What is the Jordan form of a matrix $A$ when $A$ is similar to $-A$?

Question

I've been trying to solve this problem from an old exam. Since $A$ and $-A$ are similar, they have the same Jordan form, say $J$. Then $A=P^{-1}JP$ and $B=Q^{-1}JQ$ for some invertible matrices $P,Q$. Using only this, how can I say something about the the matrix $J$? Is there something else about the matrices $P$ and $Q$ I'm missing? The original exam question also specified that $A$ is $3\times 3$, but I'd also like to know in a general case if possible.

Answer 1

Negating a matrix will preserve the eigenspaces but negate each eigenvalue. More precisely, negating a single Jordan block produces $$ \begin{pmatrix}-\lambda&-1\\&-\lambda&-1\\&&-\lambda&\smash{\ddots} \\&&&\smash\ddots\end{pmatrix} $$ where you can flip the $-1$s on the off diagonal to $1$ by conjugating with $\operatorname{Diag}(1,-1,1,-1,\ldots)$.

Thus, in $J$ each Jordan block with a non-zero eigenvalue needs to be accompanied with a Jordan block of the same size with the opposite eigenvalue.

For a $3\times 3$ matrix, this only leaves a few possibilities for $J$ (up to reordering of the Jordan blocks):

$$ \begin{pmatrix}0\\&\lambda\\&&-\lambda\end{pmatrix}, \begin{pmatrix}0&1\\&0&1\\&&0\end{pmatrix}, \begin{pmatrix}0&1\\&0\\&&0\end{pmatrix} $$