Prove given A and B ∈ n(ℝ) , and A and B are symmetric then AB is symmetric

Question

I have problems trying to prove this

Prove given A and B ∈ n(ℝ) , and A and B are symmetric then AB is symmetric and AB <=> BA

So with the first given A=[a b;b c] and B=[e f;f g] , i have the following:

 AB =[ae + bf, af+bg]
     [be + cf, bf+cg]

So af+bg = be +cf

And

BA = [ a*e + b*f, b*e + c*f]
     [ a*f + b*g, b*f + c*g]

be + cf = af + bg

So solving this system , i get that f = (b*e)/(a - c) and g=0 , but this is not a prove is only for a couple of cases.

And also for this example:

A= [1 2 ; 2 4], 
B= [2 3 ; 3 6] 

i get AB=[8 15;16 30] and BA=[8 16;15 30] so this no way is symmetric at all.

The statement is from a problem from uni , so i guess is correct ,am i misunderstanding something ?

Answer 1

You probably misread the assignment, that should be:

Prove that, if $A,B\in M_n(\mathbb{R})$, $A$ and $B$ symmetric, then $AB$ is symmetric if and only if $AB=BA$.

If $AB$ is symmetric, then $$ AB=(AB)^T=B^TA^T=BA $$ so $AB=BA$.

Conversely, if $AB=BA$, then $$ (AB)^T=B^TA^T=BA=AB $$ and so $AB$ is symmetric.

Answer 2

Hint $AB$ is symmetric if and only if $AB=(AB)^T$. But

$$(AB)^T=B^TA^T=BA$$

Answer 3

We have to clarify the computations with your example. With $A=\begin{pmatrix} 1 & 2 \cr 2 & 3 \end{pmatrix}$ and $B=\begin{pmatrix} 2 & 3 \cr 3 & 6 \end{pmatrix}$ we obtain $$ AB=\begin{pmatrix} 8 & 15 \cr 13 & 24 \end{pmatrix},\quad BA=\begin{pmatrix} 8 & 13 \cr 15 & 24 \end{pmatrix}, $$ so indeed we have $(AB)^T=BA$ and everything is right. But in any case we obtain $(AB)^T=B^TA^T=BA$, because of $A^T=A$ and $B^T=B$.

So the result is: If $A$ and $B$ are symmetric, then $AB$ is symmetric if and only if $AB=BA$. So the "and" in your statement is wrong.