Change of basis in Mathematica












2














I'm trying to set up a change of basis matrix from the base V -> W with the RowReduce function that's built into Mathematica. I am then supposed to verify the result by taking the product of the change of basis matrix and multiplying it with the co-ordinate vector for x. You can see the Bases and the vector x below.



So, my question is, how do I setup the change of basis matrix and then verify it, with just the RowReduce function? Thanks in advance.



V = {{1, 3}, {4, 6}}

W = {{4, 6}, {2, 5}} 



x = {6, 6}





matrix mathematical-optimization linear-algebra







share|improve this question
























  • Pleeeease. Don't use MatrixForm in computations. See here why. Moreover, it would be appreciated if you would post copyable Mathematica code instead of images.
    – Henrik Schumacher
    Dec 27 '18 at 13:29






  • 1




    @HenrikSchumacher I'll edit the post then :)
    – wznd
    Dec 27 '18 at 13:31
















2














I'm trying to set up a change of basis matrix from the base V -> W with the RowReduce function that's built into Mathematica. I am then supposed to verify the result by taking the product of the change of basis matrix and multiplying it with the co-ordinate vector for x. You can see the Bases and the vector x below.



So, my question is, how do I setup the change of basis matrix and then verify it, with just the RowReduce function? Thanks in advance.



V = {{1, 3}, {4, 6}}

W = {{4, 6}, {2, 5}} 



x = {6, 6}





matrix mathematical-optimization linear-algebra







share|improve this question
























  • Pleeeease. Don't use MatrixForm in computations. See here why. Moreover, it would be appreciated if you would post copyable Mathematica code instead of images.
    – Henrik Schumacher
    Dec 27 '18 at 13:29






  • 1




    @HenrikSchumacher I'll edit the post then :)
    – wznd
    Dec 27 '18 at 13:31














2












2








2







I'm trying to set up a change of basis matrix from the base V -> W with the RowReduce function that's built into Mathematica. I am then supposed to verify the result by taking the product of the change of basis matrix and multiplying it with the co-ordinate vector for x. You can see the Bases and the vector x below.



So, my question is, how do I setup the change of basis matrix and then verify it, with just the RowReduce function? Thanks in advance.



V = {{1, 3}, {4, 6}}

W = {{4, 6}, {2, 5}} 



x = {6, 6}





matrix mathematical-optimization linear-algebra







share|improve this question















I'm trying to set up a change of basis matrix from the base V -> W with the RowReduce function that's built into Mathematica. I am then supposed to verify the result by taking the product of the change of basis matrix and multiplying it with the co-ordinate vector for x. You can see the Bases and the vector x below.



So, my question is, how do I setup the change of basis matrix and then verify it, with just the RowReduce function? Thanks in advance.



V = {{1, 3}, {4, 6}}

W = {{4, 6}, {2, 5}} 



x = {6, 6}





matrix mathematical-optimization linear-algebra




matrix mathematical-optimization linear-algebra






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 27 '18 at 13:33

























asked Dec 27 '18 at 13:11









wznd

315




315












  • Pleeeease. Don't use MatrixForm in computations. See here why. Moreover, it would be appreciated if you would post copyable Mathematica code instead of images.
    – Henrik Schumacher
    Dec 27 '18 at 13:29






  • 1




    @HenrikSchumacher I'll edit the post then :)
    – wznd
    Dec 27 '18 at 13:31


















  • Pleeeease. Don't use MatrixForm in computations. See here why. Moreover, it would be appreciated if you would post copyable Mathematica code instead of images.
    – Henrik Schumacher
    Dec 27 '18 at 13:29






  • 1




    @HenrikSchumacher I'll edit the post then :)
    – wznd
    Dec 27 '18 at 13:31
















Pleeeease. Don't use MatrixForm in computations. See here why. Moreover, it would be appreciated if you would post copyable Mathematica code instead of images.
– Henrik Schumacher
Dec 27 '18 at 13:29




Pleeeease. Don't use MatrixForm in computations. See here why. Moreover, it would be appreciated if you would post copyable Mathematica code instead of images.
– Henrik Schumacher
Dec 27 '18 at 13:29




1




1




@HenrikSchumacher I'll edit the post then :)
– wznd
Dec 27 '18 at 13:31




@HenrikSchumacher I'll edit the post then :)
– wznd
Dec 27 '18 at 13:31










1 Answer
1






active

oldest

votes


















4














This might give you an idea... I merge V and W into one matrix with ArrayFlatten and apply Gaussian elimination by RowReduce.



