# How to show $(Ax)^T = x^T A^T$ for square matrix $A$?

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 $$A \in \mathbb{R}^{n \times n}$$?

• 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? Jun 26, 2022 at 23:12
• $Ax$ 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 $A^T$. Jun 26, 2022 at 23:15

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