# Eigenvalues of a mean-zero symmetric random matrix

Maybe this is too obvious, but I what to be sure... Let $$Y$$ be a $$p\times p$$ symmetric random matrix (i.e. you can think about $$Y$$ as a matrix with random entries). Define $$E[Y]$$, the expectation of $$Y$$, as the matrix with entries $$(E[Y])_{ij} = E[Y_{ij}]$$. I think that the next affirmation is true:

If $$E[Y] = 0_{p\times p}$$ then $$\lambda_{\max}(Y)\geq 0$$ a.s., where $$\lambda_{\max}(Y)$$ is the greatest eigenvalue of $$Y$$ (which is real since $$Y$$ is symmetric).

My argument is as follows. Suppose that all the eigenvalues are negative. Then $$tr(Y)<0$$, which implies that $$E[tr(Y)]<0$$ and $$tr(E[Y])<0$$. This is a contradiction since $$E[Y] = 0_{p\times p}$$. Then there exist at least one non-negative eigenvalue, one of which is $$\lambda_{\max}(Y)$$.

Is my argument correct? In that case, is there a generalization of this result?

Consider $$p=1$$ and $$Y$$ equal to the 1x1 matrix 1 with probability 1/2 or the 1x1 matrix -1 with probability 1/2.

• +1, short and very clever Jan 30 '20 at 16:08
• Thanks. Good counterexample.
– RLC
Jan 30 '20 at 16:44

To add to the existing Ian's answer, the mistake in your proof is that $$tr(Y) < 0$$ does not imply that $$\mathbb{E}[tr(Y)] < 0$$.

The reason for this is that $$Y$$ is some realization of the random matrix, which may be anything within the allowed range. Consider, for example, restricting $$Y$$ to be a $$2 \times 2$$ diagonal matrix with entries $$u,v$$. Then, $$\mathbb{E}[tr(Y)] = \mathbb{E}[u+v] = 0$$, but for any particular matrix, $$tr(Y) = u+v$$ and both $$u,v$$ may end up negative in any particular realization.

For another example, consider a uniform random variable $$A \sim \mathcal{U}[-1,1]$$. Clearly, $$\mathbb{E}[A] = 0$$, but if I take a sample from this distribution, generating a sequence $$A_1, A_2, \ldots$$, a good number of them will be negative.

• Do you mean $tr(Y)<0$ does not imply $E[tr(Y)]<0$? I don't understand why not. Thanks in advance.
– RLC
Jan 30 '20 at 16:26
• @RLC please see the update Jan 30 '20 at 16:33
• Tr(Y)<0 a.s. implies E[Tr(Y)<0] but not the other way around.
– Ian
Jan 30 '20 at 16:34
• @Ian indeed, it is sufficient but not necessary :) Jan 30 '20 at 16:36
• @RLC Some confusion here, apparently. Actually, $tr(Y) < 0$ a.s. $\implies \mathbb{E}[tr(Y)]<0$, and it is true. The other way around, however, is not true. You are saying that $\mathbb{E}[tr(Y)]<0$ and are trying to imply that for almost any $Y$, you have $tr(Y) <0$, and that is false. Jan 30 '20 at 16:55