V = {{1, 3}, {4, 6}};

W = {{4, 6}, {2, 5}};

B = RowReduce[ArrayFlatten[{{V, W}}]][[All, 3 ;;]]

V.B == W



{{-3, -(7/2)}, {7/3, 19/6}}



True







share|improve this answer























  • Ah, smart. Wasn't thinking about how you could compare the two functions that easily. Just one question though. What does the ArrayFlatten, and the [[All, 3 ;;]] commands mean? Thank you very much for your answer.
    – wznd
    Dec 27 '18 at 13:34










  • ArrayFlatten can merge block matrices to a single matrix. And A[[All, 3 ;;]] reads off the columns 3 to Dimensions[A][[2]] of a matrix A. See the documentation of Part and Span for details.
    – Henrik Schumacher
    Dec 27 '18 at 13:36










  • Ah, now I get it. Thank you so much. If i were to calculate the change of basis matrix from W -> V, using the Inverse function, would I first take the inverse of W, and then multiply W^1 * V?
    – wznd
    Dec 27 '18 at 13:51










  • You're welcome. I'd rather use V.Inverse[W] or W.Inverse[V] depending on which direction you would like to have.
    – Henrik Schumacher
    Dec 27 '18 at 14:04










  • Well, I'd like to know the change of basis matrix from the base W -> V, so how would the function look like?
    – wznd
    Dec 27 '18 at 14:16











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f188455%2fchange-of-basis-in-mathematica%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














This might give you an idea... I merge V and W into one matrix with ArrayFlatten and apply Gaussian elimination by RowReduce.



V = {{1, 3}, {4, 6}};

W = {{4, 6}, {2, 5}};

B = RowReduce[ArrayFlatten[{{V, W}}]][[All, 3 ;;]]

V.B == W



{{-3, -(7/2)}, {7/3, 19/6}}



True







share|improve this answer























  • Ah, smart. Wasn't thinking about how you could compare the two functions that easily. Just one question though. What does the ArrayFlatten, and the [[All, 3 ;;]] commands mean? Thank you very much for your answer.
    – wznd
    Dec 27 '18 at 13:34










  • ArrayFlatten can merge block matrices to a single matrix. And A[[All, 3 ;;]] reads off the columns 3 to Dimensions[A][[2]] of a matrix A. See the documentation of Part and Span for details.
    – Henrik Schumacher
    Dec 27 '18 at 13:36










  • Ah, now I get it. Thank you so much. If i were to calculate the change of basis matrix from W -> V, using the Inverse function, would I first take the inverse of W, and then multiply W^1 * V?
    – wznd
    Dec 27 '18 at 13:51










  • You're welcome. I'd rather use V.Inverse[W] or W.Inverse[V] depending on which direction you would like to have.
    – Henrik Schumacher
    Dec 27 '18 at 14:04










  • Well, I'd like to know the change of basis matrix from the base W -> V, so how would the function look like?
    – wznd
    Dec 27 '18 at 14:16
















4














This might give you an idea... I merge V and W into one matrix with ArrayFlatten and apply Gaussian elimination by RowReduce.



V = {{1, 3}, {4, 6}};

W = {{4, 6}, {2, 5}};

B = RowReduce[ArrayFlatten[{{V, W}}]][[All, 3 ;;]]

V.B == W



{{-3, -(7/2)}, {7/3, 19/6}}



True







share|improve this answer























  • Ah, smart. Wasn't thinking about how you could compare the two functions that easily. Just one question though. What does the ArrayFlatten, and the [[All, 3 ;;]] commands mean? Thank you very much for your answer.
    – wznd
    Dec 27 '18 at 13:34










  • ArrayFlatten can merge block matrices to a single matrix. And A[[All, 3 ;;]] reads off the columns 3 to Dimensions[A][[2]] of a matrix A. See the documentation of Part and Span for details.
    – Henrik Schumacher
    Dec 27 '18 at 13:36










