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Cry Me a River (Justin Timberlake song)

"Cry Me a River" is a song by American singer-songwriter Justin Timberlake for his debut solo album, Justified (2002). It was thought to be inspired by Timberlake's former relationship with singer Britney Spears. Jive Records released the song to contemporary hit and rhythmic radio in the United States on November 25, 2002, as the album's second single. Accompanied by an electric piano, beatbox, guitars, synthesizers, Arabian-inspired riffs and Gregorian chants, "Cry Me a River" is a pop and R&B song about a brokenhearted man who moves on from his past.

"Cry Me a River" received acclaim from music critics, who praised Timbaland's production while calling it a standout track on Justified. The song earned several lists of best of the year and the decade (2000s), while Rolling Stone included it on their 500 Greatest Songs of All Time at 484. It won the Grammy Award for Best Male Pop Vocal Performance at the 2004 ceremony. The song peaked at number three on the US Billboard Hot 100 and Pop Songs charts and charted in the top ten in other countries. It was certified double platinum by both the Australian Recording Industry Association (ARIA) and the British Phonographic Industry (BPI).

Filmmaker Francis Lawrence directed the music video for "Cry Me a River" at 24860 Pacific Coast Hwy in Malibu, California. In the controversial video, Timberlake's character invades the home of his ex-lover and films himself having sexual activity with another woman. Spears alleged that the video was a publicity stunt, but Timberlake maintained that she did not inspire the production. In 2011, however, Timberlake admitted his breakup with Spears inspired the video. The clip won the awards for Best Male Video and Best Pop Video at the 2003 MTV Video Music Awards. Timberlake performed "Cry Me a River" on his four major concert tours: The Justified World Tour (2003–2004), Justified and Stripped Tour (2003), FutureSex/LoveShow (2007), and The 20/20 Experience World Tour (2013–2015). The song has been covered by various artists, without chart success.

Writing and production

Timberlake and Scott Storch wrote "Cry Me a River" with Timbaland, who produced the song. Storch found working with Timberlake easy because of the song's meaning. Reporters believed its lyrics were inspired by Timberlake's failing relationship with singer Britney Spears after the two split in late 2002. At the time, Spears was rumored to have had an affair with her and Timberlake's mutual friend, choreographer and songwriter Wade Robson. More than 20 years later, Spears confirmed she “made out with [Robson]” on one occasion after learning that Timberlake had cheated on her. Timberlake told MTV News, "I'm not going to specifically say if any song is about anybody. I will say writing a couple of songs on the record helped me deal with a couple of things. To me songs are songs. They can stem from things that completely happened to you personally or they can stem from ideas that you think could happen to you." In December 2011, Timberlake admitted that he had written "Cry Me a River" after an argument over the phone: "I was on a phone call that was not the most enjoyable phone call. I walked into the studio and he [Timbaland] could tell I was visibly angry." Timbaland recalled, "I was like, 'Man, don't worry about it' and he was like, 'I can't believe she did that to me' and he was like, 'You were my sun, you were my earth.'"

"Cry Me a River" was recorded at Westlake Recording Studios in Los Angeles by Senator Jimmy D, while Carlos "Storm" Martinez served as the assistant engineer. Jimmy Douglass and Timbaland mixed the song at Manhattan Center Studios in New York City. Timberlake arranged the vocals and was a backup singer alongside Timbaland, Marsha Ambrosius, Tye Tribbett and Greater Anointing. The ad lib "Cry me, cry meeee" at the end of the song was among Ambrosius's contributions. Storch compared Timberlake's vocals on "Cry Me a River" to R&B and rock singer Daryl Hall. Larry Gold provided the string arrangement and conducting, while Storch and Bill Pettaway played the clavinet and guitar, respectively.

Release and response
"Cry Me a River" was released as the second single from Justified. Jive Records serviced the song to contemporary and rhythmic radio in the United States on November 25, 2002. On December 23, three remixes were released as a 12-inch single in Canada and France. On January 5, 2003, the song was sent to urban contemporary radio stations in the US. It was released as a CD single in Germany on January 23. The single contains the album version of the song and the remixes on its 12-inch single release. A CD single, which included Johnny Fiasco's remix of the song and two additional remixes of "Like I Love You", was released on February 3 and 6 in the United Kingdom and Canada, respectively. "Cry Me a River" was released as a CD single in the US on February 18. The single features the album version of the song, its instrumental and four remixes.

Following its release, there was media speculation that Spears had written a song as a response to "Cry Me a River"; she denied the rumors, explaining, "You know, it's funny. I read that I wrote this song and I wrote these lyrics and that's not my style. I would never do that." Annet Artani, who co-wrote Spears' 2003 song "Everytime", stated that the song was written as a response to "Cry Me a River". When asked during an interview with Diane Sawyer on PrimeTime in 2003, if "Everytime" was about Timberlake, Spears responded, "I'll let the song speak for itself."

Composition and lyrical interpretation
"Cry Me a River" is a pop and R&B song with instrumentation that features clavinet, guitars, beatboxing, synthesizers, Arabian-inspired riffs and Gregorian chants. The instruments are arranged into what critics described as a graceful and mysterious melody. Jane Stevenson of Jam! said the single combines gospel and opera. Tyler Martin of Stylus Magazine enjoyed the way that the song unconventionally mixed a range of experimental sounds. According to Martin, the wave synth affects the real strings to create an unusual dissonance. The song's chorus devolves into a choral reading in which Timberlake pleads over the group. "Cry Me a River" finishes with a Timbaland vocal sample heard during the chorus.

"Cry Me a River" is written in the key of G♯ minor, in alla breve, with a tempo of 74 beats per minute. Timberlake's vocal range spans from the low note of C♯3 to the high note of B4. Billboard magazine critics called "Cry Me a River" a bittersweet song, in which the singer's "familiar tenor belting" is tempered with a soulful falsetto and a "convincingly aggressive rock-spiked baritone" rasp. David Browne of Entertainment Weekly labeled the song "a haunted, pained farewell".

Lyrically, the song is about a heartbroken man who moves on from his past. A Rolling Stone reviewer called the song a "breakup aria". According to Caroline Sullivan of The Guardian, "Cry Me a River" stands out for its "slow-building sense of drama", which highlights Timberlake at his "husky best". The song begins with the phrase "You were my sun, you were my earth", which according to Timbaland was Timberlake's inspiration to write the song. Tanya L. Edwards of MTV News observed that Timberlake was lied to and wronged, and said this is demonstrated by the lyrics: "You don't have to say whatcha did / I already know, I found out from him / Now there's just no chance." The chorus contains the lines: "Told me you loved me, why did you leave me all alone / Now you tell me you need me when you call me on the phone." Slant Magazine's Sal Cinquemani called Timberlake's 2007 single "What Goes Around... Comes Around" an ostensible sequel to "Cry Me a River" both lyrically and musically. Jason Lipshutz of Billboard stated that aside from the presence of Timbaland's "fantastically cluttered production", the difference between "Cry Me a River" and Timberlake's 2013 single "Mirrors" is clear: "10 years ago, Timberlake was broken, and now he is whole".

Reception and accolades
"Cry Me a River" received widespread critical acclaim after release. Jane Stevenson of Jam! and Sal Cinquemani of Slant Magazine called it a standout track on Justified. Stephen Thomas Erlewine of AllMusic selected the song as a highlight in his review of Justified. Peter Robinson of NME described it as a "twisted epic" and "an acidic, filthy little song teeming with spite and retribution, with Timbaland's pounding." Rolling Stone's Ben Ratliff viewed his production of "Cry Me a River" as exceptional. Denise Boyd of BBC Music felt that the song's lyrics play as large a role as Timbaland's production, unlike other songs on Justified. David Browne of Entertainment Weekly concluded that the song is "a genuine stunner" that should leave Timberlake's fellow 'N Syncers concerned that he truly may not need them any more.

John Mitchell of MTV News called "Cry Me a River" and "What Goes Around... Comes Around" "kiss-off songs", with Timberlake's revenge on Spears present throughout the lyrics. Idolator staff opined it is the sound of an artist who is "relishing ripping up people's perceptions of him as a sweet-natured teen heartthrob." VH1's Emily Exton wrote, "By stretching the boundaries of pop–and fusing it with notes of R&B, lots of electric goodness, and haunting choral vocals–Timberlake's kiss-off song instantly becomes more poignant, a pointed reaction to the hurt he's experienced by someone he once loved." According to Complex, the song "made people completely forget about 'N Sync and start asking what Timberlake would do next." The single won a Grammy Award for Best Male Pop Vocal Performance at the 2004 award ceremony. It was nominated for Song of the Year at the 2003 MTV Europe Music Awards. For Pitchfork Media, it was ranked the third-best song of 2003.

Retrospective lists
Rolling Stone ranked "Cry Me a River" at number 20 on their list of the 100 Best Songs of the 2000s; a columnist for the magazine wrote that the real inspiration behind the song was the formation of the Timberlake–Timbaland team, "a match made in pop heaven". In 2012, the magazine placed it at number 484 on their list of "The 500 Greatest Songs of All Time". VH1 ranked "Cry Me a River" at number 59 on their list of the 100 Greatest Songs of the 2000s. The Daily Telegraph listed the song at number 40 on the "100 songs that defined the Noughties". In 2015, Spotify re-ranked Rolling Stone's 500 Greatest Songs in two different lists. "Cry Me a River" ranked at number 14 on "by all Spotify users" and 10 on "streams from millennial users". In 2018, Rolling Stone ranked the song 46 on their list "The 100 Greatest Songs of the Century – So Far".

Commercial performance
"Cry Me a River" debuted at number 44 on the US Billboard Hot 100 in the issue dated December 21, 2002, earning the Hot Shot Debut honor with 29.6 million audience impressions. On February 1, 2003, the single reached a peak of number three, becoming Timberlake's first solo single to reach that position. The song debuted on the US Pop Songs chart at number 37 in the issue dated December 14, 2002 and reached a peak of three on February 1, 2003. For the issue dated December 28, 2002, "Cry Me a River" debuted at number 75 on the US Hot R&B/Hip-Hop Songs chart. It reached a peak of 11 on March 8, 2003. It also peaked at number two on the US Hot Dance Club Songs, becoming Timberlake's second top-three song, after his debut single "Like I Love You" reached number one. As of August 2003, remixes of the single have sold more than 61,000 units in the United States. As of 2018, the song has sold 1.2 million copies in the country.

The song debuted at number two on the Australian Singles Chart on March 9, 2003. It fell to number six in the following week. "Cry Me a River" became Timberlake's first top-three solo single on the chart. The song remained on the chart for 12 consecutive weeks. It was certified gold by the Australian Recording Industry Association (ARIA) for shipments of 35,000 units. The single debuted at number 44 in New Zealand on March 9, 2003. After two weeks on the chart, it reached a peak of 11. It remained on the chart for 11 weeks.

