Question

If A and B are symmetric matrices, then show that AB is symmetric if AB = BA, i.e. A and B commute.

Answer 1

Hint: We know that if the transpose of a matrix is equal to the matrix itself the matrix is known as a symmetric matrix. If we have a symmetric matrix X, then $$X^T=X$$. By using this property of a symmetric matrix we will show the required condition for AB to be symmetric.

Complete step-by-step answer:

We have been given that A and B are symmetric matrices.
Since we know the property of a symmetric matrix that, the transpose of a symmetric matrix is equal to the matrix itself.
So, $$A^T=A$$ and $$B^T=B.....(1)$$
Hence the upper case ‘T’ denotes the transpose.
We have been asked to show that AB is symmetric if AB = BA i.e. A and B commute.
Since AB matrix is symmetric, then by using the property of symmetric matrix $${\left(AB\right)}^T$$ must be equal to AB.
$$\Rightarrow {\left(AB\right)}^T=AB$$
We know that $${(x_1{x}_2{x}_3.......x_n)}^T=({x_n}^T{x_{n-1}}^T.....{x_2}^T{x_1}^T)$$
$$\Rightarrow {\left(AB\right)}^T=B^T{A}^T$$
Using (1) we get as follows:
$$\begin{array}{rl}& {\left(AB\right)}^T=BA\\ & \Rightarrow {\left(AB\right)}^T=BA=AB\end{array}$$
The above expression is true if and only if AB = BA.
Therefore, it is shown that if A and B are symmetric matrices then AB is symmetric if and only if AB = BA.

Note: Remember that two matrices A and B are said to be commute matrices if they satisfy the criteria AB = BA. Also remember the property of a matrix that is as follows:
$${(x_1{x}_2{x}_3.......x_n)}^T=({x_n}^T{x_{n-1}}^T.....{x_2}^T{x_1}^T)$$ which is a very important property to find the transpose of the matrices in multiplication form.