Prove or disprove $\ A^4 + B^4 $ can be diagonalized Let $\ A,B $ be two $\ 3\times3 $ matrices over $\ \mathbb R $ and $\ A^t = -A, B^t = B $
I need to prove or disprove that $\ A^4 + B^4 $ can be diagonalised over $\ \mathbb R $.
I really couldn't think of any direction to this. Any hints?
 A: Hint: One of your matrices is skew-symmetric, whereas the other one is symmetric. Deduce from that fact that $A^4+B^4$ is symmetric and use that fact.
A: Since
$A^T = -A, \tag 1$
we have, for any $n \in \Bbb N$
$(A^{2n})^T = (A^T)^{2n} = ((A^T)^2)^n = ((-A)^2)^n = (A^2)^n = A^{2n}; \tag 2$
thus, $A^{2n}$ is symmetric; since 
$B^T = B, \tag 3$
we also have
$(B^{2n})^T = (B^T)^{2n} = B^{2n}, \tag 4$
so $B^{2n}$ is also symmetric; therefore
$(A^{2n} + B^{2n})^T = (A^{2n})^T + (B^{2n})^T = A^{2n} + B^{2n} \tag 5$
is symmetric as well for any $n \in \Bbb N$, hence diagonalizable over $\Bbb R$.
Taking $n = 2$ yields the result required for the specific case at hand.
In the above we have used the general property that
$(C^T)^m = (C^m)^T \tag 6$
for any $m \in \Bbb N$; this is easily seen by a simple induction: when $m = 2$, we have, since $(CD)^T = D^T C^T$ for any two square matrices of the same size,
$(C^T)^2 = C^T C^T = (CC)^T = (C^2)^T; \tag 7$
now if 
$(C^T)^k = (C^k)^T, \tag 8$
then,
$(C^T)^{k + 1} = C^T (C^T)^k = C^T(C^k)^T = (C^k C)^T = (C^{k + 1})^T, \tag 9$
completing the induction and finalizing our proof of (5).
Note that the hypothesis that $A, B$ are $3 \times 3$ is not needed here; the rusult holds no matter what the size of $A$ and $B$ might be.
