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So, we have a square matrix $A=(a_{ij})_{1 \leq i,j \leq n}$ where the entries are independent random variables with the same distribution. Suppose $A = A^{*}$, where $A^{*}$ is the classical adjoint. Moreover, suppose that $E(a_{ij}) = 0$, $E(a_{ij}^{2}) < \infty$. How can I evaluate? $E(Tr A^{2})$?

Clearly, we have $E(Tr A) = 0$ and we can use linearity to get something about $E(Tr A^{2})$ in terms of the entries using simply the formula for $A^{2}$, but for instance I don't see where $A=A^{*}$ comes in... I suppose there's a clever way of handling it...

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2 Answers 2

up vote 1 down vote accepted

Clearly $$ \operatorname{Tr}(A^2) = \sum_{i,j} a_{i,j} a_{j,i} \stackrel{\rm symmetry}{=}\sum_{i,j} a_{i,j}^2 $$ Thus $$ \mathbb{E}\left(\operatorname{Tr}(A^2) \right) = \sum_{i,j} \mathbb{E}(a_{i,j}^2) = n^2 \mathbb{Var}(a_{1,1}) $$

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Identical as Sasha's reply: via Hilbert-Schmidt norms,

$$\operatorname{Tr}(A^2) = \operatorname{Tr}(A^\top A) = \|A\|_{\text HS}^2 = \sum_{i,j} a_{i,j}^2$$

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