In the United Kingdom, "Cry Me a River", debuted at number two on February 15, 2003. The next week it fell to number three, before returning to its original peak on March 1, 2003; it stayed on the chart for 13 weeks. The song has sold over 365,000 copies in the country as of 2015. It debuted at number 14 on the French Singles Chart on April 5, 2003. After three weeks, "Cry Me a River" peaked at number six, and it stayed on the chart for 21 weeks. It was less successful on the Italian Singles Chart, where it peaked at number 14 and stayed on the chart for four weeks. "Cry Me a River" also peaked at number five in Belgium (Wallonia), at number six in the Republic of Ireland and the Netherlands, at number seven in Belgium (Flanders), and at number 10 in Germany, Norway, and Sweden.

Development and release
The music video for "Cry Me a River" was directed by Francis Lawrence, and filmed in Malibu, California on October 22, 2002. Lawrence created the video's concept and told MTV News, "[Justin and I] had a conversation on the phone and all he said was he wanted to have some dancing in it, but to do my thing. He told me what the song was about, but in a [general way] as well. He just said it was a kiss-off song and so I came up with this idea and he went for it." Though the name "Britney" was not allowed in the video or on set, Lawrence alluded to Spears the best he could. He explained: "There was just this unspoken agreement between [Justin and I]. Because [Justin and Britney] were on the same label; that's part of the reason I thought the label would never, ever go for it. It was all about implying certain things, there's little elements and details that play throughout [the video] and tie it in." Timberlake enjoyed these details, or clues, which included a reference to Spears's tattoo and her newsboy hat.

Lawrence revealed that executives of Jive Records were also nervous about other aspects of the video, such as Timberlake's representation of a voyeur and the tone of some scenes he was portraying with a girl. "That's the thing I liked most about this project, was that he was coming into it with a super clean-cut image with 'N Sync and he's such a nice guy and so handsome and what I was into doing was making him be a bit scary. Lurking around the house in the rain, throwing a rock through her window, being a peeping Tom, getting revenge and doing stuff that's not really what a nice guy does." The shoot lasted three days, and involved a rain machine that gave the video a "sort of noir" feeling. To create the illusion of Timberlake dancing and floating in the house he breaks into, dancers in green-screen suits lifted, dropped, and carried Timberlake around via handles underneath Timberlake's jacket.

The music video for "Cry Me a River" premiered on MTV's TRL on November 25, 2002, the same day the single was sent to radio stations. Calling in to TRL to premiere the video, Timberlake emphasized to Carson Daly, "The video is not about her. The video is about me." The video was released onto the iTunes Store on April 28, 2003.

Synopsis
The video begins with a blonde woman, played by model and actress Lauren Hastings (portraying Spears ), walking out of her house hand-in-hand with an unidentified man. The couple leaves in the woman's car (a silver Porsche 911 Carrera Cabriolet (996)) as Timberlake rolls down the window of a black Mercedes-Benz S-Class (W220), from which he and his friend have been watching the woman. Timberlake breaks into the house by throwing a rock through a window and proceeds through the woman's house, with anti-gravity jumps and slides, not causing any other visible damage, except for kicking a picture frame of the woman across the living room. Then he searches some drawers and finds a video camera, while the driver of the car, Timbaland, signals for his female accomplice in the back of the car, played by model Kiana Bessa, to go in. She enters the house and goes with Timberlake to a bedroom, where she starts to undress and kiss Timberlake, who has a smug expression while being filmed. They stay in the bedroom for a moment; then the accomplice exits the house but a still-cocky Timberlake stays. As the blonde woman returns, he follows her around the house and hides in a closet as she showers. He gets closer to her and touches the glass surrounding the shower. The blonde woman senses someone in the room and turns around, but Timberlake is gone. She leaves the bathroom and goes into her bedroom, where the video he made with his new lover plays on the television.

Reception
After the video's TRL premiere, Carson Daly noted that Justin's character in the video has "an American Psycho feel" to it, to which Justin replied, "Hey man, I'm crazy." Peter Robinson of NME wrote that the video shows "what Justin looks like after he's had sex. Clue: he looks pretty good." According to Virginia Heffernan of The New York Times, Timberlake channels the character Neo from The Matrix film series, "pacing anxiously around wet and metallic interiors". At the 2003 MTV Video Music Awards, the video for "Cry Me a River" won in the categories of Best Male Video and Best Pop Video. It had also been nominated for Video of the Year, Best Direction in a Video and Viewer's Choice. In a 2006 issue proclaiming Timberlake the International Man of the Year, British GQ declared, "One minute Timberlake was a forgotten relic of teeny pop, the next he's making the video for 'Cry Me A River' in which he's stalking his ex and creeping up behind her to sniff her hair. And within the space of a year no one could give two hoots about his New Mickey Mouse Club beginnings. Timberlake suddenly became Trousersnake and he'd done the impossible—he'd become cool."

Following the release of the music video, Us Weekly ran a cover story titled Britney Vs. Justin: The War Is On. Timberlake continued to deny that Spears inspired the video. However, Spears told Rolling Stone in October 2003 that she received a call from Timberlake saying there would be a look-alike of her in his music video. She states that he reassured her by saying, "Don't worry about it. It's not a big deal". Spears, who had not seen the video, says she allowed him to do so but became infuriated after watching it. She recalled that when she asked why he had made a video about her, he replied, "Well, I got a controversial video." She stated that it was a great publicity stunt, commenting, "So he got what he wanted. I think it looks like such a desperate attempt, personally." After the release of Spears' video for "Toxic" (2003), Jennifer Vineyard of MTV News said her video "[made] "Cry Me a River" look like child's play".

In 2013, Billboard editor Jason Lipshutz opined the clip was "one of the more brilliant musical moments in pop music since the dawn of the century. The visual is JT's most controversial music video to date." He added, "The "Cry Me a River" video served a dual purpose for Timberlake: to court controversy, and to make its star seem more grown-up." In 2018, Billboard critics named it the 24th "greatest music video of the 21st century."

Live performances
Timberlake performed "Cry Me a River" for the first time on an episode of TRL, the same day as the release of Justified. He again performed the song at the 13th annual Billboard Music Awards, held on December 9, 2002, at the MGM Grand Garden Arena in Las Vegas. He was accompanied by a string section and a 20-member choir. He also performed the song on the Justified World Tour (2003–04), his first worldwide tour. The song was eighth on the set list of Justified/Stripped Tour (2003), his joint North American tour with Christina Aguilera. Timberlake performed "Cry Me a River" on a promotional concert held at House of Blues in West Hollywood, California on June 17, 2003. He performed a rock-inspired version of the song on Saturday Night Live on October 11, 2003; and it was fourteenth on the set list of his second worldwide tour, FutureSex/LoveShow (2007). During his world tour in Zürich, Switzerland, in August 2007, Timberlake performed "Cry Me a River" and segued into the chorus of "Rehab" by Amy Winehouse.

On October 23, 2010, while performing at the annual charity gig "Justin Timberlake and Friends" in Las Vegas, Timberlake began the show with a slow and "sultry" performance of "Cry Me a River" and segued into a cover of Bill Withers' 1971 single "Ain't No Sunshine". He later resumed "Cry Me a River" before segueing into a cover of Drake's 2010 song "Over". Jillian Mapes of Billboard described the performance as "completely lovely in an effortless sort of way." Timberlake performed "Cry Me a River" at concert he held during the 2013 Super Bowl weekend in New Orleans. Timberlake performed "Cry Me a River" in a medley with other of his songs at the 2013 MTV Video Music Awards. After the performance he accepted a Michael Jackson Video Vanguard Award at the ceremony. Timberlake included the single on the set list of his Legends of the Summer Stadium Tour with rapper Jay-Z (2013), his fifth worldwide concert tour, The 20/20 Experience World Tour (2013–15), his sixth worldwide concert tour, The Man of the Woods Tour (2018/19) and his seventh worldwide concert tour, The Forget Tomorrow World Tour (2024).

Covers
The song was first covered by Welsh alternative metal band Lostprophets, recorded in a BBC Radio One session. It appears as a B-side of the single "Last Train Home", released in 2004. American singer-songwriter Taylor Swift performed a cover of "Cry Me a River" in Memphis, Tennessee, during the Speak Now World Tour (2011–12). Justin Bieber recorded a cover of the song together with Kanye West's 2007 single "Stronger", which he posted on his YouTube account. Canadian rock band The Cliks also covered the song for their 2007 album Snakehouse. British singer Leona Lewis covered "Cry Me a River" during her debut concert tour The Labyrinth (2010). The Sheffield Star described the version as a "beautifully, sitting quietly, almost a capella". In 2011 New Zealand band Wellington International Ukulele Orchestra released a cover version on their "I Love You..." EP.

American indie pop duo Jack and White covered the song on their 2012 extended play Undercover. Recording artist Kelly Clarkson covered the song on September 1, 2012, as a fan request during her 2012 Summer Tour with The Fray. In January 2013, American singer Selena Gomez performed an acoustic version of "Cry Me a River" at the UNICEF charity concert in New York City. In February 2015, Scottish synthpop band Chvrches performed a cover of "Cry Me a River" on BBC Radio 1's Live Lounge. The New Jersey alternative metal band 40 Below Summer performed a cover of the song in their 2015 album Transmission Infrared. American Post-Hardcore vocalist Johnny Craig is also well known for his 2009 cover of the song, prior to his first solo album release.

In other media
In May 2020, Democratic presidential nominee Joe Biden released a campaign ad which took aim at Donald Trump for complaining in a 2020 interview about being "treated worse" by the media than any other president instead of focusing on the ongoing COVID-19 pandemic. The ad was soundtracked by "Cry Me a River".

Influence
"Cry Me a River" marked the first collaboration between Timberlake and Timbaland, and its critical and commercial success started "one of the most forward-thinking star-producer duos in pop music", as described by Variety's author Andrew Barker. Alternative rock band Coldplay revealed that the single was an inspiration for the drumbeat of their song "Lost!". Rock singer Marilyn Manson cited it as one of the main songs that influenced him. In order to use the song in the 2017 film Lady Bird set in 2003, director Greta Gerwig sent a letter to Timberlake, "Your album Justified was that year, and it owned that year."

Rosalía samples the chorus from "Cry Me a River" for her song "Bagdad" on her album El mal querer (2018), featuring a Barcelona children's choir. Billboard deemed the track "a transcultural re-interpretation" of "Cry Me a River". Halsey's song "Without Me" contains an interpolation of the pre-chorus of "Cry Me a River".

"Cry Me a River" was also the title of a 1953 song written by Arthur Hamilton and made famous by Julie London in 1955. According to The Wall Street Journal, Hamilton was the originator of the phrase "cry me a river" borrowed in Timberlake's song.

