17
$\begingroup$

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.

$\endgroup$
7
  • 3
    $\begingroup$ The first is true only if $A$ and $B$ commute $\endgroup$
    – Bman72
    Jul 22, 2014 at 5:48
  • 1
    $\begingroup$ In part 1 you already assumed $AB$ is symmetric. Otherwise, we don't know that we dont have $AB \neq (AB)^t$. $\endgroup$
    – jxnh
    Jul 22, 2014 at 5:50
  • $\begingroup$ 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... $\endgroup$ Jul 22, 2014 at 5:50
  • 2
    $\begingroup$ 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)? $\endgroup$ Jul 22, 2014 at 7:04
  • 7
    $\begingroup$ I think the right question is to prove that $1.\Longleftrightarrow 2.$, isn't it? $\endgroup$ Jul 22, 2014 at 8:44

2 Answers 2

53
$\begingroup$

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 $$\begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 3 \\ 5 & 5 \end{pmatrix}.$$

$\endgroup$
4
  • 6
    $\begingroup$ @nbro For symmetric matrices it's indeed true that $AB = A^T B^T = (BA)^T$ but $(BA)^\mathrm{T} \overset{\text{i.g.}}{\neq} BA$ because property 2 from the opening question does not hold. $\endgroup$
    – GDumphart
    Jan 23, 2017 at 16:32
  • $\begingroup$ @GDumphart What does the "i.g." mean over the not equals sign? $\endgroup$
    – Hank Igoe
    Jan 10, 2021 at 21:20
  • $\begingroup$ @HankIgoe: Likely it means "in general". In other words that equality holds only in special cases (namely if $AB$ is symmetric, or equivalently if $A,B$ commute). $\endgroup$
    – hardmath
    Mar 27, 2021 at 23:20
  • $\begingroup$ The simplest example is probably $\begin{pmatrix}1&\\&\end{pmatrix}\begin{pmatrix}&1\\1&\end{pmatrix}=\begin{pmatrix}&1\\&\end{pmatrix}\ne\begin{pmatrix}&\\1&\end{pmatrix}=\begin{pmatrix}&1\\1&\end{pmatrix}\begin{pmatrix}1&\\&\end{pmatrix}$ $\endgroup$
    – Kenta S
    Oct 16, 2021 at 22:07
2
$\begingroup$

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 $A_{ij}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .