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I have vector $x \in \mathbb{R}^n$ and square matrix $A \in \mathbb{R}^{n \times n}$ (doesn't depend on $x$). I believe I can make the claim $(Ax)^T = x^T A^T$.

On paper with $n = 2$, I can calculate: $(Ax)^T = x^T A^T = [a_{11} x_1 + a_{12} x_2, a_{21} x_1 + a_{22} x_2] \in \mathbb{R}^{1 \times 2}$.

This is obviously not a more general proof, as it's particular to $n = 2$.

In general, what is a nice way to show that $(Ax)^T = x^T A^T$, knowing vector $x \in \mathbb{R}^n$ and square matrix $A \in \mathbb{R}^{n \times n}$?

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    $\begingroup$ Have you tried writing out the $(i.j)$th element of both the LHS and the RHS with sigma (summation) notation? And maybe doing a few re-namings of variables if you can't see it at first? $\endgroup$
    – JonathanZ
    Jun 26, 2022 at 23:12
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    $\begingroup$ $Ax$ can be seen as the matrix multiplication of each row in $A$ by $x$. $x^TA^T$ can be seen as the matrix multiplication of $x^T$ by each column of $A^T$. $\endgroup$
    – layabout
    Jun 26, 2022 at 23:15

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That's the proof for the most general case, were $A$ is a $m\times n$ matrix and $B$ is a $n\times p$ matrix.

$(AB)_{ji} = (AB)^T_{ij}$, hence $$(AB)^T_{ij} = (AB)_{ji} = \sum\limits_{k=1}^nA_{jk}B_{ki},$$ and $$(B^TA^T)_{ij}= \sum\limits_{k=1}^nB^T_{ik}A^T_{kj}= \sum\limits_{k=1}^nB_{ki}A_{jk}= \sum\limits_{k=1}^nA_{jk}B_{ki},$$ so, since $(AB)^T_{ij} = (B^TA^T)_{ij}$ for all $i=1,...,p$ and $j=1,...,m$ we have $$(AB)^T = B^TA^T.$$

Just apply to $A$ a $n\times n$ matrix and the vector $x$, a $n\times 1$ matrix

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