  • Ah, now I get it. Thank you so much. If i were to calculate the change of basis matrix from W -> V, using the Inverse function, would I first take the inverse of W, and then multiply W^1 * V?
    – wznd
    Dec 27 '18 at 13:51










  • You're welcome. I'd rather use V.Inverse[W] or W.Inverse[V] depending on which direction you would like to have.
    – Henrik Schumacher
    Dec 27 '18 at 14:04










  • Well, I'd like to know the change of basis matrix from the base W -> V, so how would the function look like?
    – wznd
    Dec 27 '18 at 14:16














4












4








4






This might give you an idea... I merge V and W into one matrix with ArrayFlatten and apply Gaussian elimination by RowReduce.



V = {{1, 3}, {4, 6}};

W = {{4, 6}, {2, 5}};

B = RowReduce[ArrayFlatten[{{V, W}}]][[All, 3 ;;]]

V.B == W



{{-3, -(7/2)}, {7/3, 19/6}}



True







share|improve this answer














This might give you an idea... I merge V and W into one matrix with ArrayFlatten and apply Gaussian elimination by RowReduce.



V = {{1, 3}, {4, 6}};

W = {{4, 6}, {2, 5}};

B = RowReduce[ArrayFlatten[{{V, W}}]][[All, 3 ;;]]

V.B == W



{{-3, -(7/2)}, {7/3, 19/6}}



True








share|improve this answer














share|improve this answer



share|improve this answer








edited Dec 27 '18 at 13:36

























answered Dec 27 '18 at 13:32









Henrik Schumacher

48.8k467139




48.8k467139












  • Ah, smart. Wasn't thinking about how you could compare the two functions that easily. Just one question though. What does the ArrayFlatten, and the [[All, 3 ;;]] commands mean? Thank you very much for your answer.
    – wznd
    Dec 27 '18 at 13:34










  • ArrayFlatten can merge block matrices to a single matrix. And A[[All, 3 ;;]] reads off the columns 3 to Dimensions[A][[2]] of a matrix A. See the documentation of Part and Span for details.
    – Henrik Schumacher
    Dec 27 '18 at 13:36










  • Ah, now I get it. Thank you so much. If i were to calculate the change of basis matrix from W -> V, using the Inverse function, would I first take the inverse of W, and then multiply W^1 * V?
    – wznd
    Dec 27 '18 at 13:51










  • You're welcome. I'd rather use V.Inverse[W] or W.Inverse[V] depending on which direction you would like to have.
    – Henrik Schumacher
    Dec 27 '18 at 14:04










  • Well, I'd like to know the change of basis matrix from the base W -> V, so how would the function look like?
    – wznd
    Dec 27 '18 at 14:16


















  • Ah, smart. Wasn't thinking about how you could compare the two functions that easily. Just one question though. What does the ArrayFlatten, and the [[All, 3 ;;]] commands mean? Thank you very much for your answer.
    – wznd
    Dec 27 '18 at 13:34










  • ArrayFlatten can merge block matrices to a single matrix. And A[[All, 3 ;;]] reads off the columns 3 to Dimensions[A][[2]] of a matrix A. See the documentation of Part and Span for details.
    – Henrik Schumacher
    Dec 27 '18 at 13:36










  • Ah, now I get it. Thank you so much. If i were to calculate the change of basis matrix from W -> V, using the Inverse function, would I first take the inverse of W, and then multiply W^1 * V?
    – wznd
    Dec 27 '18 at 13:51










  • You're welcome. I'd rather use V.Inverse[W] or W.Inverse[V] depending on which direction you would like to have.
    – Henrik Schumacher
    Dec 27 '18 at 14:04










  • Well, I'd like to know the change of basis matrix from the base W -> V, so how would the function look like?
    – wznd
    Dec 27 '18 at 14:16
















Ah, smart. Wasn't thinking about how you could compare the two functions that easily. Just one question though. What does the ArrayFlatten, and the [[All, 3 ;;]] commands mean? Thank you very much for your answer.
– wznd
Dec 27 '18 at 13:34




