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Let's say I have these 2 matrices:

$$A = \begin{bmatrix} c \\ d \end{bmatrix} $$ and

$$B = \begin{bmatrix} e \\ f \end{bmatrix}$$

$A'B = ce + df $ and $B'A = ec + fd$

As shown above, $A'B = B'A$. But is there a matrix product property that could've told me this without having to manually check? I haven't been able to find any matrix properties that prove that $A'B = B'A$

QuantumSuperfield
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ATP
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1 Answers1

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A real number can be considered a $1\times 1$ real matrix, which is trivially symmetric. Therefore if $A^TB = C\in\mathbb{R}$, then $A^TB = C = C^T = B^TA$.

whpowell96
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    It does not follow. I proved that $C=C^T$ if $C\in\mathbb{R}^{1\times 1}$. $(A^TB)^T = B^TA$ is a separate fact about matrix transposes. – whpowell96 Feb 17 '24 at 21:44