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Let $A$ and $B$ be symmetric matrices. Prove:

  1. $AB=BA$
  2. $AB$ is a symmetric matrix

As for 1. due to the axiom $(AB)^T=B^T A^T$ so $AB=BA$
As for 2. I did not find any axiom that can support the claim, but from test I found that it is true for symmetric matrices when the entries on the diagonal are equal.

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The first is true only if $A$ and $B$ commute – Bman72 Jul 22 '14 at 5:48
In part 1 you already assumed $AB$ is symmetric. Otherwise, we don't know that we dont have $AB \neq (AB)^t$. – JHance Jul 22 '14 at 5:50
I think the issue is whether 1) asks you to prove whether $(AB)^{T}=BA$ or $AB=BA$. The first is definitely true, the second... – Semiclassical Jul 22 '14 at 5:50
Maybe you are supposed to prove that if $AB=BA$ then $AB$ is a symmetric matrix? That is, prove 1) implies 2), rather than prove 1) and 2)? – Gerry Myerson Jul 22 '14 at 7:04
@GerryMyerson so 2 implies 1. AB symmetric we know that $(AB)^t=B^t*A^t$ so AB=BA – gbox Jul 22 '14 at 7:46
up vote 18 down vote accepted

Both claims are false and almost any $A$ and $B$ are counterexamples. For a specific example, you can see $$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 3 & 5 \\ 3 & 5 \end{pmatrix}$$ while $$\begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 3 \\ 5 & 5 \end{pmatrix}.$$

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Recall that by definition the product of two matrices (with components $A_{ij}$, $B_{ij}$) has components $(AB)_{ik}=\sum_j A_{i j} B_{jk}$. What about the components of ${BA}$? You'll also need to see what being symmetric implies about the components $A_{ij}$.

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