Mdthbs

Why is `abs()` implemented differently?

Roll Dice to get a random number between 1 and 150

What word means "to make something obsolete"?

Pressure inside an infinite ocean?

A non-technological, repeating, phenomenon in the sky, holding its position in the sky for hours

To customize a predefined symbol with different colors

60s (or earlier) SF short story with FTL Travel using electron psychology aka addiclenendar technology

Would glacier 'trees' be plausible?

When and why did journal article titles become descriptive, rather than creatively allusive?

Is there a legal ground for stripping the UK of its UN Veto if Scotland and/or N.Ireland split from the UK?

Why is B♯ higher than C♭ in 31-ET?

Why was the battle set up *outside* Winterfell?

Quoting Yourself

How can I close a gap between my fence and my neighbor's that's on his side of the property line?

Is induction neccessary for proving that every injective mapping of a finite set into itself is a mapping onto itself?

What happens to matryoshka Mordenkainen's Magnificent Mansions?

What happens if I start too many background jobs?

Did we get closer to another plane than we were supposed to, or was the pilot just protecting our delicate sensibilities?

For a benzene shown in a skeletal structure, what does a substituent to the center of the ring mean?

Besides the up and down quark, what other quarks are present in daily matter around us?

In Avengers 1, why does Thanos need Loki?

Is it cheaper to drop cargo than to land it?

Can fracking help reduce CO2?

Was Unix ever a single-user OS?

Transpose of product of matrices

Assistance with proof of $(AB)^T=B^T A^T$What is the geometric interpretation of the transpose?If $A$ is a real skew-symmetric matrix, why is $(I-A)(I+A)^{-1}$ orthogonal?Product of nilpotent matrices invertibleSymplectic matrices transposehow to count the number of integer matrices with inverse equal to transpose?Transpose applications and matrices propertiesgradient of product of matricesMatrices similar to their inverse or transposehow do you transpose more than 2 matricesTranspose of matrices

3

2

$begingroup$

How do you prove the following fact about the transpose of a product of matrices? Also can you give some intuition as to why it is so.

$(AB)^T = B^TA^T$

linear-algebra

share|cite|improve this question

asked 2 hours ago

feli andrésfeli andrés

161

New contributor

feli andrés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  While I have seen this asked many time before on Math.SE, I have not been able to find a link to a duplicate. For now, you may find this article helpful.
  $endgroup$
  – Brian
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We need a good answer to this question, and in this case Ted Shifrin has answered, so I hope this question is not closed.
  $endgroup$
  – littleO
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note: the same fact holds for matrix inverses
  $endgroup$
  – J. W. Tanner
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Assistance with proof of $(AB)^T=B^T A^T$
  $endgroup$
  – Brian
  18 mins ago
  

  
  
  
  

add a comment | 

3

2

$begingroup$

How do you prove the following fact about the transpose of a product of matrices? Also can you give some intuition as to why it is so.

$(AB)^T = B^TA^T$

linear-algebra

share|cite|improve this question

asked 2 hours ago

feli andrésfeli andrés

161

New contributor

feli andrés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  While I have seen this asked many time before on Math.SE, I have not been able to find a link to a duplicate. For now, you may find this article helpful.
  $endgroup$
  – Brian
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We need a good answer to this question, and in this case Ted Shifrin has answered, so I hope this question is not closed.
  $endgroup$
  – littleO
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note: the same fact holds for matrix inverses
  $endgroup$
  – J. W. Tanner
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Assistance with proof of $(AB)^T=B^T A^T$
  $endgroup$
  – Brian
  18 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

How do you prove the following fact about the transpose of a product of matrices? Also can you give some intuition as to why it is so.

$(AB)^T = B^TA^T$

linear-algebra

share|cite|improve this question

asked 2 hours ago

feli andrésfeli andrés

161

New contributor

feli andrés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked 2 hours ago

feli andrésfeli andrés

161

New contributor

feli andrés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 2 hours ago

feli andrésfeli andrés

161

New contributor

feli andrés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 hours ago

feli andrésfeli andrés

161

New contributor

feli andrés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 hours ago

feli andrésfeli andrés

161

3 Answers
3

active

oldest

votes

3

$begingroup$

Here's an alternative argument. The main importance of the transpose (and this in fact defines it) is the formula
$$Axcdot y = xcdot A^top y.$$
(If $A$ is $mtimes n$, then $xin Bbb R^n$, $yinBbb R^m$, the left dot product is in $Bbb R^m$ and the right dot product is in $Bbb R^n$.)

