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The typical matrix product is as follows: $$ (\mathbf{A}\mathbf{B})_{ij} = \sum_{k=1}^m A_{ik}B_{kj}\,. $$

Is there a name or characterization for one such as $$(\mathbf{A}\mathbf{B})_{ij} = \sum_{k=1}^m A_{ki}B_{jk}\,? $$ Furthermore, what can be said about the matrix vector product $$ (\mathbf{A}\mathbf{b})_{i} = \sum_{k=1}^m A_{ki}b_{k}\,? $$ Is there any way to express the above matrix-vector product in terms of traditional linear algebra?

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    $\begingroup$ Great questions. The beginner's mind is a precious thing. $\endgroup$
    – orangeskid
    Jul 11, 2015 at 0:24
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    $\begingroup$ Agreed, this is a really good question! It is not trivial but amazing that your formulas turn out to be $A^TB^T$ and $A^TB$. $\endgroup$
    – Eli Rose
    Jul 11, 2015 at 1:00
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    $\begingroup$ @EliRose why is this not trivial? It's pretty much by definition of transposition. $\endgroup$
    – Ruslan
    Jul 11, 2015 at 11:04

3 Answers 3

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Your second formula is just $A^tB^t$ and your third one $A^tb$, where the $(\hskip1ex)^t$ means transposed

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    $\begingroup$ (Also, note that $\;\;\; A^tB^t \: = \: (BA)^t \:\:\:\:$.) $\;\;\;\;\;\;\;\;\;$ $\endgroup$
    – user57159
    Jul 11, 2015 at 3:13
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There is not much new, unfortunately, if you know what the transpose of matrix is.

Given a matrix $\mathbf{A} = (A)_{ij}$ the transpose of a matrix $\mathbf{A}^{\intercal}$ is defined by $\mathbf{A}^{\intercal} = (A)_{ji}$. Intuitively the transpose of a matrix is found by reflecting the matrix across the line through the diagonal coefficients $i = j$. See https://en.wikipedia.org/wiki/Transpose.

With this in mind your first alternate version of multiplication can be written as $\sum_{k=1}^m A_{ki}B_{jk} = \mathbf{A}^\intercal\mathbf{B}^\intercal $.

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$$ \sum_{k=1}^n A_{ki} B_{jk} = \sum_{k=1}^n B_{jk} A_{ki} = (BA)_{ji} = (BA)^\intercal = A^\intercal B^\intercal $$ and $$ \sum_{k=1}^m A_{ki}b_{k} = \sum_{k=1}^m A_{ik}^{\intercal} b_{k} = A^{\intercal} b $$

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