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# 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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Matrix transformations
asked 2020-12-25
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 $$AB=BA.$$

## Answers (1)

2020-12-26
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 $$AB \neq BA$$

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