$AB=BA$ if $A$ is symmetric can you show that $B$ is symmetric $A$ and $B$ are square matrices, $A$ is symmetric and $AB=BA$.
I am trying to prove that $B$ is symmetric, my best try is as
$(AB)^t=(BA)^t \rightarrow B^tA^t=A^tB^t \rightarrow B^tA=AB^t$ 
can it be done? if so, is it the correct approach?
 A: This is not necessarily true. For example, if $A=0$ (which is symmetric), then $AB=BA$ for any $B$ including non-symmetric $B$. 
A: The statement of the OP is false. Let’s prove in which case it fails. 
I claim that, given a symmetric $n \times n$ matrix $A$, then: 
for all $B$ such that $AB=BA$, $B$ is symmetric $\Leftrightarrow$ $A$ has no multiple eigenvalues. 
Proof: write $A=UDU^T$, where $U$ is orthogonal and $D$ is diagonal. Let $B$ be a square matrix and $B’=U^TBU$. 
It is easy to see that $B$ commutes with $A$ iff $B’$ commutes with $D$, and $B$ is symmetric iff $B’$ is.
If $D$ has distinct diagonal entries (ie $A$ has no multiple eigenvalue), then a simple computation shows that $B’D=DB’$ iff $B’$ is diagonal. Therefore, if $B’D=DB’$, $B’$ is symmetric and we are done. 
Assume now that $A$ has one multiple eigenvalue. We can select $U$ and $D$ such that $D_{1,1}=D_{2,2}$. Consider the matrix $B=UB’U^T$ such that the only nonzero entry of $B’$ is $[B’]_{1,2}=1$. It is easy to check that $B’D=DB’$, thus $AB=BA$, but $B’$ is not symmetric, so neither is $B$. 