Track listings

 * 12-inch single
 * 1) "Cry Me a River" (Dirty Vegas vocal mix) – 8:11
 * 2) "Cry Me a River" (Bill Hamel Justinough vocal remix) – 7:43
 * 3) "Cry Me a River" (Johny Fiasco mix) – 7:56

 * Germany maxi single
 * 1) "Cry Me a River" – 4:48
 * 2) "Cry Me a River" (Dirty Vegas vocal mix) – 8:11
 * 3) "Cry Me a River" (Bill Hamel Justinough vocal remix) – 7:43
 * 4) "Like I Love You" (Basement Jaxx vocal mix) – 6:04

 * UK maxi single
 * 1) "Cry Me a River" – 4:48
 * 2) "Cry Me a River" (Dirty Vegas vocal mix) – 8:11
 * 3) "Cry Me a River" (Bill Hamel vocal remix) – 7:43

 * Canada maxi single
 * 1) "Cry Me a River" (Johny Fiasco mix) – 7:56
 * 2) "Like I Love You" (Basement Jaxx vocal mix) – 6:04
 * 3) "Like I Love You" (Deep Dish Zigzag remix) – 9:40

 * US maxi single
 * 1) "Cry Me a River" – 4:48
 * 2) "Cry Me a River" (instrumental version) – 4:48
 * 3) "Cry Me a River" (Dirty Vegas vocal mix) – 8:11
 * 4) "Cry Me A River" (Junior's Vasquez Earth club mix) – 6:43
 * 5) "Like I Love You" (Basement Jaxx vocal mix) – 6:04
 * 6) "Like I Love You" (Deep Dish Zigzag remix) – 9:40

Credits and personnel
Credits are adapted from the liner notes of Justified.

Recording and mixing
 * Recorded at Westlake Recording Studios (Los Angeles, California)
 * Strings recorded at The Studio (Philadelphia, Pennsylvania)
 * Mixed at Manhattan Center Studios (New York City)

Personnel

 * Songwriting – Justin Timberlake, Timothy Mosley, Scott Storch
 * Production – Timbaland
 * Recording – Senator Jimmy D
 * Assistant engineer – Carlos "Storm" Martinez
 * Mixing – Jimmy Douglass, Timbaland
 * Mixing assistant – "Steamy"
 * String arrangement and conducting – Larry Gold
 * Background vocals – Justin Timberlake, Timbaland, Marsha Ambrosius, Tyrone Tribbett, Greater Anointing
 * Clavinet – Scott Storch
 * Guitar – Bill Pettaway
 * Vocal arrangement – Justin Timberlake

title: Cry Me a River (Justin Timberlake song)


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Talk:Zoodio

Category:Playground songs
I removed Category:Playground songs from this article and replaced it with Category:Children's songs instead. According to the Florida State Library, the game was "often taught at Bible Schools". That's the opposite of what a "playground song" from children's street culture is supposed to be. Playground songs are the songs children learn from other children, not from adults. They tend to be "inappropriate" in some way, like Miss Susie, which is based on getting away with saying "naughty" words like "ass" and "hell"; Comet (song), which refers to eating a poisonous cleaning product; Beans, Beans, the Musical Fruit, which celebrates flatulence; The Burning of the School, which glorifies a school burning down; and K-I-S-S-I-N-G, which mocks the alleged romances of other children. --Metropolitan90 (talk) 20:35, 17 December 2016 (UTC)

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

title: Talk:Zoodio


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Snaffle bit

A snaffle bit is the most common type of bit used while riding horses. It consists of a bit mouthpiece with a ring on each side and acts with direct pressure. A bridle utilizing only a snaffle bit is often called a "snaffle bridle", particularly in the English riding disciplines. A bridle that carries two bits, a curb bit and a snaffle, or "bradoon", is called a double bridle.

A snaffle is not necessarily a bit with a jointed bit mouthpiece, as is often thought. A bit is a snaffle because it creates direct pressure without leverage on the mouth. It is a bit without a shank. Therefore, a single- or double-jointed mouthpiece, though the most common designs for snaffle bits, does not make a bit a snaffle. Even a mullen mouth (a solid, slightly curved bar) or a bar bit is a snaffle.

Action
The snaffle bit works on several parts of the horse's mouth; the mouthpiece of the bit acts on the tongue and bars, the lips of the horse also feel pressure from both the mouthpiece and the rings. The rings also serve to act on the side of the mouth, and, depending on design, the sides of the jawbone.

A snaffle is sometimes mistakenly thought of as "any mild bit". While direct pressure without leverage is milder than pressure with leverage, nonetheless, certain types of snaffle bits can be extremely harsh when manufactured with wire, twisted metal or other "sharp" elements. A thin or rough-surfaced snaffle, used harshly, can damage a horse's mouth.

Curb chains or straps have no effect on a true snaffle because there is no leverage to act upon. English riders do not add any type of curb strap or curb chain to a snaffle bit. While some riders in western disciplines do add a curb strap to the rings, it is merely a "hobble" for the rings, has no leverage effect and is there only as a safety feature to prevent the rings from being pulled through the mouth of the horse, should the animal gape open its mouth in an attempt to avoid the bit, an outcome prevented in an English bridle by the presence of a cavesson noseband.

Difference from a curb
The snaffle differs from the pelham bit, the curb bit, and the kimberwicke in that it is a non-leverage bit, and so does not amplify the pressure applied by the reins. With a snaffle, one ounce of pressure applied by the reins to a snaffle mouthpiece will apply one ounce of pressure on the mouth. With a curb, one ounce of pressure on the reins will apply more – sometimes far more – than one ounce of pressure on the horse's mouth.

There are many riders (and a remarkable number of tack shops) who do not know the true definition of a snaffle: a bit that is non-leverage. This often results in a rider purchasing a jointed mouthpiece bit with shanks, because it is labeled a "snaffle," and believing that it is soft and kind because of the connotation the snaffle name has with being mild. In truth, the rider actually bought a curb bit with a jointed mouthpiece, which actually is a fairly severe bit due to the combination of a nutcracker effect on the jaw and leverage from the shanks.

A true snaffle does not have a shank like a pelham or curb bit. Although the kimberwicke appears to have a D-shaped bit ring like a snaffle, the bit mouthpiece is not centered on the ring, and thus applying the reins creates leverage; in the Uxeter kimberwicke, there are slots for the reins placed within the bit ring, which allows the reins to create additional leverage. Both are used with a curb chain, thus the ring acts like a bit shank and creates a slight amount of leverage, making it a type of curb bit.

A true snaffle also will not be able to slide up and down the rings of the bit or cheekpieces of the bridle, as this would place it in the gag bit category.

The mouthpiece
The mouthpiece is the more important part of a snaffle, as it controls the severity of the bit. Thinner mouthpieces are more severe, as are those that are rougher.

 * Jointed mouthpiece: applies pressure to the tongue, lips, and bars with a "nutcracker" action. This is the most common mouthpiece found on a snaffle.
 * Mullen mouth: made of hard rubber or a half-moon of metal, it places even pressure on the mouthpiece, lips, and bars. It is a very mild mouthpiece.
 * French mouth: a double-jointed mouthpiece with a bone-shaped link in the middle. It reduces the nutcracker action and encourages the horse to relax. Very mild.
 * Dr. Bristol: a double-jointed mouthpiece with a thin rectangular link in the middle that is set at an angle, creating a pressure point. It is a fairly severe bit.  The French link is similar but much gentler because the link in the middle is flat against the tongue, lips, and bar and has no pressure points.  Neither the Dr. Bristol nor the French Link nutcracker, but their severity is totally opposite.
 * Slow twist: a single-jointed mouthpiece with a slight twist in it. Stronger and more severe.
 * Corkscrew: Many small edges amplifies the pressure on the mouth. Severe.
 * Single- and double-twisted wire: two of the most severe mouthpieces, as they are not only thin, but they also have a "nutcracker" action from the single joint and the mouthpiece concentrates pressure due to its severe twisting.
 * Roller mouthpieces: tend to make horses relax their mouth and activate the tongue, encouraging salivation and acceptance of the bit. This may also focus tense or nervous horses to the bit.
 * Hollow mouth: usually single-jointed with a thick, hollow mouthpiece which spreads out the pressure and makes the bit less severe. May not fit comfortably in some horses' mouths if they are a little small.

The snaffle rings

There are several types of rings that affect the action of the bit.

 * Loose ring: slides through the mouthpiece. Tends to make the horse relax his jaw and chew the bit. May pinch the corners of the horse's mouth if the holes in the mouthpiece are large, in which case a bit guard should be used.
 * Egg butt/barrel head: mouthpiece does not rotate, and is so more fixed in the horse's mouth, which some horses prefer. Will not pinch the lips.
 * D-ring/ racing snaffle: ring in the shape of a "D" which does not allow the bit to rotate and so the bit is more fixed. The sides of the D provide a lateral guiding effect.
 * Full cheek: has long, extended arms above and below the mouthpiece on either side of the lips of the horse, with a ring attached to it. The cheeks have a lateral guiding effect, and also prevent the bit from sliding through the mouth. The full cheek is often used with bit keepers to prevent the cheeks from getting caught on anything, and to keep the bit in the right position inside the mouth.
 * Half-cheek: has only an upper or, more commonly, lower cheek, as opposed to both seen in a full cheek snaffle. Often used in racing, as there is less chance of the cheek being caught on the starting gate, or in driving as there is less chance of getting caught on harness straps.
 * Baucher (hanging cheek): has a ring on the side of the mouthpiece, with a smaller ring above to attach the cheekpiece of the bridle. Tends to concentrate pressure on the bars. It is very fixed in the mouth.
 * Fulmer: a full cheek bit with a loose ring attached, so that it not only has the lateral guiding effect, but can also move freely as with a loose ring.

title: Snaffle bit


text:
Kanzeon Zen Center

Kanzeon Zen Center was a Zen Buddhist center located in Salt Lake City, Utah. It was an affiliate of the White Plum Asanga, an association of Zen centers stemming from the tradition of Taizan Maezumi. The founder and Abbot of Kanzeon Zen Center was Dennis Genpo Merzel Roshi, who resigned in 2011 amidst controversy. Kanzeon Zen Center was the home temple and the hub of Kanzeon Sangha International, founded by Genpo Roshi in 1984, with affiliate teachers, centers and groups in the US and seven European countries. The center was housed at 1274 E. South Temple, a historic building listed as a contributing property in the South Temple Historic District. It closed in the wake of the sex scandals involving Merzel. News reports stated that the center was deeply financially in debt to Merzel.

title: Kanzeon Zen Center


text:
Boskin Commission

The Boskin Commission, formally called the "Advisory Commission to Study the Consumer Price Index", was appointed by the United States Senate in 1995 to study possible bias in the computation of the Consumer Price Index (CPI), which is used to measure inflation in the United States. Its final report, titled "Toward A More Accurate Measure Of The Cost Of Living" and issued on December 4, 1996, concluded that the CPI overstated inflation by about 1.1 percentage points per year in 1996 and about 1.3 percentage points prior to 1996.

