# If the sum $A_1 + A_2 + · · · + A_n$ is equal to the negative of the identity operator on $V$ , show that $dim_{\mathbb{R}} V$ is even.

Let $V$ be a finite-dimensional vector space over the real numbers $\mathbb{R}$. Suppose $A_1,A_2,....,A_n$ are finitely many pairwise commuting linear operators on $V$. Assume that none of the operators $A_i$ has a negative real eigenvalue. If the sum $A_1 + A_2 + · · · + A_n$ is equal to the negative of the identity operator on $V$ , show that $dim_{\mathbb{R}} V$ is even.

## 2 Answers

Suppose $dim_{\mathbb{R}}(V)$ is odd, so for each $i$ there exist a real eigenvalue, $r_i$ of $A_i$, and hence exist eigenspaces $W_i$'s corresponding to $r_i$'s for all those $A_i$'s. Now since $A_i$'s pairwise commutes, so for each fixed $i$ the $W_i$ is invariant under all of $\{A_j \mid j\in \{1,2,...n\}\}$

Now if each of $W_i \mid i\in \{1,2,..,n\}$ is equals to $V$, then we've $\sum_{i}^{n} r_i =-1$, which is a contradiction, since none of the operators $A_i$ has a negative real eigenvalue. So for some $W_i$ we've $0 \subsetneq W_i \subsetneq V$. Say $W=W_i$. Now $V/W$ is also invariant under all of $A_i$'s, so now all of $A_i|_{W}$ and $A_i|_{V/W}$ has same property as $A_i$'s (I mean the same given properties as given in the problem). So, by induction hypothesis $dim_{\mathbb{R}}(W),dim_{\mathbb{R}}(V/W)$ are even, forces $dim_{\mathbb{R}}(V)$ is even.

• Why can you choose $W'$ to be $A_j$ (or even $A_i$) invariant? – levap Jun 17 '16 at 12:55

we can use induction on $n.$
Clearly $n\ne1,$ as $-I$ itself has negative real eigenvalues.
For $n=2,$ we find $A_2=-I-A_1.$ Moreover, if $\lambda$ is an eigenvalue of $A_1,$ then $-1-\lambda$ is one of $A_2.$
Then find an invertible matrix $V$ such that $V^{-1}A_1V$ is the Jordan form for $A_1.$ It follows that $D:=V^{-1}A_2V=-I-V^{-1}A_1V$ is also the Jordan form of $A_2,$ and that the diagonals of $D$ are of the form $-1-\lambda$ where $\lambda$ is a (maybe complex) eigenvalue of $A_1.$
If the dimension is odd, then $A_1$ must have a real eigenvalue, which is $\ge0,$ hence $-1-\lambda\lt0$ is an eigenvalue of $A_2,$ a contradiction. Thus the dimension must be even.
Now suppose it holds for $n\ge2$ matrices. And suppose $A_1,\cdots,A_{n+1}$ satisfy the conditions.
For any negative real eigenvalue $\lambda$ of $A:=A_n+A_{n+1},$ consider its eigen-space $E_\lambda.$ We observe that $$A\mid_{E\lambda}=\lambda I,$$ so $(A_n+A_{n+1})/(-\lambda)=-I$ and neither of $A_n/(-\lambda)$ and $A_{n+1}/(-\lambda)$ has negative real eigenvalues by hypotheses. It follows that $\operatorname{dim}E_\lambda$ is even by the above.
After this, we can decompose $V=K\oplus L$ where $K$ is the direct sum of generalized eigenspaces corresponding to real negative eigenvalues, and $L$ that corresponding to other eigenvalues. As $A_i$ are pairwise commuting, both $K$ and $L$ are invariant under $A_i.$ Then the linear operators $A_1,\cdots,A_{n-1},A$ restricted to $L,$ are $n$ linear operators satisfying the hypotheses, so by induction, $\operatorname{dim}L$ is even.
Finally $\operatorname{dim}V=\operatorname{dim}K+\operatorname{dim}L$ is even.

Sorry this is messy, and hope this helps.