# How do you prove that AB = BA if and only if AB is also symmetric?

Jul 29, 2016

This statement is false.

#### Explanation:

$\boldsymbol{3 \times 3}$ matrices

Let:

$A = B = \left(\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{matrix}\right)$

Then:

$A B = B A = \left(\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{matrix}\right) \left(\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{matrix}\right) = \left(\begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{matrix}\right)$

which is not symmetric.

$\textcolor{w h i t e}{}$
$\boldsymbol{2 \times 2}$ matrices

Let:

$A = B = \left(\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}\right)$

Then:

$A B = B A = \left(\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}\right) \left(\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}\right) = \left(\begin{matrix}1 & 2 \\ 0 & 1\end{matrix}\right)$

which is not symmetric.

Jul 29, 2016

See below

#### Explanation:

A corret proposition could be:

If $A$ is symmetric $A B = B A \iff B$ is symmetric

Suppose that $A , B$ are non null matrices and $A B = B A$ and $A$ is symmetric but $B$ is not

then

$A B = {\left(A B\right)}^{T} = {B}^{T} {A}^{T} = B A$

but $A = {A}^{T}$

so

${B}^{T} {A}^{T} - B A = 0 \to \left({B}^{T} - B\right) A = 0 \to {B}^{T} = B$ which is an absurd. So $B$ must be also symmetric.

Note.

There are matrices $A , B$ not symmetric such that verify

$A B = B A$. Example

$A = \left(\begin{matrix}4 & - 1 \\ \frac{1}{2} & 3\end{matrix}\right)$
$B = \left(\begin{matrix}1 & 2 \\ - 1 & 3\end{matrix}\right)$

$A B = B A = \left(\begin{matrix}5 & 5 \\ - \frac{5}{2} & 10\end{matrix}\right)$