The report was important because inflation, as calculated by the Bureau of Labor Statistics, is used to index the annual payment increases in Social Security and other retirement and compensation programs. This implied that the federal budget had increased by more than it should have, and that projections of future budget deficits were too large. The original report calculated that the overstatement of inflation would add $148 billion to the deficit and $691 billion to the national debt by 2006.

The report highlighted four sources of possible bias:
 * Substitution bias occurs because a fixed market basket fails to reflect the fact that consumers substitute relatively less for more expensive goods when relative prices change.
 * Outlet substitution bias occurs when shifts to lower price outlets are not properly handled.
 * Quality change bias occurs when improvements in the quality of products, such as greater energy efficiency or less need for repair, are measured inaccurately or not at all.
 * New product bias occurs when new products are not introduced in the market basket, or included only with a long lag.

The members of the Boskin Commission were:
 * Michael Boskin, Stanford University (Chair)
 * Ellen R. Dulberger
 * Robert J. Gordon
 * Zvi Griliches
 * Dale Jorgenson

The Boskin Commission was the first extensive evaluation of inflation measurement since the Stigler Commission in 1961. Griliches was also on that commission.

title: Boskin Commission


text:
User talk:Bliksem

Welcome!

Hello,, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are a few good links for newcomers: I hope you enjoy editing here and being a Wikipedian! Please sign your name on talk pages using four tildes (~~); this will automatically produce your name and the date. If you need help, check out Where to ask a question, ask me on my talk page, or place  on your talk page and someone will show up shortly to answer your questions. Again, welcome!
 * The five pillars of Wikipedia
 * How to edit a page
 * Help pages
 * Tutorial
 * How to write a great article
 * Manual of Style

Cheers, --R.Koot 00:25, 5 August 2005 (UTC)

Your WP:PAC assignment
Since you have not edited your WP:PAC assignments, Quilava and Typhlosion, within the last week, you have been un-assigned from the articles. Unfortunately, we have to enforce this rule if we wish to finish the remaining articles within a reasonable amount of time.

If you wish, you are welcome to request a new assignment. --Pidgeot (t) (c) (e) 13:55, 9 September 2005 (UTC)

Michuru81 was here
Michuru81 has vandalized your discussion page! That's okay. You'll forgive him for this most heinous crime. You're just that nice of a person. --Michuru 19:45, 19 September 2005 (UTC)

title: User talk:Bliksem


text:
Picasso's Blue Period

The Blue Period (Período Azul) comprises the works produced by Spanish painter Pablo Picasso between 1901 and 1904. During this time, Picasso painted essentially monochromatic paintings in shades of blue and blue-green, only occasionally warmed by other colors. These sombre works, inspired by Spain and painted in Barcelona and Paris, are now some of his most popular works, although he had difficulty selling them at the time.

This period's starting point is uncertain; it may have begun in Spain in the spring of 1901 or in Paris in the second half of the year. In choosing austere color and sometimes doleful subject matter—prostitutes, beggars and drunks—Picasso was influenced by a journey through Spain and by the suicide of his friend Carles Casagemas, who took his own life at the L'Hippodrome Café in Paris, France on February 17, 1901. Although Picasso himself later recalled, "I started painting in blue when I learned of Casagemas's death", art historian Hélène Seckel has written: "While we might be right to retain this psychologizing justification, we ought not lose sight of the chronology of events: Picasso was not there when Casagemas committed suicide in Paris ... When Picasso returned to Paris in May, he stayed in the studio of his departed friend, where he worked for several more weeks to prepare his exhibition for Vollard". The works Picasso painted for his show at Ambroise Vollard's gallery that summer were generally characterized by a "dazzling palette and exuberant subject matter". Picasso's psychological state worsened as 1901 continued.

In the latter part of 1901, Picasso sank into a severe depression and blue tones began to dominate his paintings. Picasso's painting La mort de Casagemas, completed early in the year following his friend's suicide, was done in hot, bright hues. The painting considered the first of his Blue Period, Casagemas in His Coffin, was completed later in 1901 when Picasso was sinking into a major depression. Picasso, normally an outgoing socializer, withdrew from his friends. Picasso's bout of depression was to last several years. Picasso's career had been promising before 1901 and early in that year he was making "a splash" in Paris. However, as he moved towards subject matter such as society's poor and outcast, and accented this with a cool, anguished mood with blue hues, the critics and the public turned away from his works. Members of the public were uninterested in displaying the Blue Period works in their homes. Picasso continued his output, but his financial situation suffered:"His pictures, not merely melancholy but profoundly depressed and cheerless, inspired no affection in the public or in buyers. It was not poverty that led him to paint the impoverished outsiders of society, but rather the fact that he painted them that made him poor himself."

From 1901 to 1903, he painted several posthumous portraits of Casagemas, culminating in the gloomy allegorical painting La Vie, painted in 1903 and now in the Cleveland Museum of Art. The same mood pervades the well-known etching The Frugal Repast (1904) which depicts a blind man and a sighted woman, both emaciated, seated at a nearly bare table. Blindness is a recurrent theme in Picasso's works of this period, also represented in The Blindman's Meal (1903, the Metropolitan Museum of Art) and in the portrait of Celestina (1903).

Infrared imagery of Picasso's 1901 painting The Blue Room reveals another painting beneath the surface.

Other frequent subjects include female nudes and mothers with children. Solitary figures dominate his Blue Period works. Themes of loneliness, poverty and despair pervade the works as well. Possibly his most well known work from this period is The Old Guitarist. Other major works include Portrait of Soler (1903) and Las dos hermanas (1904).

Picasso's Blue Period was followed by his Rose Period. Picasso's bout with depression gradually ended, and as his psychological state improved, he moved towards more joyful, vibrant works, and emphasized the use of pinks ("rose" in French) and other warm hues to express the shift in mood and subject matter.

The painting Portrait of Suzanne Bloch (1904), one of the final works from this period, was stolen from the São Paulo Museum of Art (MASP) on December 20, 2007, but retrieved on January 8, 2008.

title: Picasso's Blue Period


text: User talk:69.214.22.252

title: User talk:69.214.22.252


text:
Ashland School District

There are several school districts in the United States called Ashland School District, including:

 * Ashland Independent School District — Ashland, Kentucky
 * Ashland District School — Ashland, Maine
 * Ashland City School District — Ashland, Ohio
 * Ashland School District (Oregon) — Ashland, Oregon
 * Ashland School District (Wisconsin) — Ashland, Wisconsin

title: Ashland School District


text:
Louis Svitek

Louis Svitek is an American musician and former guitarist for a number of bands including Zoetrope, M.O.D., Ministry, Pigface, Mind Funk, The Hollow steps, Project .44, and The Joy Thieves. His work can also be found in many movie soundtracks such as The Matrix and Blue Hill Avenue.

WuLi Records
Louis Svitek and partner Ryan McGuire founded their own recording studio and independent record label, "WuLi Records", in Chicago, IL. WuLi Records is responsible for discovering Chicago's acts such as The Redd, Young Inno, The Waking, and American Idol 2010 winner Lee DeWyze. Prior to Idol, Louis Svitek was part of the Lee DeWyze Band that also included Ryan McGuire and Jeff Henderson.

title: Louis Svitek


text:
Wikipedia:Articles for deletion/Ultimate Dominion

 This page is an archive of the proposed deletion of the article below. Further comments should be made on the appropriate discussion page (such as the article's talk page or on a Votes for Undeletion nomination). No further edits should be made to this page.

The result of the debate was Delete --Allen3 talk 13:32, August 4, 2005 (UTC)

Ultimate Dominion
spam Y0u (Y0ur talk page) (Y0ur contributions) 20:10, July 24, 2005 (UTC)
 * Delete Try 'advertising'. Sonic Mew | talk to me 20:40, July 24, 2005 (UTC)
 * Keep. I cleaned it up.  It's neutral and verifiable. --Tony Sidaway|Talk 05:35, 26 July 2005 (UTC)
 * Delete. Advertising. --Carnildo 08:31, 3 August 2005 (UTC)
 * The above discussion is preserved as an archive of the debate. Please do not modify it. Subsequent comments should be made on the appropriate discussion page (such as the article's talk page or in an undeletion request). No further edits should be made to this page.


text:
Talk:Matrix exponential

Example (homogeneous)
Could someone check the result for the matrix exponential
 * $$e^{tM}=\begin{bmatrix}

2e^t - 2te^{2t} & -2te^{2t}   & 0 \\ -2e^t + 2(t+1)e^{2t} & 2(t+1)e^{2t} & 0 \\ 2te^{2t} & 2te^{2t}    & 2e^t\end{bmatrix}$$ The answer that Mathematica gives is quite different:
 * $$e^{tM}=\begin{bmatrix}

\frac12 e^{2t}(1+e^{2t}-2t) & -te^{2t}   & e^{3t}\sinh t \\ -\frac12 e^{2t}(-1+e^{2t}-2t & (t+1)e^{2t} & -e^{3t}\sinh t \\           -\frac12 e^{2t}(-1+e^{2t}+2t & te^{2t}     & e^{3t}\cosh t\end{bmatrix}$$ Obviously, this would give another solution to the system as well...

 * Well, the stated answer is wrong on its face, as $$e^{0M}=I$$. — Arthur Rubin |  (talk) 12:28, 22 August 2006 (UTC)

 * So, should we change it to the Mathematica-approved answer that I wrote a few lines above?