Ah, smart. Wasn't thinking about how you could compare the two functions that easily. Just one question though. What does the ArrayFlatten, and the [[All, 3 ;;]] commands mean? Thank you very much for your answer.
– wznd
Dec 27 '18 at 13:34












ArrayFlatten can merge block matrices to a single matrix. And A[[All, 3 ;;]] reads off the columns 3 to Dimensions[A][[2]] of a matrix A. See the documentation of Part and Span for details.
– Henrik Schumacher
Dec 27 '18 at 13:36




ArrayFlatten can merge block matrices to a single matrix. And A[[All, 3 ;;]] reads off the columns 3 to Dimensions[A][[2]] of a matrix A. See the documentation of Part and Span for details.
– Henrik Schumacher
Dec 27 '18 at 13:36












Ah, now I get it. Thank you so much. If i were to calculate the change of basis matrix from W -> V, using the Inverse function, would I first take the inverse of W, and then multiply W^1 * V?
– wznd
Dec 27 '18 at 13:51




Ah, now I get it. Thank you so much. If i were to calculate the change of basis matrix from W -> V, using the Inverse function, would I first take the inverse of W, and then multiply W^1 * V?
– wznd
Dec 27 '18 at 13:51












You're welcome. I'd rather use V.Inverse[W] or W.Inverse[V] depending on which direction you would like to have.
– Henrik Schumacher
Dec 27 '18 at 14:04




You're welcome. I'd rather use V.Inverse[W] or W.Inverse[V] depending on which direction you would like to have.
– Henrik Schumacher
Dec 27 '18 at 14:04












Well, I'd like to know the change of basis matrix from the base W -> V, so how would the function look like?
– wznd
Dec 27 '18 at 14:16




Well, I'd like to know the change of basis matrix from the base W -> V, so how would the function look like?
– wznd
Dec 27 '18 at 14:16


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematica Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f188455%2fchange-of-basis-in-mathematica%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Сан-Квентин

Image
Сан-Квентин Местоположение Сан-Квентин, Калифорния Координаты 37°56′13″ с. ш. 122°29′20″ з. д. H G Я O L Текущий статус Действует Режим безопасности минимальный и максимальный Количество мест 3 302 Открытие Июнь 1852 Находится в ведомстве California Department of Corrections and Rehabilitation Начальник Mike Martel, начальник  Сан-Квентин на Викискладе Сан-Квентин (англ.  San Quentin ) — тюрьма, располагающаяся на 432 акрах (около 1,7 км²), на мысе Сан-Квентин, в округе Марин, штат Калифорния, США. Тюрьма штата Калифорния Сан-Квентин была открыта в июле 1852 года и является старейшей в штате. Она была построена заключёнными, которые проживали в течение строительства на тюремном судне Вабан. В Сан-Квентине отбывали наказание и мужчины, и женщины до 1934 года, когда построили женскую тюрьму в Техачапи и туда перевели всех заключённых женщин. В Сан-Квентине приводятся в исполнение смертные казни, для этого имеется газовая камера, но...
Read more

8-я гвардейская общевойсковая армия

Image
8-я гвардейская общевойсковая ордена Ленина армия Награды: Войска: РККА → Советская армия → СВ России Род войск: Формирование: 1943 Расформирование (преобразование): 1992 Предшественник: 62-я армия (1942—43) → 8-я гвардейская общевойсковая армия (1943—92) → 8-й гвардейский армейский корпус (1992—98) Боевой путь Изюм-Барвенковская операция Донбасская операция Запорожская операция Битва за Днепр Днепропетровская операция Березнеговато-Снигирёвская операция Одесская операция Люблин-Брестская операция Варшавско-Познанская операция Восточно-Померанская операция Берлинская наступательная операция У этого термина существуют и другие значения, см. 8-я армия. Фотография Гвардейского Красного Знамени (лицевая сторона) 8-й гвардейской армии , образца 1943 года. Фотография Гвардейского Красного Знамени (оборотная сторона) 8-й гвардейской армии , образца 1943 года. 8-я гвардейская общевойсковая ордена Ленина армия  — гвардейское фор...
Read more

Алькесар

Image
Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове...
Read more