# Prove that $(\mathbf{AB})^{T} = \mathbf B^{T}\mathbf A^{T}$ where $\mathbf A$ and $\mathbf B$ are matrices

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 $$\tag!\sum_{k=1}^{n}(a_{j,k}b_{k,i})^{T} = \sum_{k=1}^{n}a_{k,j}b_{i,k}$$

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

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 $${\bf A}_{j,k} {\bf B}_{k,i} = ({\bf A}^T)_{k,j} ({\bf B}^T)_{i,k} = ({\bf B}^T)_{i,k} ({\bf A}^T)_{k,j}$$
$$\sum_k (a_{j,k}b_{k,i})^T$$.