 * well, acutally, I get a different form:
 * $$e^{tM} = \begin{bmatrix}

e^{3t} \cosh t - t e^{2t} & -t e^{2t} & e^{3t} \sinh t \\ -e^{3t} \sinh t + t e^{2t} & (t+1) e^{2t}& -e^{3t} \sinh t \\ e^{3t} \sinh t + t e^{2t} & t e^{2t} & e^{3t} \cosh t \end{bmatrix}$$
 * Same values, except for a couple of typos, but it looks simpler to me. — Arthur Rubin |  (talk) 00:27, 31 August 2006 (UTC)

 * Perhaps it's better to use the same matrix that is mentioned further up in the article? -- Jitse Niesen (talk) 02:50, 31 August 2006 (UTC)

It looks like the two examples were copied from somewhere - it refers to examples "earlier in the article" where the exponential matrix is calculated, but this example doesn't exist. I can't tell if this is referring to a part of the wiki that was removed, or whether the example was just copied in it's entirety from another source. 130.215.95.157 (talk) 18:45, 5 March 2008 (UTC)
 * I see this still hasn't been fixed. See  below.  — Arthur Rubin  (talk) 21:29, 23 October 2014 (UTC)

Column method
It seems to me that the column method is the same as the method based on the Jordan decomposition, but explained less clearly. Hence, I am proposing to remove that section. -- Jitse Niesen (talk) 20:11, 24 July 2005 (UTC)

I now removed the section. -- Jitse Niesen (talk) 11:21, 3 August 2005 (UTC)

 * I think there is indeed A Point to doing it like that, but I need to have a close look over it again as I'm a little unfamiliar on the material and need to get acquanted with it again, and I haven't had a chance to do this. On a cursory look the removal looks okay, though... Dysprosia 09:32, 4 August 2005 (UTC)

Continuity, etc.
For any two matrices X and Y, we have
 * $$ \| e^{X+Y} - e^X \| \le \|X\| e^{\|X\|} e^{\|Y\|}, $$

is clearly incorrect -- just look at the case X = 0. Perhaps the equation should be
 * $$ \| e^{X+Y} - e^X \| \le e^{\|X\|} e^{\|Y\|}, $$

(which I changed it to), but I'm not sure that's correct, either. Arthur Rubin | (talk) 21:12, 1 February 2006 (UTC)

 * Whoops, I'm pretty sure I put that in. According to H&J, Corollary 6.2.32, we have
 * $$ \| e^{A+E} - e^A \| \le \|E\| e^{\|E\|} e^{\|A\|}, $$
 * or, using X and Y,
 * $$ \| e^{X+Y} - e^X \| \le \|Y\| e^{\|X\|} e^{\|Y\|}. $$
 * Apparently I made a mistake while renaming the variables. Thanks a lot! -- Jitse Niesen (talk) 21:36, 1 February 2006 (UTC)

Thank you
Hi, I just wanted to say thank you for writing this article so clearly. I needed to quickly look up how to do matrix exponentials again and I thought I'd try Wikipedia instead of MathWorld first this time. Nice simple explainations here, and written very clearly. This thanks also goes out to all deticated wikipedians who are updating the math pages.--Johnoreo 02:21, 9 February 2006 (UTC)

Arbitrary field?
The article mentions calculations over an arbitrary field. I think this should be changed since it gives the impression that there exists an exponential map over arbitrary fields.T3kcit (talk) 19:55, 16 December 2007 (UTC)

 * Agreed and changed. Solian en (talk) 14:37, 23 April 2008 (UTC)

Transition matrix
The transition matrix used here does not jive with what I understand is a matrix of eigenvectors. The rows need not sum to one. Is this a different matrix? If so, I haven't found an entry for the transition matrix used as a product with the diagonal matrix of eigenvalues and the inverse transition matrix to solve for the matrix exponential. The internal link does not clearly address this usage.John (talk) —Preceding comment was added at 02:31, 26 February 2008 (UTC)

 * The term transition matrix here means the matrix associated with the similarity transform that puts the matrix A into Jordan form; it has little to do with the meaning explained in transition matrix. I doubt this term is used very often, so I reformulated the text to avoid this term. Hope this helps. -- Jitse Niesen (talk) 14:52, 29 February 2008 (UTC)

Computing the matrix exponential (general case, arbitrary field)
There is no reference for the X = A + N decomposition over an arbitrary field. I couldn't find one in my textbooks (except for the Jordan decomposition in C). The French version points to the Dunford decomposition, which requires the matrix's minimal polynomial to have all its roots in the field. I believe that the decomposition does not hold in general. If no one objects, I will specify the conditions under which it exists. Solian en (talk) 15:04, 2 April 2008 (UTC)

 * Unfortunately, we've now obscured the fact that a complex matrix always has a unique A + N decomposition. I think we should drop consideration of arbitrary fields altogether until someone wants to handle them properly (probably in another section). At any rate it isn't at all clear when the article switches from the complex case to more general fields. -- Fropuff (talk) 16:30, 23 April 2008 (UTC)

Commutativity
Does anybody know if the equation $$\exp(A+B)=\exp(A) \exp(B)$$ implies $$AB=BA$$? Franp9am (talk) 10:18, 20 June 2008 (UTC)

 * It does not. A counterexample (from Horn & Johnson) is given by
 * $$ A = \begin{bmatrix} 0 & 0 \\ 0 & 2\pi\textrm{i} \end{bmatrix} \quad\text{and}\quad B = \begin{bmatrix} 0 & 1 \\ 0 & 2\pi\textrm{i} \end{bmatrix}. $$
 * $$A$$ and $$B$$ do not commute, but $$\exp(A+B) = \exp(A) \exp(B) = \text{the identity matrix}$$. I added something to the article. -- Jitse Niesen (talk) 10:59, 20 June 2008 (UTC)

- Note added; sorry if I'm not following conventions here, I haven't commented before. --Tom

It's also not even correct in general to say that $$\exp(A+B)=\exp(A)\exp(B)$$. For a simple counterexample, take nilpotent matrices $$A$$ and $$B$$ that sum to a nonsingular $$A+B$$, say

$$ A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \quad\text{and}\quad B = \begin{bmatrix} 0 & 0 \\ -1 & 0 \end{bmatrix}. $$

The power series for $$\exp(A)=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$ and $$\exp(B)=\begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}$$ terminate after the second terms, but $$\exp(A+B)$$ never terminates at all: $$\exp(A+B)=\begin{bmatrix} \cos(1) & \sin(1) \\ -\sin(1) & \cos(1) \end{bmatrix}$$.

Horn and Johnson give a similar example in Topics in Matrix Analysis, p. 434-435. Automatic Tom (talk) 22:18, 9 August 2008 (UTC)

 * The above appears to be correct. Michael Hardy (talk) 21:46, 9 August 2008 (UTC)
 * Side-note: $$ A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \quad\text{and}\quad B = \begin{bmatrix} 0 & 0 \\ -1 & 0 \end{bmatrix}$$ do not commute. Therefore they are not covered by the theorem. TN (talk) 11:55, 13 August 2020 (UTC)

The reference to the Horn and Johnson Topics in Matrix Analysis book quotes a theorem that says if $$ A $$ and $$ B $$ contain only algebraic entries then you can only say that $$ e^Ae^B = e^B\,e^A $$ if and only if $$ A $$ and $$ B $$ commute. This is different than saying $$ e^{A+B} = e^Ae^B $$ if and only if $$ A $$ and $$ B $$ commute. — Preceding unsigned comment added by Robleroble (talk • contribs) 08:23, 6 March 2013 (UTC)

Matrix Exponential via Laplace transform
I don't know how to properly code the math behind this, but the page is missing a section on solving for the matrix exponential via the Laplace transform.

 * $$ \mathcal{L}^{-1} \{ e^{At} \} = (sI-A)^{-1} $$

Substituting t=1 in the above equation will yield the matrix exponential of A.

99.236.42.178 (talk) 01:29, 3 November 2008 (UTC)

 * I am afraid this is correct but aggressively meaningless, as is section 2.5 which I propose to eliminate. That section is a circular restatement of true but jubilantly unusable facts. Unless you provided an explicit situation where a generalization of the Laplace transform for matrices provided explicit answers not available to the Taylor expansion of the exponential, or, in the truly astounding case, the Magnus expansion mentioned earlier, I don't see what the point of that section could possibly be.  Cuzkatzimhut (talk) 19:55, 1 November 2013 (UTC)

 * I'm not sure how this is "aggressively meaningless". If the point of section 2 is to give methods for the computation of the matrix exponential, then you can do it by the Laplace transform in just the way it was written. This isn't a generalization of the Laplace transform for matrices, merely the element-wise application of the inverse Laplace to $$(sI - A)^{-1}$$.  That being said, I don't know if this page should aim to provide every possible method for computing the matrix exponential. Zfeinst (talk) 01:54, 2 November 2013 (UTC)

 * OK, proceed to justify your statement by illustrating how the compact bottom line formula of the LT yields the above trivial formula of 2.4.1  for a general 2x2 matrix. It could make that section meaningful. As I indicated above, all this LT rigmarole is a circular way of rewriting the matrix exponential in terms of its power expansion, which is how it was defined in the first place, accepting application of the LT and its convergence peculiarities on the space of matrices. The connection to the differential equation is already detailed in section 1. You need a LT to handle an exponential? What would you believe the student of matrix exponentials would have lost, had she not read this subsection (2.5)? Cuzkatzimhut (talk) 11:55, 2 November 2013 (UTC)

I proceed to eliminate the LT tautology, and effectively assign its section number to the preceding subsection, 2.4.1, which evidently merits promotion to a self-standing section. I furthermore stuck the above s=1 formal identification in the Properties section further up in the article. Cuzkatzimhut (talk) 00:58, 5 November 2013 (UTC)

proof
It would be useful to sketch how some of the properties (eg. exp(M+N)=exp(M)exp(N) if ..) are proved. Cesiumfrog (talk) 07:59, 13 March 2010 (UTC)

Exponential of Sums Error & Lie Product Formula
I have a couple issues with this statement: "The converse is false: the equation eX + Y = eXeY does not necessarily imply that X and Y commute. However, the converse is true if X and Y contain only algebraic numbers and their size is at least 2×2 (Horn & Johnson 1991, pp. 435–437)"
 * "Horn & Johnson" should be omitted in favor of a link to the actual reference at the bottom. The reference is actually to "Topics in Matrix Analysis", whereas lots of people are used to thinking of "Matrix Analysis" when they see just "Horn & Johnson".
 * I read that section of Topics in Matrix Analysis this morning and what it actually says is that, if X and Y are NxN matrices, N >= 2, containing only algebraic numbers, then e^Xe^Y = e^Ye^X if and only if X and Y commute.  Nothing to do with e^{X+Y}.  It's also more of a side note in the H&J book; it's not proved or even stated as a theorem, lemma, or even a numbered equation.

On a related note, it may be helpful to some people (I know it would have been helpful to me!) to put a link somewhere here to the Lie product formula article, which says lim_{m\rightarrow\infty} (e^{A/m}e^{B/m})^m = e^{A+B} for ANY A and B.

I don't want to just make these changes because a) I don't really have time and b) I'm hoping someone who can take care to double check and integrate it with the rest of the article can do it.

