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I am asked to prove following:

Let $\mathbf A$ and $\mathbf B$ be matrices. Prove that $(\mathbf{AB})^{T} = \mathbf B^{T}\mathbf A^{T}$

My attempt:

Consider arbitrary entry of the $(\mathbf A \mathbf B)^{T}$, namely $((\mathbf A \mathbf B)^{T})_{i,j}$

$$((\mathbf A \mathbf B)^{T})_{i,j} = (\mathbf A \mathbf B_{j,i})^{T} =\sum_{k=1}^{n}(a_{j,k}b_{k,i})^{T} = \sum_{k=1}^{n}a_{k,j}b_{i,k} =\sum_{k=1}^{n} b_{i,k}a_{k,j} = (B^{T}A^{T})_{i,j}$$

Since we considered arbitrary entry, we conclude that $(\mathbf{AB})^{T} = \mathbf B^{T}\mathbf A^{T}$ $\Box$

Is it correct?


Although I can't tell for sure, I believe that something is wrong with the proof above. The step that concerns me the most (perhaps because of the notation involved) is $$\tag!\sum_{k=1}^{n}(a_{j,k}b_{k,i})^{T} = \sum_{k=1}^{n}a_{k,j}b_{i,k} $$

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    $\begingroup$ $(AB)^T_{i,j}=(AB)_{j,i}$. To put into words, the $(i,j)^{th}$ entry of $(M)^T$ is the $(j,i)^{th}$ entry of $M$, where $M=AB$. $\endgroup$ Sep 2, 2019 at 11:29
  • $\begingroup$ Think about what each step means. E.g., what does $(AB_{j,i})^T$ mean? Is that really what you want to say? $\endgroup$
    – BallBoy
    Sep 2, 2019 at 11:32
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    $\begingroup$ @BallBoy I meant transposition of the arbitrary entry, but now I see that it is incorrect. $\endgroup$ Sep 2, 2019 at 11:44

2 Answers 2

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It's incorrect, you cannot transpose a specific entry of a matrix (unless you treat it as 1x1 matrix, but that's not going to get you anywhere).

To do it correctly you need to write $$ \big(({\bf AB})^T\big)_{i,j} = ({\bf AB})_{j,i}$$ and later after using the formula for the entries of product of matrices, you'll go back with $$ {\bf A}_{j,k} {\bf B}_{k,i} = ({\bf A}^T)_{k,j} ({\bf B}^T)_{i,k} = ({\bf B}^T)_{i,k} ({\bf A}^T)_{k,j}$$

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Note that the transposition of matrices is well-defined but not the transposition of entries:

$\sum_k (a_{j,k}b_{k,i})^T$.

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