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If $A$ and $B$ are two square matrices of the same order, then prove that matrix $B'AB$ is symmetric or skew symmetric according as $A$ is symmetric or skew symmetric.

I took two $2\times 2$ matrices and verified the above result but how to prove this result I don't know. So anyone please try to prove this.
Thanks!!!

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  • $\begingroup$ Note that $(AB)^T=B^TA^T$ which implies $(ABC)^T=C^TB^TA^T$ (here T represents transposition) $\endgroup$
    – JMoravitz
    Commented Mar 17, 2017 at 18:39
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    $\begingroup$ Note that $$(B'AB)'=(B)'(A)'(B')'=B'A'B$$ Hence, if $A=A'$, then $$(B'AB)'=B'AB$$And if $A'=-A$, then ... $\endgroup$
    – Mark Viola
    Commented Mar 17, 2017 at 18:41

2 Answers 2

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Note that for any two matrices $A$ and $B$, we have

$$(AB)_{ij}=A_{ik}B_{kj}$$

Hence, the transpose of $AB$ is given by

$$\begin{align} (AB)'_{ij}&=(AB)_{ji}\\\\ &=A_{jk}B_{ki}\\\\ &=(B')_{ik}(A')_{kj}\\\\ &=(B'A')_{ij} \end{align}$$

Therefore, we find that $(AB)'=B'A'$. Using this relationship, it is easy to see that

$$(B'AB)'=B'A'B$$

If $A=A'$, then $(B'AB)'=B'AB$.

If $A=-A'$, then $(B'AB)'=-B'AB$.

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$$(B'AB)'=B'A'B$$

If $A=A'$ then $(B'AB)'=B'AB$.

If $A=-A'$, then $(B'AB)'=-B'AB$.

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