# Is $AX = X^TA^T$ if $A$ and $X$ are square matrices? Could this be extended to non-square matrices?

I think $$AX = X^TA^T$$, if $$A$$ and $$X$$ are $$n\times n$$ matrices, since for $$i = 1, 2, \dots,n$$, the row vector $$\begin{bmatrix} a_{i1} & a_{i2} & \dots & a_{in} \end{bmatrix}$$ of $$A$$ is equal to the column vector $$\begin{bmatrix} a_{1i} \\ a_{2i} \\ \vdots \\ a_{ni} \end{bmatrix}$$ of $$A^T$$ (same with $$X$$), and since $$AX$$ means to perform the matrix multiplication on the $$ith$$ row of $$A$$ with the $$ith$$ column of $$X$$, then $$X^TA^T$$ means to perform the matrix multiplication on the $$ith$$ row of $$X^T$$ with the $$ith$$ column of $$A^T$$, and since the $$ith$$ row of $$X^{T}$$ is equal to the $$ith$$ column of $$X$$ and the $$ith$$ column of $$A^T$$ is equal to the $$ith$$ row of $$A$$, then the operation $$X^TA^T = AX$$.

Can this be generalized to non-square matrices?

In general, no, $$AX=X^TA^T$$ is not true. For example, take

$$X=\begin{bmatrix}1&0\\0&1\end{bmatrix}\\A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$$

Since $$X$$ is the identity matrix, you can easily see that $$AX=A$$, while $$X^TA^T=A^T$$, and since $$A$$ is not symmetric ($$A\neq A^T$$), you have $$A=AX\neq X^TA^T=A^T$$.

It is, however, true in general that $$(AB)^T=B^TA^T.$$

This equality holds for all matrices, square and non-square.

• And that shows that $AX=A^TX^T$ is $AX$ is symmetric. Apr 12, 2021 at 9:32
• @MichaelHoppe You probably meant $AX=X^TA^T$.
– 5xum
Apr 19, 2021 at 7:56
• Very probably, indeed! Apr 19, 2021 at 8:20