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Suppose that $A \in M_{m,p}(F)$ and $B \in M_{n,p}(F)$. Let $a_k$ be the $k$th column of $A$ and let $b_k$ be the $k$th column of $B$. Then $AB^T = \sum_{k=1}^n a_k b_k^T$: each summand is an $m$-by-$n$ matrix, the outer product of $a_k$ and $b_k$

I'm struggling in understanding how to get to the result. Thanks for your help.

Me La Ria
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    I suggest you write $A = \sum_{j=1}^n \mathbf {a_je_j}^T$ and $B^T = \sum_{k=1}^n \mathbf {e_k b_k}^T$ with standard basis vectors $\mathbf e_i$ and multiply out these two sums. Please show us your attempt. – user8675309 Apr 15 '23 at 15:46

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As suggested by @user8675309, here's my attempt:

$A = \sum_{j=1}^n \mathbf{a_j} \mathbf{e_j}^T$

$B^T = \sum_{k=1}^n \mathbf{e_k} \mathbf{b_k}^T$

$AB^T = \sum_{j=1}^n \mathbf{a_j} \mathbf{e_j}^T\sum_{k=1}^n \mathbf{e_k} \mathbf{b_k}^T = \sum_{j=1}^n \sum_{k=1}^n \mathbf{a_j} \mathbf{e_j}^T\mathbf{e_k} \mathbf{b_k}^T$

since $\mathbf{e_j}^T\mathbf{e_k} = 1$ if $j=k$ and $0$ otherwise, we have that the above equals:

$\sum_{k=1}^n \mathbf{a_k}\mathbf{b_k}^T$

Me La Ria
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