Question

a) Let A and B be symmetric matrices of the same size.
Prove that AB is symmetric if and only \(AB=BA.\)
b) Find symmetric \(2 \cdot 2\)
matrices A and B such that \(AB=BA.\)

Answer 1

a) \(\Rightarrow\) Suppose that AB is symmetric.
This means that \((AB)^{T} = AB\)
Since \((AB)^{T} = B^{T} A^{T} = BA\)
(because A and B are symmetric), we get \(AB = BA\)
as required
\(\Leftarrow\)
Suppose that \(Ab = BA\)
To prove that AB is symmetric, we will prove that \(AB = (AB)^{T}.\)
Since \((AB)^{T} = B^{T} A^{T} = BA = AB\)
where we used that A and B are symmetric in the second equality and the assumption \(Ba = Ab\)
in the last equation and the assumption \(BA = BA\) in the last equality,
we get that \((AB)T = AB\)
as required
b) Let
\(A=\begin{bmatrix}1 & 1 \\1 & 0 \end{bmatrix}\ and\ B=\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}\)
Then \(AB=\begin{bmatrix}1 & 1 \\1 & 0 \end{bmatrix}\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}=\begin{bmatrix}0 & 1 \\0 & 0 \end{bmatrix}\)
and \(AB=\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}\begin{bmatrix}1 & 1 \\1 & 0 \end{bmatrix}=\begin{bmatrix}0 & 0 \\1 & 0 \end{bmatrix}\)
so \(AB \neq BA\)