Prove that $\dim(V)$ is even Let $V$ be a finite dimensional vector space. Let $A_1,A_2: V\rightarrow V$ be commuting linear operators such that $A_1+A_2=-I$ where $I$ is the identity operator. Also $A_1,A_2$ have no negative eigenvalues. Prove that $\dim(V)$ is even.


How to approach this problem?
(Updated)

The assumption that $A_1$ and $A_2$ commute is superfluous. This already follows from $A_1+A_2=-I$.

It is tacitly assumed that the ground field is ${\mathbb R}$; for otherwise $A_1=[i]$, $A_2=[-1-i]$ would be a counterexample.

Now for the hint: Show that $A_1$ and $A_2$ can have no real eigenvalues at all, and think about the degree of the characteristic polynomial: it has to be even.
$A_1 + A_2 = -I$ => $A_2 = -I - A_1$

$ A_1A_2 - A_2A_1 = A_1(-I-A_1) - (-I-A_1)A_1 = 0 $

So for every $A_1$ there is an $A_2$ to commute with.

If $\lambda &gt; 0$ is an eigenvalue of $A_1$ and s is eigenvector:

$A_2s = (-I-A_1)s = -(1+\lambda)s $
So if $A_1$ has only positive eignevalues, $A_2$ has only negative ones.