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In the definition of Wigner Matrix (a certain type of random Matrices) we take to independent family of i.i.d zero mean distributions $\{Z_{i,j}\}_{1<i<j}$ and $\{Y_{i}\}_{1\leq i}$ and then the entries of Wigner Matrix $X$ define as below:

$$\forall i,j; 1\leq i<j\leq N : X_N(i,j) = Z_{i,j}/\sqrt{n}$$ $$\forall i,j; 1\leq j<i\leq N : X_N(i,j) = Z_{j,i}/\sqrt{n}$$ $$\forall i; 1\leq i \leq N : : X_N(i,i) = Y_{i}/\sqrt{n}$$

After this task we will have a self-adjoint matrix that its eigenvalues are all real and we have Wigner's Semicircle Law about it.

My question is this: Is there any intuition or short proof that we should normalize the entries with $\frac{1}{\sqrt{N}}$ for controlling eigenvalues from being large.

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  • $\begingroup$ You are probably assuming that all expectations are zero. For a very rough explanation, you can consider any of the common matrix norms. $\endgroup$ Mar 24 '15 at 16:27
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On the one hand, you can compute the average of the trace of $X^2$ by using the eigenvalues, $\langle {\rm Tr}(X^2)\rangle=\sum_{i}\langle \lambda_{i}\rangle$. If the eigenvalues must be $O(1)$, this must be $O(N)$.

On the other hand, you may compute this average using matrix elements, $\langle {\rm Tr}(X^2)\rangle=\sum_{i,j}\langle X_{i,j}X_{j,i}\rangle$. If you don't normalize the entries with $1/\sqrt{N}$, this won't be $O(N)$ as required.

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