Commutativity of matrix and its transpose If a matrix is symmetric or skew-symmetric it commutes in the obvious way with its transpose.
(For symmetric: $SS^T=S^2$ and $S^TS =S^2$)
The less obvious is the case of commutativity  for orthogonal matrix but such matrix also commutes with its transpose because $RR^T=RR^{-1}= I= R^{-1}R=R^TR$.
Questions:


*

*Are there  other cases when a matrix commutes with its transpose ?   

*What is the general property of such matrix which allows it to
commute with its transpose ?

 A: Matrices for which $S^*S=SS^*$ holds are called normal. Here $*$ denotes the conjugate transpose. In the case of a real matrix this is equivalent to $SS^T=S^TS$. 
Normal matrices are precisely the ones that are diagonalisable by a unitary transformation, i.e
$$ S = U^* D U $$ for $U$ unitary and $D$ diagonal. 
Unitarity is in general a key property. Even if $S$ is real, an orthogonal transformation will not do in general. Take any non-trivial rotation in the plane for example. It is certainly normal, but has no real eigenvalues.
They can be unitarily diagonalised though.
On the other hand, (skew-) symmetric matrices are an example of real matrices that can be orthogonally diagonalised.
Several equivalent characterisations can be found for example on the Wikipedia page.
A: This is not an answer, but might be fun to read. Nephente's comment on normal matrices is an answer.
Suppose that a matrix $A$ commutes with it's transpose $A^T$. Let's calculate $(AA^T)_{ij}=\sum_{k=1}^n a_{ik}(a_{kj}^t)=\sum_{k=1}^n a_{ik}a_{jk}$. View $A$ as a collection of rows $A_1, \dots ,A_n$. Then $(AA^T)_{ij}=\left\langle A_i,A_j \right\rangle$ where $\left\langle \cdot, \cdot \right\rangle$ denotes the standard inner product. In the same fashion $(A^TA)_{ij}=\left\langle C_i, C_j \right\rangle$ where $C_i$ is the $i$-th column of $A$. Hence $AA^T=A^TA$ implies that the inner product of the $i$-th row with the $j$-th row is the same of for the corresponding columns. In fact this characterizes normal matrices.
