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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?

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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.

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

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