Symmetric matrix multiplication

Question

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

$AB=BA$
$AB$ 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.

In part 1 you already assumed $AB$ is symmetric. Otherwise, we don't know that we dont have $AB \neq (AB)^t$. — jxnh, Jul 22 '14 at 05:50
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 05:50
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 07:04
@GerryMyerson so 2 implies 1. AB symmetric we know that $(AB)^t=B^t*A^t$ so AB=BA — gbox, Jul 22 '14 at 07:46
I think the right question is to prove that $1.\Longleftrightarrow 2.$, isn't it? — Omran Kouba, Jul 22 '14 at 08:44

score 54 · Accepted Answer · answered Jul 22 '14 at 05:46

54

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

answered Jul 22 '14 at 05:46

user165670

926

6

@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. – GDumphart Jan 23 '17 at 16:32
@GDumphart What does the "i.g." mean over the not equals sign? – Hank Igoe Jan 10 '21 at 21:20
@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). – hardmath Mar 27 '21 at 23:20
The simplest example is probably $\begin{pmatrix}1&\&\end{pmatrix}\begin{pmatrix}&1\1&\end{pmatrix}=\begin{pmatrix}&1\&\end{pmatrix}\ne\begin{pmatrix}&\1&\end{pmatrix}=\begin{pmatrix}&1\1&\end{pmatrix}\begin{pmatrix}1&\&\end{pmatrix}$ – Kenta S Oct 16 '21 at 22:07

score 2 · Answer 2 · answered Jul 22 '14 at 05:45

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 $A_{ij}$.

Symmetric matrix multiplication

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