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 ${{\left( {{x}_{1}}{{x}_{2}}{{x}_{3}}.......{{x}_{n}} \right)}^{T}}=\left( {{x}_{n}}^{T}{{x}_{n-1}}^{T}.....{{x}_{2}}^{T}{{x}_{1}}^{T} \right)$
$\Rightarrow {{\left( AB \right)}^{T}}={{B}^{T}}{{A}^{T}}$
Using (1) we get as follows:

& {{\left( AB \right)}^{T}}=BA \\\ & \Rightarrow {{\left( AB \right)}^{T}}=BA=AB \\\ \end{aligned}$$ The above expression is true if and only if AB = BA. Therefore, it is shown that if A and B are symmetric matrices then AB is symmetric if and only if AB = BA. Note: Remember that two matrices A and B are said to be commute matrices if they satisfy the criteria AB = BA. Also remember the property of a matrix that is as follows: $${{\left( {{x}_{1}}{{x}_{2}}{{x}_{3}}.......{{x}_{n}} \right)}^{T}}=\left( {{x}_{n}}^{T}{{x}_{n-1}}^{T}.....{{x}_{2}}^{T}{{x}_{1}}^{T} \right)$$ which is a very important property to find the transpose of the matrices in multiplication form.