# 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.

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 $AB=BA.$
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Layton
a) $⇒$ Suppose that AB is symmetric.
This means that $\left(AB{\right)}^{T}=AB$
Since $\left(AB{\right)}^{T}={B}^{T}{A}^{T}=BA$
(because A and B are symmetric), we get $AB=BA$
as required
$⇐$
Suppose that $Ab=BA$
To prove that AB is symmetric, we will prove that $AB=\left(AB{\right)}^{T}.$
Since $\left(AB{\right)}^{T}={B}^{T}{A}^{T}=BA=AB$
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 $\left(AB\right)T=AB$
as required
b) Let

Then $AB=\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$
and $AB=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$
so $AB\ne BA$