Now note that
$$(AB)xcdot y = A(Bx)cdot y = Bxcdot A^top y = xcdot B^top(A^top y) = xcdot (B^top A^top)y.$$
Thus, $(AB)^top = B^top A^top$.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Ted ShifrinTed Shifrin

65.4k44792

$endgroup$

add a comment | 

2

$begingroup$

When you multiply $A$ and $B$, you are taking the dot product of each ROW of $A$ and each COLUMN of $B$.

The resulting dimension is $A_{#col}times B_{#row}$, and after transposing, you have $B_{#row}times A_{#col}$.

When you multiply $B^T$ and $A^T$, you take the dot product of each row of $B^T$ (column of B) and column of $A^T$, or row of $A$.

Your resulting dimension is $B^T_{#col}times A^T_{#row}$ which is just $B_{#row}times A_{#col}$

This formula ensures that each entry is correct, and that the dimensions are identical.

share|cite|improve this answer

answered 2 hours ago

Saketh MalyalaSaketh Malyala

7,8881535

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just to make up some notation to express your first + third sentence: let $operatorname{row}_i(M)$ and $operatorname{col}_j(M)$ denote the $i^{text{th}}$ row and $j^{text{th}}$ column of $M$, respectively. Then $(AB)_{ij} = operatorname{row}_i(A) cdot operatorname{col}_j(B)$, and $(B^T A^T)_{ji} = operatorname{row}_j(B^T) cdot operatorname{col}_i(A^T) = operatorname{col}_j(B) cdot operatorname{row}_i(A)$, so $(AB)_{ij} = (B^T A^T)_{ji}$.
  $endgroup$
  – Misha Lavrov
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

If you know about dual spaces and maps, a conceptual proof can be obtained by observing that $A^T$ corresponds to the dual map of $A$ and that taking the dual is contravariant with respect to composition. That is, $(T circ S)^* = S^* circ T^*$.

share|cite|improve this answer

answered 2 hours ago

Aniruddh AgarwalAniruddh Agarwal

706

New contributor

Aniruddh Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

feli andrés is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3208939%2ftranspose-of-product-of-matrices%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

Here's an alternative argument. The main importance of the transpose (and this in fact defines it) is the formula
$$Axcdot y = xcdot A^top y.$$
(If $A$ is $mtimes n$, then $xin Bbb R^n$, $yinBbb R^m$, the left dot product is in $Bbb R^m$ and the right dot product is in $Bbb R^n$.)

Now note that
$$(AB)xcdot y = A(Bx)cdot y = Bxcdot A^top y = xcdot B^top(A^top y) = xcdot (B^top A^top)y.$$
Thus, $(AB)^top = B^top A^top$.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Ted ShifrinTed Shifrin

65.4k44792

$endgroup$

add a comment | 

3

$begingroup$

Here's an alternative argument. The main importance of the transpose (and this in fact defines it) is the formula
$$Axcdot y = xcdot A^top y.$$
(If $A$ is $mtimes n$, then $xin Bbb R^n$, $yinBbb R^m$, the left dot product is in $Bbb R^m$ and the right dot product is in $Bbb R^n$.)

Now note that
$$(AB)xcdot y = A(Bx)cdot y = Bxcdot A^top y = xcdot B^top(A^top y) = xcdot (B^top A^top)y.$$
Thus, $(AB)^top = B^top A^top$.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Ted ShifrinTed Shifrin

65.4k44792

$endgroup$

add a comment | 

$begingroup$

Here's an alternative argument. The main importance of the transpose (and this in fact defines it) is the formula
$$Axcdot y = xcdot A^top y.$$
(If $A$ is $mtimes n$, then $xin Bbb R^n$, $yinBbb R^m$, the left dot product is in $Bbb R^m$ and the right dot product is in $Bbb R^n$.)

Now note that
$$(AB)xcdot y = A(Bx)cdot y = Bxcdot A^top y = xcdot B^top(A^top y) = xcdot (B^top A^top)y.$$
Thus, $(AB)^top = B^top A^top$.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Ted ShifrinTed Shifrin

65.4k44792

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Ted ShifrinTed Shifrin

65.4k44792

answered 2 hours ago

Ted ShifrinTed Shifrin

65.4k44792

answered 2 hours ago

Ted ShifrinTed Shifrin

65.4k44792

2

$begingroup$

When you multiply $A$ and $B$, you are taking the dot product of each ROW of $A$ and each COLUMN of $B$.

The resulting dimension is $A_{#col}times B_{#row}$, and after transposing, you have $B_{#row}times A_{#col}$.

When you multiply $B^T$ and $A^T$, you take the dot product of each row of $B^T$ (column of B) and column of $A^T$, or row of $A$.

