Suppose $C_n$ is a product of $n$ $d\times d$ matrices with IID entries coming from standard normal. The following appears to be true. Is there an elementary proof?


This follows from discussion on math.SE on the moment method, but unclear how to adapt it to this, since the moment method requires fixing $n$.


1 Answer 1


This follows from the fact that $\mathbb{E}[A^\dagger A]=d I$ (with $A^\dagger$ the conjugate transpose of $A$ and $I$ the $d\times d$ identity matrix). Hence $$\mathbb{E}[\|C_n\|_F^2]=\operatorname{tr}\mathbb{E}[A_1A_2\cdots A_nA_n^\dagger\cdots A_2^\dagger A_1^\dagger]$$ $$=\operatorname{tr}\mathbb{E}[A_2\cdots A_nA_n^\dagger\cdots A_2^\dagger]\mathbb{E}[ A_1^\dagger A_1]$$ $$=d\operatorname{tr}\mathbb{E}[A_2\cdots A_nA_n^\dagger\cdots A_2^\dagger]$$ $$=d\operatorname{tr}\mathbb{E}[A_3\cdots A_nA_n^\dagger\cdots A_3^\dagger]\mathbb{E}[ A_2^\dagger A_2]$$ $$=d^2\operatorname{tr}\mathbb{E}[A_3\cdots A_nA_n^\dagger\cdots A_3^\dagger]$$ $$=d^{n-1}\operatorname{tr}\mathbb{E}[A_n^\dagger A]=d^n\operatorname{tr} I=d^{n+1}.$$

  • 1
    $\begingroup$ Ah, that's a lot simpler than the moment method, thanks! $\endgroup$ Dec 3, 2023 at 19:42

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