If A and B are symmetric matrices of the same order, show that (AB – BA) is a skew symmetric matrix.
If A and B are symmetric matrices of the same order, show that (AB – BA) is a skew symmetric matrix.

Solution:

We have A and B are symmetric matrices. Therefore A^{T}=A and B^{T}=B
The transpose of the matrix is an operation of making interchange of elements by the rule on positioned element a_{j i} shifted to new position a_{j i}.
The symmetric matrix is defined as similarity of transpose of matrix with it self. i.e, A^{T}=A.

The skew-symmetric matrix is defined as similarity of transpose of matrix with it self. i.e, A^{T}=-A.
We have some properties of transpose of matrices:

(i) (A+B)^{T}=A^{T}+B^{T} (ii) (A B)^{T}=B^{T} A^{T}
Let’s check
\begin{array}{l} (A B-B A)^{T}=(A B)^{T}-(B A)^{T}=B^{T} A^{T}-A^{T} B^{T}=B A-A B \\ =-(A B-B A) \end{array}
Therefore we are having (A B-B A) is skew symmetric.