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