If A and B are symmetric matrices of the same order, then A B-B A is a :

If $A$A and $B$B are symmetric matrices of the same order, then $AB-BA$AB−BA is a skew-symmetric matrix.

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Recall that a symmetric matrix $X$X has the property that $X = X^T$X=XT, where $X^T$XT is the transpose of $X$X.

A skew-symmetric matrix $Y$Y has the property that $Y = -Y^T$Y=−YT.

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Let's find the transpose of $AB - BA$AB−BA:

$$(AB - BA)^T = (AB)^T - (BA)^T$$(AB−BA)T=(AB)T−(BA)T

For any matrices $P$P and $Q$Q, we know that $(PQ)^T = Q^T P^T$(PQ)T=QTPT. So:

$$(AB)^T = B^T A^T$$(AB)T=BTAT

$$(BA)^T = A^T B^T$$(BA)T=ATBT

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Since $A$A and $B$B are symmetric matrices, $A = A^T$A=AT and $B = B^T$B=BT.

Substituting into our transpose expression:

$$(AB - BA)^T = B^T A^T - A^T B^T = BA - AB = -(AB - BA)$$(AB−BA)T=BTAT−ATBT=BA−AB=−(AB−BA)

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Since $(AB - BA)^T = -(AB - BA)$(AB−BA)T=−(AB−BA), by definition $AB - BA$AB−BA is skew-symmetric.

Note: This means all diagonal elements will be zero, and corresponding off-diagonal elements will be negatives of each other.