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This question already has an answer here:

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.

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

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marked as duplicate by darij grinberg, воитель, Xander Henderson, José Carlos Santos linear-algebra Jun 12 at 20:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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 May 1 at 0:58
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    $\begingroup$ 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 May 1 at 1:06
  • $\begingroup$ Note: the same fact holds for matrix inverses $\endgroup$ – J. W. Tanner May 1 at 1:31
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Here's an alternative argument. The main importance of the transpose (and this in fact defines it) is the formula $$Ax\cdot y = x\cdot A^\top y.$$ (If $A$ is $m\times n$, then $x\in \Bbb R^n$, $y\in\Bbb R^m$, the left dot product is in $\Bbb R^m$ and the right dot product is in $\Bbb R^n$.)

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

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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.

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    $\begingroup$ 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 May 1 at 1:15
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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, $(T \circ S)^* = S^* \circ T^*$.

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  • $\begingroup$ But then you're just delaying the actual argument until you prove that taking duals is a contravariant functor $\endgroup$ – FreeSalad May 31 at 23:26
  • $\begingroup$ Actually, my bad, the fact that $ (-)^* = \mathrm{Hom}(-, k) $ is enough. But it still is a lot of work (the term "corresponds" actually hiding equivalences of categories). $\endgroup$ – FreeSalad May 31 at 23:29
  • $\begingroup$ Well, proving that taking the dual corresponds to transposing a matrix only takes 3--4 lines. Let $T : V \rightarrow W$ be a linear map and $(v_i)$ and $(w_i)$ be basis for $V$ and $W$ respectively. Let $A$ be the matrix for $T$ and $A'$ be the matrix for $T^*$. It is enough to show that $A_{ij} = A'_{ji}$. Well $A_{ij} = w_i^*(T(v_j))$ and similarly $A'_{ji} = v_j^{**}(w_j^* \circ T)$ so it is enough to show that $v_j^{**}(w_j^* \circ T) =w_i^*(T(v_j))$. But this calculation is very simple. $\endgroup$ – Aniruddh Agarwal Jun 2 at 20:32
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I marked this as community wiki since it so close to Saketh Malyala's answer.


For the intuition/background, please read this site answer.

We will now prove the assertion.

For any matrix $C$ let $\text{Row}(C,i)$ denote the $i^\text{th}$ row of $C$ represented in a natural way as vector.

For any matrix $C$ let $\text{Col}(C,j)$ denote the $j^\text{th}$ column of $C$ represented in a natural way as vector.

The $(i,j)^\text{th}$ entry of $AB$ is equal to $\langle \text{Row}(A,i), \text{Col}(B,j)\rangle$

The $(j,i)^\text{th}$ entry of $B^tA^t$ is equal to $\langle \text{Row}(B^t,j), \text{Col}(A^t,i)\rangle$

But $\text{Row}(B^t,j) = \text{Col}(B,j)$ and $\text{Col}(A^t,i) = \text{Row}(A,i)$, so indeed,

$$ {(AB)}^t = B^t A^t$$

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