--M0nstr42 (talk) 01:50, 30 September 2010 (UTC)

Exponential skew symmetric matrices
I added the section on skew symmetric matrices. My proofs are my own independent work - I have seen the results, however, given by a few different authors (only after I went through all the work did I bother to look anything up, derp). None of the sources prove my results. Does anyone know of anything more elegant than what I did? Lhibbeler (talk) 02:38, 4 November 2010 (UTC)

@ User:Arthur_Rubin, see Jean Gallier and Dianna Xu, "Computing Exponentials of Skew Symmetric Matrices And Logarithms of Orthogonal Matrices", International Journal of Robotics and Automation, 2000. My results are *not* original research. —Preceding unsigned comment added by Lhibbeler (talk • contribs) 04:19, 4 November 2010 (UTC)

 * Ahem. You said they were original research.  As for the polynomial equations for the exponentials of the skew-symmetric matrices, it's not that interesting a result.  It follows from some of the things which should be in [[matrix polynomial]; as a two-dimensional skew-symmetric matrix W has eigenvalues $$\pm i \alpha$$, and exp(W) has eigenvalues $$\exp(\pm i \alpha)$$.  Hence, if
 * $$P(x) = \cos \alpha + \frac {sin(\alpha)}{\alpha} x$$,
 * then P(W) = exp(W). — Arthur Rubin  (talk) 04:54, 4 November 2010 (UTC)

 * I said my proofs are original as far as I know- I have seen *the results* but not the derivations. I was trying to explain where the formulas come from rather than just throw up a formula. And who are you to say what is interesting? The results are interesting and important to mechanicians- using that formula is more than 10 times faster on my laptop than using Pade approximants. When you have a large number of skew matrices and you need to calculate their exponentials (rotation updates in crystal plasticity, for example), an order of magnitude decrease in computation time is, in fact, very interesting. It sounds like you have more beef with the matrix polynomial page than with my edit. If you still object to my revision, then get rid of the collapsible windows with the proofs. Besides, I have put two dents in the statement that opens the section "Finding reliable and accurate methods to compute the matrix exponential is difficult, and this is still a topic of considerable current research in mathematics and numerical analysis." Lhibbeler (talk) 05:19, 4 November 2010 (UTC)

 * You could list those formulas as examples in the "Alternative" section. Perhaps that section could be promoted, but, you really haven't provided anything else.  — Arthur Rubin  (talk) 05:49, 4 November 2010 (UTC)
 * Rechecking the 3D formula:
 * $$P(x) = 1 + \frac{\sin \alpha}{\alpha} x + \frac{1-\cos \alpha}{\alpha^2} x^2 $$
 * $$P(0) = 1 = \exp(0)$$
 * $$P(\pm i \alpha) = \cos \alpha \pm i \sin \alpha = \exp(\pm i \alpha)$$
 * Yep, it's the equation from the "Alternative" section. — Arthur Rubin  (talk) 05:55, 4 November 2010 (UTC)

What about this
For a 2x2 matrix A one can write the exponent as follows:

 * $$ e^A = \sqrt{e^{tr A}} (\frac{sinh(q)}{q}A + (cosh(q) - \frac{tr(A) sinh(q)}{2q}) I) $$

where
 * $$ q = \sqrt{-det(A-\frac{tr A}{2} I)} $$

and I is the 2x2 identity matrix, and sinh(0)/0 is set as 1.

This doesn't seem to be in the article and might be interesting for those looking for a closed expression. —Preceding unsigned comment added by 95.117.203.24 (talk) 12:15, 22 January 2011 (UTC)
 * Is that correct if A has its two eigenvalues equal, but is not a multiple of the identity? If it is, it seems notable, but is sufficiently complex as to require a source.  — Arthur Rubin  (talk) 15:07, 22 January 2011 (UTC)
 * Never mind, it's accurate. Still, a better expression might be:
 * $$ e^A = e^{\frac{\operatorname{tr} A}{2}} \left(\left(\cosh q\right) I +\frac{\sinh q }{q}\left(A-\frac{\operatorname{tr} A}{2}I\right) \right) $$
 * where
 * $$ q = \sqrt{-\det\left(A-\frac{\operatorname{tr} A}{2} I\right)} $$
 * Which can be seen by
 * $$e^\lambda \approx e^{\frac{\operatorname{tr} A}{2}}\left (\cosh q + \frac{\sinh q}{q} \left(\lambda - \frac{\operatorname{tr} A}{2}\right)\right) $$
 * at the eigenvalues λ of A. — Arthur Rubin  (talk) 16:31, 22 January 2011 (UTC)

 * Hello. Yes, that case leads to q = 0.  I don't have a source for this equation but the proof is not hard:  one writes $$ A =  \begin{bmatrix}

a & b \\ c & d \end{bmatrix} = \begin{bmatrix} \frac{a+d}{2} & 0 \\ 0 & \frac{a+d}{2} \end{bmatrix} + \begin{bmatrix} \frac{a-d}{2} & b \\ c & \frac{d-a}{2} \end{bmatrix} = B + C $$. Since B and C commute we have e^A = (e^B)(e^C), and since B and C^2 are diagonal matrices, both e^B and e^C can be calculated directly.

Unfortunately this doesn't work so for 3x3 and bigger matrices. Maybe your reasoning will lead to a similar formula for bigger matrices, I'll have to think about it! —Preceding unsigned comment added by 95.117.203.24 (talk) 16:50, 22 January 2011 (UTC)

 * Included, now, as an alternative to the alternative section for 2×2 matrices. — Arthur Rubin  (talk) 17:22, 22 January 2011 (UTC)

Applications to quantum computing?
I didn't see anything in this article about its applications to quantum computing. In particular, with Hermitian matrices, Pauli operators, and rotations about the Bloch Sphere. There's a bunch about this in the book I'm reading: An Introduction to Quantum Computing (Phillip Kaye, Raymond Laflamme, and Michelle Mosca - 2007). Just thought I'd point this out in case anyone feels it's important to mention.

24.212.140.98 (talk) 21:55, 22 January 2012 (UTC)

A method I was taught, but cannot find external references
Hello, I was taught a method by a Hungarian mathematician and I'm not even sure what it is called. The problem is to find $$ e^{A t} $$. If you do not care about $$ t $$ then replace it with a 1 when done. I'm not sure if anyone else is aware of this method or if it is even useful. So I'm putting it here under the talk section of the article.

Anyways I will include an example and steps below. Given $$A = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & -1/8 \\ 0 & 0 & 1/2 & 1/2 \end{pmatrix} $$

First, compute the eigenvalues. $$ \lambda_1 = 3/4 $$ and $$ \lambda_2 = 1 $$ each with an algebraic multiplicity of two.

Now, take the exponential of each eigenvalue multiplied by $$ t $$: $$ e^{\lambda_i t}$$. Multiply by an unknown matrix $$ B_i $$. If the eigenvalues have an algebraic multiplicity greater than 1, then repeat the process, but multiply by a factor of $$ t $$ for each repetition. If one eigenvalue had a multiplicity of three, then there would be the terms: $$ B_{i_1} e^{\lambda_i t}, B_{i_2} t e^{\lambda_i t}, B_{i_3} t^2 e^{\lambda_i t} $$. Sum all terms.

In our example we get:

$$ e^{A t} = B_{1_1} e^{\lambda_1 t} + B_{1_2} t e^{\lambda_1 t} + B_{2_1} e^{\lambda_2 t} + B_{2_2} t e^{\lambda_2 t} $$

$$ e^{A t} = B_{1_1} e^{3/4 t} + B_{1_2} t e^{3/4 t} + B_{2_1} e^{1 t} + B_{2_2} t e^{1 t} $$.

So how can we get enough equations to solve for all of the unknown matrices? Differentiate with respect to $$t$$.

$$ e^{A t} = B_{1_1} e^{3/4 t} + B_{1_2} t e^{3/4 t} + B_{2_1} e^{1 t} + B_{2_2} t e^{1 t} $$

$$ A e^{A t} = 3/4 B_{1_1} e^{3/4 t} + \left( 3/4 t + 1 \right) B_{1_2} e^{3/4 t} + 1 B_{2_1} e^{1 t} + \left(1 t + 1 \right) B_{2_2} e^{1 t} $$

$$ \begin{align} A^2 e^{A t} =& (3/4)^2 B_{1_1} e^{3/4 t} + \left( (3/4)^2 t + ( 3/4 + 1 \cdot 3/4) \right) B_{1_2} e^{3/4 t} + B_{2_1} e^{1 t} + \left(1^2 t + (1 + 1 \cdot 1 )\right) B_{2_2} e^{1 t} \\ =& (3/4)^2 B_{1_1} e^{3/4 t} + \left( (3/4)^2 t + 3/2 \right) B_{1_2} e^{3/4 t} + B_{2_1} e^{t} + \left(t + 2\right) B_{2_2} e^{t} \end{align} $$

$$ \begin{align} A^3 e^{A t} =& (3/4)^3 B_{1_1} e^{3/4 t} + \left( (3/4)^3 t + ( (3/4)^2 + (3/2) \cdot 3/4) ) \right) B_{1_2} e^{3/4 t} + B_{2_1} e^{1 t} + \left(1^3 t + (1 + 2) \cdot 1 \right) B_{2_2} e^{1 t} \\ =& (3/4)^3 B_{1_1} e^{3/4 t} + \left( (3/4)^3 t + 27/16 ) \right) B_{1_2} e^{3/4 t} + B_{2_1} e^{t} + \left(t + 3 \cdot 1 \right) B_{2_2} e^{t} \end{align} $$

Since the these equations must be true, regardless the value of $$t$$, we set $$t=0$$. Then we can solve for the unknown matrices.

$$ \begin{align} I =& B_{1_1} + B_{2_1} \\ A =& 3/4 B_{1_1} + B_{1_2} + B_{2_1} + B_{2_2} \\ A^2 =& (3/4)^2 B_{1_1} + (3/2) B_{1_2} + B_{2_1} + 2 B_{2_2} \\ A^3 =& (3/4)^3 B_{1_1} + (27/16) B_{1_2} + B_{2_1} + 3 B_{2_2} \end{align} $$

This can be solved using linear algebra (don't let the fact the variables are matrices confuse you). Once solved using linear algebra you have:

$$ \begin{align} B_{1_1} =& 128 A^3 - 366 A^2 + 288 A - 80 I \\ B_{1_2} =& 16 A^3 - 44 A^2 + 40 A - 12 I \\ B_{2_1} =&-128 A^3 + 366 A^2 - 288 A + 80 I\\ B_{2_2} =& 16 A^3 - 40 A^2 + 33 A - 9 I \end{align} $$

Plugging in the value for $$A$$ gives:

$$ \begin{align} B11 =& \begin{pmatrix}0 & 0 & 48 & -16\\ 0 & 0 & -8 & 2\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}\\ B12 =& \begin{pmatrix}0 & 0 & 4 & -2\\ 0 & 0 & -1 & \frac{1}{2}\\ 0 & 0 & \frac{1}{4} & -\frac{1}{8}\\ 0 & 0 & \frac{1}{2} & -\frac{1}{4}\end{pmatrix}\\ B21 =& \begin{pmatrix}1 & 0 & -48 & 16\\ 0 & 1 & 8 & -2\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}\\ B22 =& \begin{pmatrix}0 & 1 & 8 & -2\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix} \end{align} $$

