or else never. Your thoughts? Duoduoduo (talk) 15:08, 17 May 2013 (UTC)

 * In the literature both $$F_X$$ and $$F$$ are used, the latter for ease of notation, and only if there is no confusion about the random variable. In the examples you mention there is not always an inconsistency. In the first one you're right, but if it reads: If the CDF F of X ..., it merely states $$F_X=F$$, where $$F$$ is some specified function. Nijdam (talk) 06:11, 18 May 2013 (UTC)

 * But there's no confusion anywhere in the article regardless of which is used. So why not use the same one everywhere? Duoduoduo (talk) 12:30, 18 May 2013 (UTC)

Continuous Random Variables
Elsewhere on Wikipedia, and in many published books, a continuous random variable has an absolutely continuous c.d.f., not merely continuous as stated in the properties section. I suggest that this page should also state that the c.d.f. is absolutely continuous so that there is a p.d.f. Paulruud (talk) 17:26, 30 March 2015 (UTC)

Definition as expectation value
I found this in the introduction of Characteristic function:

''The characteristic function provides an alternative way for describing a random variable. Similarly to the cumulative distribution function''
 * $$F_X(x) = \operatorname{E} \left [\mathbf{1}_{\{X\leq x\}} \right],$$

( where 1{X ≤ x} is the indicator function — it is equal to 1 when X ≤ x, and zero otherwise), which completely determines behavior and properties of the probability distribution of the random variable X, the characteristic function
 * $$   \varphi_X(t) = \operatorname{E} \left [ e^{itX} \right ]$$

also completely determines behavior and properties of the probability distribution of the random variable X''. The two approaches are equivalent in the sense that knowing one of the functions it is always possible to find the other, yet they both provide different insight for understanding the features of the random variable.''

The notation $$F_X(x) = \operatorname{E} \left [\mathbf{1}_{\{X\leq x\}} \right]$$ is so confusing I ask the community a clarification in this page (Characteristic function).

Kind of reciprocity
There is a theorem stating that for each F nondecreasing right-continuous with limits 0 in -∞ and 1 in +∞, there exists a unique probability mesure μ such that F is a the cdf of a random variable X distributed according to μ. I think this important theorem should be displayed in this page.

GizTwelve (talk) 12:28, 5 October 2017 (UTC)

Colors in the first graph are too similar
Please use clearly distinguishable colors, that work also with color vision deficiency. — Preceding unsigned comment added by 88.219.233.164 (talk) 12:31, 11 December 2018 (UTC)

Connection to measure theory
I would like to know the full connection to measure theory. It seems if I have a probabilistic measure space $$(A,\mathcal{A},P_A)$$ and a measurable function $$X\colon A\to \mathbb{R}$$ it might be the pushforward measure $$X_* P_A= P_A \circ X^{-1}$$ which I guess would be a mapping $$ (\mathbb{R}, \mathcal{B}(\mathbb{R}), ||\cdot||_2)\mapsto \mathbb{R}$$ which for some reason is then restricted to half open sets $$F_X \overset{?}{:=} (X_* P_A)\restriction [-\infty, r)\times \mathbb{R}$$. But this is original research and feels a bit patchy (unclear to me how to generalize to random variables with values in $$\mathbb{R}^n$$) so please if someone who has connected the dots could add the connection. — Preceding unsigned comment added by Rostspik (talk • contribs) 05:58, 27 April 2019 (UTC)

h — Preceding unsigned comment added by 2.183.8.104 (talk) 05:39, 16 November 2019 (UTC)

Check Required
Hello! I am new editor of Wikipedia and I have tried to do some edits and give some citations I hope I have done things right if not pls let me know my mistakes. This would help me a lot Thank you. Stobene45 (talk) 09:49, 2 July 2021 (UTC)

 * Hello! ALong with your addition to my talk page I have taken a look at some of your edits. I would like to direct you to the teahouse to actually ask a question about what to add to the article and what not to add to articles (most of which can be found in the Manual of Style). I have reverted your edits however please note that I do see your edits as being in good faith considering you contacted me first saying that you are a new user wanting to learn. Otherwise, I probably would've given you warnings. Blaze The Wolf | Proud Furry and Wikipedia Editor (talk) 18:17, 2 July 2021 (UTC)
 * this is mostly a matter of Wikipedia's accepted style conventions. We don't artificially underline subheadings, but rather use the pre-defined styles generated by the staggered classes of headings; and while italics can be used for emphasis, this should be done very sparingly (see Manual_of_Style/Text_formatting). As for the citations, references to Google results are not suitable anywhere on Wikipedia - the cited source must be the specific work used, not just a search result containing the work's title. Cheers! -- Elmidae (talk · contribs) 16:26, 3 July 2021 (UTC)

title: Talk:Cumulative distribution function