Xjyuk

Which industry am I working in? Software development or financial services?

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

Returning the outputs of a nested structure

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

Identifying my late father's D&D stuff found in the attic

What happens to matryoshka Mordenkainen's Magnificent Mansions?

Why is `abs()` implemented differently?

In a vacuum triode, what prevents the grid from acting as another anode?

Transpose of product of matrices

Sub query result is 0

Is Cola "probably the best-known" Latin word in the world? If not, which might it be?

Can fracking help reduce CO2?

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

If Earth is tilted, why is Polaris always above the same spot?

Casual versus formal jacket

How to reply this mail from potential PhD professor?

Would glacier 'trees' be plausible?

Why is ‘eines’ used in the sentence ‘Eines gleich vorweg’?

Why is B♯ higher than C♭ in 31-ET?

Alias to source .bashrc after it's been edited?

Accidentally deleted the "/usr/share" folder

Was Unix ever a single-user OS?

Python password manager

My ID is expired, can I fly to the Bahamas with my passport?

Transpose of product of matrices

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 1 hour 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
  1 hour 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
  55 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note: the same fact holds for matrix inverses
  $endgroup$
  – J. W. Tanner
  30 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 1 hour 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
  1 hour 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
  55 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note: the same fact holds for matrix inverses
  $endgroup$
  – J. W. Tanner
  30 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 1 hour 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 1 hour 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 1 hour 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 1 hour 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 1 hour ago

feli andrésfeli andrés

161

3 Answers
3

active

oldest

votes

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 $A_#coltimes B_#row$, and after transposing, you have $B_#rowtimes A_#col$.

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 $B^T_#coltimes A^T_#row$ which is just $B_#rowtimes A_#col$

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

share|cite|improve this answer

answered 1 hour ago

Saketh MalyalaSaketh Malyala

7,8881535

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just to make up some notation to express your first + third sentence: let $operatornamerow_i(M)$ and $operatornamecol_j(M)$ denote the $i^textth$ row and $j^textth$ column of $M$, respectively. Then $(AB)_ij = operatornamerow_i(A) cdot operatornamecol_j(B)$, and $(B^T A^T)_ji = operatornamerow_j(B^T) cdot operatornamecol_i(A^T) = operatornamecol_j(B) cdot operatornamerow_i(A)$, so $(AB)_ij = (B^T A^T)_ji$.
  $endgroup$
  – Misha Lavrov
  47 mins ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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 1 hour ago

Ted ShifrinTed Shifrin

65.4k44792

$endgroup$

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 1 hour ago

Aniruddh AgarwalAniruddh Agarwal

586

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

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 $A_#coltimes B_#row$, and after transposing, you have $B_#rowtimes A_#col$.

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 $B^T_#coltimes A^T_#row$ which is just $B_#rowtimes A_#col$

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

share|cite|improve this answer

answered 1 hour ago

Saketh MalyalaSaketh Malyala

7,8881535

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just to make up some notation to express your first + third sentence: let $operatornamerow_i(M)$ and $operatornamecol_j(M)$ denote the $i^textth$ row and $j^textth$ column of $M$, respectively. Then $(AB)_ij = operatornamerow_i(A) cdot operatornamecol_j(B)$, and $(B^T A^T)_ji = operatornamerow_j(B^T) cdot operatornamecol_i(A^T) = operatornamecol_j(B) cdot operatornamerow_i(A)$, so $(AB)_ij = (B^T A^T)_ji$.
  $endgroup$
  – Misha Lavrov
  47 mins 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 $A_#coltimes B_#row$, and after transposing, you have $B_#rowtimes A_#col$.

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 $B^T_#coltimes A^T_#row$ which is just $B_#rowtimes A_#col$

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

share|cite|improve this answer

answered 1 hour ago

Saketh MalyalaSaketh Malyala

7,8881535

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just to make up some notation to express your first + third sentence: let $operatornamerow_i(M)$ and $operatornamecol_j(M)$ denote the $i^textth$ row and $j^textth$ column of $M$, respectively. Then $(AB)_ij = operatornamerow_i(A) cdot operatornamecol_j(B)$, and $(B^T A^T)_ji = operatornamerow_j(B^T) cdot operatornamecol_i(A^T) = operatornamecol_j(B) cdot operatornamerow_i(A)$, so $(AB)_ij = (B^T A^T)_ji$.
  $endgroup$
  – Misha Lavrov
  47 mins 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 $A_#coltimes B_#row$, and after transposing, you have $B_#rowtimes A_#col$.

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 $B^T_#coltimes A^T_#row$ which is just $B_#rowtimes A_#col$

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

share|cite|improve this answer

answered 1 hour ago

Saketh MalyalaSaketh Malyala

7,8881535

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Saketh MalyalaSaketh Malyala

7,8881535

answered 1 hour ago

Saketh MalyalaSaketh Malyala

7,8881535

answered 1 hour ago

Saketh MalyalaSaketh Malyala

7,8881535

2

$begingroup$

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 1 hour ago

Ted ShifrinTed Shifrin

65.4k44792

$endgroup$

add a comment | 

2

$begingroup$

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 1 hour 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 1 hour ago

Ted ShifrinTed Shifrin

65.4k44792

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Ted ShifrinTed Shifrin

65.4k44792

answered 1 hour ago

Ted ShifrinTed Shifrin

65.4k44792

answered 1 hour ago

Ted ShifrinTed Shifrin

65.4k44792

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 1 hour ago

Aniruddh AgarwalAniruddh Agarwal

586

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 1 hour ago

Aniruddh AgarwalAniruddh Agarwal

586

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 1 hour ago

Aniruddh AgarwalAniruddh Agarwal

586

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 1 hour ago

Aniruddh AgarwalAniruddh Agarwal

586

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 1 hour ago

Aniruddh AgarwalAniruddh Agarwal

586

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 1 hour ago

Aniruddh AgarwalAniruddh Agarwal

586