So the final answer would be:

$$ {e}^{t\,A}=\begin{pmatrix}{e}^{t} & t\,{e}^{t} & \left( 8\,t-48\right) \,{e}^{t}+\left( 4\,t+48\right) \,{e}^{\frac{3\,t}{4}} & \left( 16-2\,t\right) \,{e}^{t}+\left( -2\,t-16\right) \,{e}^{\frac{3\,t}{4}}\\ 0 & {e}^{t} & 8\,{e}^{t}+\left( -t-8\right) \,{e}^{\frac{3\,t}{4}} & -\frac{4\,{e}^{t}+\left( -t-4\right) \,{e}^{\frac{3\,t}{4}}}{2}\\ 0 & 0 & \frac{\left( t+4\right) \,{e}^{\frac{3\,t}{4}}}{4} & -\frac{t\,{e}^{\frac{3\,t}{4}}}{8}\\ 0 & 0 & \frac{t\,{e}^{\frac{3\,t}{4}}}{2} & -\frac{\left( t-4\right) \,{e}^{\frac{3\,t}{4}}}{4}\end{pmatrix} $$

This method was taught as it is nearly the same procedure to calculate $$A^t$$. Except that instead of taking the derivative to generate more equations, you can increment $$t$$ (which is useful for discrete difference equations). Also, replace using $$e^{(\lambda_i t)}$$ with $$\lambda_i^t$$. — Preceding unsigned comment added by 150.135.222.237 (talk) 18:57, 6 December 2013 (UTC)

 * I confess I don't know a name for it, and I have not stumbled on it in just this form in matrix analysis books. You might register to Wikipedia and clean it up and add it as a subsection near the end. It is quite tasteful in its immediacy, as it does not require diagonalizability, as manifest in your example of a neat defective matrix. The ansatz is a most general one for the generic polynomial dictated by the Cayley-Hamilton theorem, in your case a cubic; the "Wronskian" trick for multiple roots ensures linear independence, and is equivalent to the analysis of the Matrix_exponential subsection in the article. In fact, I wonder if you wished to adapt the example to a shorter section in the same notation, i.e. the ansatz exp(tA) = Bα exp (tα) + Bβ exp(tβ), yielding the same expression after solving this expression and its first derivative at t=0 for the Bs in terms A and I, arguably faster. It is a pretty method, and you'd be quite welcome to contribute it to the article (be bold, etc). Cuzkatzimhut (talk) 23:22, 15 December 2013 (UTC)
 * Actually, for distinct eigenvalues, it is just Sylvester's formula for the exponential, where the simplicity of the l.h.s. allows for ready indirect evaluation of the Bi s, the Frobenius covariants projecting onto the eigenspace corresponding to eigenvalues λi . For the general, non-diagonalizable case you are addressing, it is just about Buchheim's generalization. It' is well-worth covering in the article. Cuzkatzimhut (talk) 19:36, 16 December 2013 (UTC)

 * I am actually registered (I didn't sign in when I originally posted this), but following wikipedia's best practices, I cannot post "original research." Thus I cannot in good conscience post this as a subsection until someone can give this method a name or at least reference it elsewhere. Maybe this will help; this method is derived from the method of undetermined coefficients for linear differential equations. How to handle repeated roots (eigenvalues) follows from the same explanation here (https://www.khanacademy.org/math/differential-equations/second-order-differential-equations/complex-roots-characteristic-equation/v/repeated-roots-of-the-characteristic-equation). If you think this is enough support, then we (or I) can post it. Alternatively, if you can dig up the resources that discuss this being Bucheim's generalization of Sylvester's formula, then we can post it. .Mouse7mouse9 22:22, 29 December 2013 (UTC) — Preceding unsigned comment added by Mouse7mouse9 (talk • contribs)

 * I also wanted to mention there is a discrete variant (from solving discrete difference equations), allowing one to take symbolic matrix powers (raising a square matrix to an unknown scalar variable). I mention it here (https://en.wikipedia.org/wiki/Talk:Matrix_multiplication#Matrix_powers_and_discrete_difference_equations). Both the above method and the linked method are very similar. The symbolic matrix power is derived from solving discrete linear difference equations (again using the method of undetermined coefficients), rather than continuous differential equations, as in the case of the matrix exponential. — Preceding unsigned comment added by Mouse7mouse9 (talk • contribs) 23:26, 29 December 2013 (UTC)

I am sorry, I do not have a name for it, but, as I said, it is a less hidebound expression of 2.5. Different communities of users often employ different names. Please, do go ahead and post it--hopefully the names and the references will accrue later; but nobody would accuse you of posting original research, to the extent this is self-evident, a linear determination of the Frobenius covariants. So, for instance, you could call it subsection 2.6, something like: "A rapid implementation of Sylvester's formula", and work out first the trivial 2x2 example above, and then your neat defective matrix 4x4 example, in some semblance of uniform notation, as a segue to 2.5 (note its last case!). Ideally, somebody could provide a nice reference for it, but its logic is what is already covered in the exposition preceding it. I could help with the wp-linking after you were done. As an in itialreference, maybe this could do: Rinehart, R. F. (1955). "The equivalence of definitions of a matric function". The American Mathematical Monthly, 62 (6), 395-414. Cuzkatzimhut (talk) 00:03, 30 December 2013 (UTC)

 * OK, I put it in the page as stated. Lots of room for improvement, especially by the father thereof, User:Mouse7mouse9. A big thanks. Cuzkatzimhut (talk) 15:34, 30 December 2013 (UTC)

The new section: Rotation case
The new section Matrix exponential seems to be overly convoluted, and is unreferenced; I suspect that it is OR. It also semi-restricts itself to three dimensions, and uses perpendicular unit vectors unnecessarily. It also assumes Euclidean space without explicitly saying so – yet matrices and rotations apply in many contexts in which one does not (and in cases cannot) have an orthonormal basis. —Quondum 15:50, 25 September 2014 (UTC)
 * A simple rotation (which can be defined from a pair of vectors) in any number of dimensions should be expressible as a simple exponential $$R(\theta)=\exp(\theta G)$$ – no need for projections.
 * The "main article" Rodrigues' rotation formula is hardly appropriate. It is a way of handling rotation only in three dimensions involving quantities such as an axis of rotation that is not mentioned in this section, and is in a about sidestepping matrix exponentiation via vector algebra and trigonometry; thus hardly the subject of this article or this section.
 * The restriction to perpendicular unit vectors is unnecessary, though it does perhaps simplify normalization if one wishes to have only pure rotation (no dilations). A more general treatment (i.e. allowing any pair of vectors) should be hardly more complex.

 * I only cleaned it up and copyedited it, but it is anodyne and obvious... I have not seen the Weyl book adduced by the original author, but I trusted it is OK. In any case, the final link I adduced at the bottom of the section covers the same material, less abstractly. It cannot be OR, any more than the Pythagorean theorem can be!
 * The perpendicular vectors ensure the Projection operators work fine. You are right this is basically plane rotations embedded in an n-dim space, and all extra dimensions behave like 1 in the Rodrigues formula. Indeed, it deals with Euclidean space, and that fact should be thrown in, incidentally.
 * Indeed, any rotation is matrix exponential, as it is here, but the point is this exponential is explicitly calculated through the projection ops, which is why I suspect this subsection was put in this week. Perhaps the original author can argue for it. In any case calculation leads to the Rodrigues formula, and for the same reasons it has that form: the structure is evidently 3-dim.
 * I am befuddled by proposed avoidance of the Rodrigues formula. The answer is the Rodrigues formula, in its modern, matrix form, and it is, of course living effectively in 3d, as any direction perpendicular to the ab-plane is treated the same, and no rotations around different axes are composed. I think insistence on vector language for the Rodrigues is excessively antique: the Rodrigues formula is effectively the rotation matrix expressed as a quadratic polynomial in 3×3 spin matrices! The algebra is the same for n×n matrices, here, as asserted, and very different indeed from the generalization of the Rodrigues formula for higher dimensional representations of SO(3). I suspect any further discussion of this in the section cannot fail to confuse the reader, but if you had a pithy summary of the explanation, that would not be out of place.
 * I am not sure what you are proposing... This is not a deep geometrical point, it is the expansion of the exponential in powers of G with the obvious rearrangement of the self-evident structure. If the link of the last line is clearer, perhaps you might undertake to blend the two treatments. Cuzkatzimhut (talk) 17:01, 25 September 2014 (UTC)

PS on the clarification requested for the diagonalizable case: the exponential is but a sum of powers: a similarity transform of a power and of a sum of powers is but the same sum of the same powers of the similarity transform of the argument. Cuzkatzimhut (talk) 19:12, 25 September 2014 (UTC)

PSPS to User:Jbergquist: it is best to leave questions here, rather than in comments in the text code. I had the same reaction myself, early on, yesterday, i.e. that maybe this belongs elsewhere, in one of the handful of articles on rotations. However, what is illustrated here is purely the technical evaluation of the matrix exponential, and it is perfectly position to segue from the previous, projector section, to the following one. This one illustrates the cyclic structure of G2=−I, i.e. that G behaves like i, superposed on the projector structure. It is a memorable derivation of the Rodrigues rotation formula, and illustrates its projective character. Cuzkatzimhut (talk) 20:44, 25 September 2014 (UTC)

 * It is probably best to keep the section as simple as possible. The definition of G comes from the derivation of R for a rotation of a vector in an arbitrary n-plane defined by a and b. Any two normal unit vectors in the plane will do. You get the same rotation matrix. A summary derivation of the exponential formula starting with the exponential expansion could be included. We could use a good reference for that. R(θ) = I cos(θ) + G sin(θ) only works for vectors in the plane so the projection operators are needed to split an arbitrary vector into two parts for the more general rotation matrix. The exponential expression in the definition of the rotation comes from a modification of plane rotation taking into account the Pab. My presentation here is somewhat backwards. The assumption is that the generator and rotation are derived elsewhere. The location indicated by the generator pipe would be a good place. I derived the results from scratch a short while ago and a simple method for finding a rotation matrix certainly deserves backup. I would be interested in hearing if you find this published elsewhere any time in the last 100 years or so. Sometimes mathematicians tend to be rather cryptic. --Jbergquist (talk) 22:14, 25 September 2014 (UTC)

I am in full agreement that this section should be kept breezy and short. It is self-explanatory to me, but then I am not the intended audience. I don't think more refs are needed, since, up to obvious changes of notation, it is a re-write of both WP links now provided at the end of the section. I think these links are enough. The subsection looks stable and informative.

