Anticommuting matrices and their eigenvalues Let $A,B\in \mathcal{M}_n(\mathbb{C})$. 
It is known that if $AB=BA$ and $\lambda_1, \lambda_2, \dots, \lambda_n $ are the eigenvalues of $A$ and $\beta_1, \beta_2, \dots, \beta_n$ are the eigenvalues of $B$ then for any polynomial $P(x, y)$, the eigenvalues of $P(A, B)$ are $P(\lambda_i, \beta_i)$.
My question is, does the theorem also hold true if $AB=-BA$ instead of $AB=BA$?
 A: There are several things to be said here, but first: a similar statement cannot hold for anti-commuting matrices.  For example, take
$$ A=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, B=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. $$
Then $A$ and $B$ anti-commute and they both have $1$ and $-1$ for eigenvalues.  Take $P(x,y)=xy$.  Then $P(A,B)=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ has $i$ and $-i$ for eigenvalues, which cannot be obtained by evaluating $xy$ at $\pm 1$.

Now, even if we wanted a statement for anti-commuting matrices, we would need more information.  In the commuting case, the proof of the result relies on the fact that $A$ and $B$ can be simultaneously triangularized, i.e. there is an invertible matrix $P$ such that $PAP^{-1}$ and $PBP^{-1}$ are both triangular, with their eigenvalues on the diagonal.  This gives us an ordering on the eigenvalues $\lambda_1,\ldots, \lambda_n$ and $\beta_1,\ldots, \beta_n$ which is implicit in your question.  In the anti-commuting case, we do not have this, so the (implicit) ordering used in the statement does not exist anymore.  Thus, without more information on the ordering of the eigenvalues, the statement is not completely well-defined.
That said, the example above shows that the statement cannot hold, whatever the ordering of the eigenvalues is.
