Question:
If A and B are symmetric matrices, prove that AB − BA is a skew symmetric matrix.
Solution:
It is given that A and B are symmetric matrices. Therefore, we have:
$A^{\prime}=A$ and $B^{\prime}=B$ ....(1)
Now, $(A B-B A)^{\prime}=(A B)^{\prime}-(B A)^{\prime} \quad\left[(A-B)^{\prime}=A^{\prime}-B^{\prime}\right]$
$\begin{array}{ll}=B^{\prime} A^{\prime}-A^{\prime} B^{\prime} & {\left[(A B)^{\prime}=B^{\prime} A^{\prime}\right]} \\ =B A-A B & {[U \operatorname{sing}(1)]} \\ =-(A B-B A) & \end{array}$
$\therefore(A B-B A)^{\prime}=-(A B-B A)$
Thus, (AB − BA) is a skew-symmetric matrix.