To be more explicit as to why this is "essentially" SO(3): As it stands, of course it is an SO(2) rotation in n-dim space, i.e. generators of 2x2 blocks of a trivial nxn matrix. A further rotation, if in the same 2x2 block, would lead to trivial SO(2) composition. If in a completely disjoint 2x2 block, to an SO(2)xSO(2) structure. Only if it is in a 2x2 block overlapping this one, i.e. in a space involving, e.g. b and a different orthogonal c, but not a, do we have a nontrivial structure composing such rotations, to wit an SO(3), which makes the formula useful. (B.t.w., I would use the terminology "n-space" and "2-plane" embeddings in it, never "n-plane", but that's just personal taste.) You might take a look at Rotation_matrix to see what I mean. Cuzkatzimhut (talk) 22:59, 25 September 2014 (UTC) Cuzkatzimhut (talk) 22:59, 25 September 2014 (UTC)

 * Heh-heh. Do I see OR in progress?  Not that I'm against it, but don't ignore me when I point out missed opportunities for simplification. I'm liable to make certain statements that challenge some of the complexities and assumptions here.
 * Browsing the source (Weyl) did not show me a derivation as given using projections.
 * There is no "essentially SO(3)" about rotations in any number of dimensions, once the plane(s) of rotation are specified. Once a plane of rotation is fixed (only one plane in our example), it is "essentially SO(2)".
 * While projection is a neat way of deriving the result, it should produce a universal result in any number of dimensions: $R(θ) = exp(θG)$. This is easily seen: if the rotation is continuous, it can be subdivided into progressively smaller identical rotations that compose to the larger rotation.  Basically, given that the rotation matrix in any number of dimensions is $R(θ)$, we can find a matrix $G$ such that $R(θ) = lim_{n→∞}(I + θG/n)^{n} = exp(θG)$.
 * I'd actually be in favour of removing the projections in order to keep it short, simply stating the result, giving $G$ in terms of two vectors, though it would be just as easy to expand this to any number of mutually orthogonal vectors, paired to define planes, giving a general rotation in $n$ dimensions. —Quondum 00:16, 26 September 2014 (UTC)

I did not ignore you--- I did not imagine the author considered is as OR. Of course, as I keep indicating, it is not, as per the two WP pages linked at the end of the subsection. It is SO(3), though. You see the collapse to a trivial SO(2) in n=2, which does not tell anyone anything interesting, since it is the dumb rotation of a 2-vector not the Rodrigues formula. But the interesting part of the formula, as I indicate above, is precisely the SO(3), 3-plet representation (spin 1) part, hence the Rodrigues formula, precisely due to the projection. So this is the instructive part of the subsection. It is, of course, given in Axis–angle_representation but here it is produced compactly and directly. I did not appreciate your middle bullet: I cannot see how it has any bearing on the result, by which I mean an expansion of the exponential to be spanned by I, G and G2. This will not be the case for other reps of SO(3), such as spin 3/2, or generic SO(n) transformations, since higher powers of G would then survive in the expansion. My sense is that, as it stands, the subsection makes sense, and throwing the baby with the bathwater might exactly be not be salutary. Cuzkatzimhut (talk) 00:51, 26 September 2014 (UTC)
 * Sorry, I came across all wrong there. I did not intend to suggest that you did ignore me, only only that since I'm shooting from the hip, as it were, I may make several errors, but that there may be some thoughts worth sifting out of what I say. ;-)  —Quondum 04:37, 26 September 2014 (UTC)

 * On checking the expression for R(θ) I see that it is equal to eGθ. Thanks for pointing that out. I'll make the simplifications. --Jbergquist (talk) 01:54, 26 September 2014 (UTC)
 * There's a similar version of eGθ involving hyperbolic functions. The reference is Bjorken and Drell, Relativistic Quantum Mechanics, 1964, McGraw-Hill, p.22. --Jbergquist (talk) 02:25, 26 September 2014 (UTC)

 * I think a worked example may be needed. How about this?

 * $$\begin{align}

& a=\left( \begin{matrix}  1  \\   0  \\   0  \\ \end{matrix} \right)\quad b=\left( \begin{matrix}   0  \\   1  \\   0  \\ \end{matrix} \right) \\ & I=\left( \begin{matrix}  1 & 0 & 0  \\   0 & 1 & 0  \\   0 & 0 & 1  \\ \end{matrix} \right)\quad G=\left( \begin{matrix}   0 & -1 & 0  \\   1 & 0 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right)\quad {{G}^{2}}=\left( \begin{matrix}   -1 & 0 & 0  \\   0 & -1 & 0  \\   0 & 0 & 0  \\ \end{matrix} \right) \\ & \\  & \theta =30^\circ \quad \Rightarrow \quad {{e}^{G\theta }}=\left( \begin{matrix}   0.866 & 0.5 & 0  \\   0.5 & 0.866 & 0  \\   0 & 0 & 1  \\ \end{matrix} \right) \\ \end{align}$$

--Jbergquist (talk) 03:40, 26 September 2014 (UTC)

 * Looking good. I still need to wrap my head around some of this. I've been transferring most of my thoughts from geometric algebra, and some of my inferences might not hold.  The translation from a rotor in GA to a rotation matrix is tricky. In particular, the equivalent of $G$ in GA, being $a ∧ b$, is generally invertible in the algebra in any dimension, unlike the matrix $G$. Something new for me. —Quondum 04:37, 26 September 2014 (UTC)
 * I was running checks on the formulas using Mathcad which doesn't check the mathematics behind the formulas. That's how the projection operators came to be involved and why the reduction to an exponential wasn't complete. I was bothered by that since the determinant |R(θ)| = 1 but I knew that the magnitude of the vector wasn't changed by the formulas posted. Mathtype is useful for doing examples but there's a sign missing above which I caught in time. --Jbergquist (talk) 04:51, 26 September 2014 (UTC)

 * I think I got off on the wrong foot by putting the cart before the horse which is why I decided to place the derivation of G in the generator article. I think initially we were working at cross purposes but I needed so show that the expression derived for R was related to an exponential function. That's why I chose to do it in this article and the projection operator P was needed for readers of the other article to connect R with eGθ. For n-dimensions matrix methods produce simpler results that tensors for rotations. Part of my motivation for doing all this stems from a lecture in the Mathematics department of a local university about 30 years ago in which it was mentioned that one of their students had claimed that you couldn't define an angle in four dimensions. (I collect Dover books too.) --Jbergquist (talk) 00:17, 28 September 2014 (UTC)
 * Do you have a particular interest in the claim? (It may be a confusion: assuming Euclidean space an angle between two vectors is defined in any number of dimensions, angles of rotation in 4 d are a well-defined composite of two independent angles in two fully orthogonal 2-d subspaces, and angles between planes are messy.) —Quondum 05:08, 28 September 2014 (UTC)

I tweaked the presentation to start from R=exp(θG) in the longer eqn, linking to the underlying point of the subsection, the exponential map involved in rotations, by wikilinking to the relevant section of rotation matrix; perhaps you'd choose to call that the "main article". And reorganized the expansion in powers of G to highlight the Rodrigues formula it is basically illustrating, without delving in a surfeit of notation.

My sense is that, here, no geometry or Lie theory is warranted---they are more than adequately covered in two handfuls (!) of relevant articles---but, instead, a compact illustration of evaluating simple matrix exponentials and the structure of the answers, and their relation to the evaluations of the more elaborate, non degenerate, real deal, matrix exponentials in the rest of the article. I sense that this has been achieved: Things appear stable now, but too much additional talk about rotations instead of just matrix exponentials might be out of place here. Cuzkatzimhut (talk) 10:37, 26 September 2014 (UTC)

 * I have no objection to your simplifications. We don't really need to talk about n-space at the beginning of the section. It's probably better left as a footnote at the end so we don't loose the majority of the readers. It's left out in the proof in the generator section. One should note that eG1θ1 and eG2θ2 don't commute so the order of their multiplication is significant when the two G's are not the same. This would be important for compound rotations where the matrix exponentials can't be added to produce a combined matrix exponential. --Jbergquist (talk) 21:06, 26 September 2014 (UTC)

I see I've unwittingly undone an intentional edit. Use of extra blank lines to partition text is a symptom of the need for more structure. In this case, the examples are pretty voluminous compared to the rest of the section. In this case, perhaps we need to place the examples in a collapsed expandable box? —Quondum 02:51, 27 September 2014 (UTC)

I've reduced the mention of higher dimensions to a footnote. The derivation of the expression for eGθ just depends on the antisymmetric properties of G which leads to the two dimensional properties of P and we can leave it at that. There's less clash present now. As is, article should be comprehensible to anyone with an elementary understanding of matrix algebra. --Jbergquist (talk) 20:43, 27 September 2014 (UTC)

Linear differential equation examples
We have a problem here. Neither of the recent versions satisfies $e^{0A} = I$. I don't have time to solve the equation at the moment. — Arthur Rubin (talk) 20:44, 23 October 2014 (UTC)
 * At least for the equation given right now (where the disputed tag is placed) it does satisfy $$e^{0A} = I$$. That doesn't necessarily mean this is the correct equation (I don't have the time to solve the equation at the moment either) but this quick complaint does not seem accurate to me. Zfeinst (talk) 21:38, 23 October 2014 (UTC)
 * I have now checked it, using the method of section 2.7: exp (tA) = B exp2t + Ct exp2t +D exp4t; it follows that B= A−A2/4,  D = 1−A +A2/4,  and C= −4 +3A −  A2/2. It all checks in the disputed formula, so I believe you should take down the unwarranted tag. Cuzkatzimhut (talk) 13:59, 24 October 2014 (UTC)

Matrix-matrix exponential
Why need the base be normal? As noted in matrix logarithm, any non-singular complex matrix has a logarithm.

— Although the paper states in this way, I actually agree with you that for the base matrix, just being non-singular suffice.Xuancong (talk) 03:17, 5 September 2015 (UTC)

Also, I would write $X^{Y}$ is similar to $^{Y}X$, rather than that they have the same eigenvalues. — Arthur Rubin (talk) 18:20, 4 September 2015 (UTC)

— Sure, you can change to 'similar' that if you wish.Xuancong (talk) 03:17, 5 September 2015 (UTC)

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Nomenclature
Please include explanation/definition/link for $$\oplus$$ and anything else not obvious to undergraduate mathematics students. —DIV (120.17.110.164 (talk) 12:48, 27 March 2018 (UTC))
 * Direct sum not obvious to math undergrads? Rather alarming. Cuzkatzimhut (talk) 21:08, 27 March 2018 (UTC)

Matrix-vector product
For iterative solvers (e.g. GMRES) one does not need to compute explicitly the matrix exponential $$e^M$$ but rather the action of the exponential on a vector: $$e^M \vec{b}$$. One can use Krylov subspace methods to approximate the action which is for large sparse systems more efficient.

Linking to existing packages such as http://www.maths.uq.edu.au/expokit/ would be helpful. HerrHartmuth (talk) 09:04, 6 December 2019 (UTC)

title: Talk:Matrix exponential


text:
User:Moomint

This is me.

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title: User:Moomint