Vtyjkyuk

Let $ A,B $ be two $ 3times3 $ 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?

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.

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.