Your resulting dimension is $B^T_{#col}times A^T_{#row}$ which is just $B_{#row}times A_{#col}$

This formula ensures that each entry is correct, and that the dimensions are identical.

share|cite|improve this answer

answered 2 hours ago

Saketh MalyalaSaketh Malyala

7,8881535

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just to make up some notation to express your first + third sentence: let $operatorname{row}_i(M)$ and $operatorname{col}_j(M)$ denote the $i^{text{th}}$ row and $j^{text{th}}$ column of $M$, respectively. Then $(AB)_{ij} = operatorname{row}_i(A) cdot operatorname{col}_j(B)$, and $(B^T A^T)_{ji} = operatorname{row}_j(B^T) cdot operatorname{col}_i(A^T) = operatorname{col}_j(B) cdot operatorname{row}_i(A)$, so $(AB)_{ij} = (B^T A^T)_{ji}$.
  $endgroup$
  – Misha Lavrov
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

When you multiply $A$ and $B$, you are taking the dot product of each ROW of $A$ and each COLUMN of $B$.

The resulting dimension is $A_{#col}times B_{#row}$, and after transposing, you have $B_{#row}times A_{#col}$.

When you multiply $B^T$ and $A^T$, you take the dot product of each row of $B^T$ (column of B) and column of $A^T$, or row of $A$.

Your resulting dimension is $B^T_{#col}times A^T_{#row}$ which is just $B_{#row}times A_{#col}$

This formula ensures that each entry is correct, and that the dimensions are identical.

share|cite|improve this answer

answered 2 hours ago

Saketh MalyalaSaketh Malyala

7,8881535

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just to make up some notation to express your first + third sentence: let $operatorname{row}_i(M)$ and $operatorname{col}_j(M)$ denote the $i^{text{th}}$ row and $j^{text{th}}$ column of $M$, respectively. Then $(AB)_{ij} = operatorname{row}_i(A) cdot operatorname{col}_j(B)$, and $(B^T A^T)_{ji} = operatorname{row}_j(B^T) cdot operatorname{col}_i(A^T) = operatorname{col}_j(B) cdot operatorname{row}_i(A)$, so $(AB)_{ij} = (B^T A^T)_{ji}$.
  $endgroup$
  – Misha Lavrov
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

When you multiply $A$ and $B$, you are taking the dot product of each ROW of $A$ and each COLUMN of $B$.

The resulting dimension is $A_{#col}times B_{#row}$, and after transposing, you have $B_{#row}times A_{#col}$.

When you multiply $B^T$ and $A^T$, you take the dot product of each row of $B^T$ (column of B) and column of $A^T$, or row of $A$.

Your resulting dimension is $B^T_{#col}times A^T_{#row}$ which is just $B_{#row}times A_{#col}$

This formula ensures that each entry is correct, and that the dimensions are identical.

share|cite|improve this answer

answered 2 hours ago

Saketh MalyalaSaketh Malyala

7,8881535

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Saketh MalyalaSaketh Malyala

7,8881535

answered 2 hours ago

Saketh MalyalaSaketh Malyala

7,8881535

answered 2 hours ago

Saketh MalyalaSaketh Malyala

7,8881535

1

$begingroup$

If you know about dual spaces and maps, a conceptual proof can be obtained by observing that $A^T$ corresponds to the dual map of $A$ and that taking the dual is contravariant with respect to composition. That is, $(T circ S)^* = S^* circ T^*$.

share|cite|improve this answer

answered 2 hours ago

Aniruddh AgarwalAniruddh Agarwal

706

New contributor

Aniruddh Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

1

$begingroup$

If you know about dual spaces and maps, a conceptual proof can be obtained by observing that $A^T$ corresponds to the dual map of $A$ and that taking the dual is contravariant with respect to composition. That is, $(T circ S)^* = S^* circ T^*$.

share|cite|improve this answer

answered 2 hours ago

Aniruddh AgarwalAniruddh Agarwal

706

New contributor

Aniruddh Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

If you know about dual spaces and maps, a conceptual proof can be obtained by observing that $A^T$ corresponds to the dual map of $A$ and that taking the dual is contravariant with respect to composition. That is, $(T circ S)^* = S^* circ T^*$.

share|cite|improve this answer

answered 2 hours ago

Aniruddh AgarwalAniruddh Agarwal

706

New contributor

Aniruddh Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Aniruddh AgarwalAniruddh Agarwal

706

New contributor

Aniruddh Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 hours ago

Aniruddh AgarwalAniruddh Agarwal

706

New contributor

Aniruddh Agarwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 hours ago

Aniruddh AgarwalAniruddh Agarwal

706