# Are the eigenvectors of real Wigner matrices made of independent random variables with zero-mean?

I am trying to understand a portion of this paper [p. 3] and got stuck in the following statement, which sounded kind of trivial, but has been deceiving me for a while. Would you help me understand?

Let $$X$$ be an $$n \times n$$ real symmetric random matrix with independent entries of zero mean, $$x_1, \dots, x_n$$ its (unit) eigenvectors, $$A = \{1, \dots, n/2\}$$, $$B = \{n/2 + 1, \dots, n\}$$ and let $$u$$ be the vector where $$u_i = \frac{1}{\sqrt{n}}$$ if $$i \in A$$, and $$u_i = \frac{-1}{\sqrt{n}}$$ if $$i \in B$$.

On page 3, left column, the authors claim:

(...) we note that since $$X$$ is a random matrix, its eigenvectors are also random, so that cross terms cancel in the quantity $$(u^T x_i)^2$$ and the average value is simply $$| x_i |^2/n = 1/n$$.

Since

\begin{align}n(u^T x_i)^2 &= \sum_{k=1}^n (x_i)_k^2 + \left( \sum_{k \neq l; (k, l) \in (A \times A) \cup (B \times B)} (x_i)_k (x_i)_l - \sum_{(k, l) \in (A \times B) \cup (B \times A)} (x_i)_k (x_i)_l \right)\\ &= |x_i|^2 + Z_i,\end{align}

we want to show that $$\mathbb{E}[Z_i] = 0$$.

Now, it seems the authors are suggesting that since $$X$$ is such a matrix, then when $$n \rightarrow \infty$$ (implied throughout the paper) we may see the coordinates within each of its eigenvectors as independent random variables of zero mean. I can't see how, since they are so tangled by the elements of the matrix (as by the spectral theorem we have $$X_{ij} = \sum_{k=1}^n \lambda_k (x_i)_k (x_j)_k$$).

Maybe the answer is not as general as it seems and relies on features of the specific distribution of this $$X$$. The $$X_{ij}$$ are independent Bernoulli-distributed variables centered at the mean, with

$$X_{ij} \sim \begin{cases} Be(p) - p \text{, if } \{i,j\} \in (A \times A) \cup (B \times B) \\ Be(q) - q \text{, if } \{i,j\} \in (A \times B) \cup (B \times A). \end{cases}$$

The argument they refer to [p. 507] solves a similar case, where they were considering a vector $$v$$ drawn uniformly from the unit sphere and were concerned with $$|v|^2$$ instead of $$(u^T x_i)^2$$. After some research (e.g. this survey [§2 of Ch. 6, on p. 15] on eigenvectors of random matrices), it seems that for Wigner matrices like the one I am considering, it is expected that the eigenvectors follow a distribution close to uniform from the unit sphere (a... Haar measure?), so maybe this is the missing link?

But then how do Bernoulli-distributed variables centered at the mean relate to, for example, the Gaussian Orthogonal Ensemble or anything like that? (I could not find anything, since the first four moments don't match the criteria for direct comparisons). I have been trying to see if there is something about the adjacency matrix of random graphs on this matter as well, but it has been unfruitful up to now.

Am I on the right track, even?

If you read this up to here, thank you for reading! (:

• Is the author talking about a symmetric distribution? If not, the statement isn't true. E.g. when $n=2$ and the entries of $X$ are i.i.d. $Unif\{3,-1,-2\}$, we have $E((u^Tx_1)^2)=0.60846\ne\frac12$. – user1551 Nov 26 '18 at 20:15
• @user1551 Hey! Thanks for pointing that out! (: Checked your example and it is as you say. So the authors were assuming more things than they led me to believe... I guess it is the "as $n \to \infty$" that is implied throughout the paper (as I mentioned in the text), but I am not sure yet... – Felix Liu Nov 27 '18 at 2:40