Let $M$ be a real symmetric matrix of size $N$ with its components $M_{ij}$ following a normal distribution centered around 0.

Let $x\in\mathbb{R}^N$ be an eigenvector of $M$ with eigenvalue $\lambda\in\mathbb{R}$: $$\sum_j M_{ij}x_j=\lambda x_i$$

I know that in that case the eigenvectors of different eigenvalues are independent from one another. However:

  • Is there any universality property regarding the distribution of the elements of the eigenvectors? If we want normalised eigenvectors I suppose that each entry of the eigenvector must have a variance that scales with $1/N$?
  • Is there any known relation between the matrix elements and the corresponding eigenvector elements? I am particularly interested in the correlation $\langle M_{ij}x_j\rangle$?

1 Answer 1


• For large $N$ the elements of an eigenvector have independent Gaussian distributions with zero mean and variance $1/N$.

• To find $\mathbb{E}[M_{ij}x_j]$ I decompose the matrix $M$ into eigenvalues and eigenvectors, $$M_{ij}=\sum_{k} O_{ki}\lambda_k O_{kj},$$ with an orthogonal matrix $O$. Then $$\mathbb{E}[M_{ij}x_j]=\sum_{k} \mathbb{E}[O_{ki}\lambda_k O_{kj}O_{1j}].$$ This vanishes because eigenvectors and eigenvalues are independent, and $\mathbb{E}[\lambda_k]=0$.

  • $\begingroup$ How can I show the first point? Is it still the case when the elements are not independent anymore but have some correlation? $\endgroup$
    – Matt
    Commented Mar 16, 2021 at 5:13
  • $\begingroup$ If the elements of the matrix are correlated, can the eigenvectors and eigenvalues still be independent? $\endgroup$
    – Matt
    Commented Mar 16, 2021 at 5:14
  • 1
    $\begingroup$ (1) The first point is the Porter-Thomas distribution of the Gaussian Orthogonal Ensemble (GOE), it holds only for $N\gg 1$ and only if you are considering the joint distribution of a small subset $n\ll N$ of the elements. (2) The second point holds exactly for any $N$ in the GOE, so if the elements of $M$ are uncorrelated; if they are correlated, it holds for $N\gg 1$ because the GOE is reached in that limit. $\endgroup$ Commented Mar 16, 2021 at 7:36

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