Let $\mathbf{H}_{N,K}$ be a random matrix whose entries are i.i.d complex Gaussian random variables with variance $1$. Then, we know from the law of large number that if $N,K\rightarrow\infty$, we have $$\frac{1}{N}\mathbf{H}^\mathrm{H}\mathbf{H}\rightarrow\mathbf{I},$$ where $\mathbf{I}$ is the identity matrix. Now, I have some questions regarding this large dimension analysis: Let define $\mathbf{A}=\mathbf{H}^\mathrm{H}\mathbf{H}$. If $f$ be a continues function, is it true to write? $$f(\mathbf{A})\rightarrow f(\mathbf{\mathbb{E}[\mathbf{A}]})=f(N\mathbf{I}).$$ More over, can we write as follows? $$\mathbf{H}^{\mathrm{H}}f(\mathbf{A})\mathbf{H}\rightarrow \mathbf{H}^{\mathrm{H}}f(N\mathbf{I})\mathbf{H},$$ and for special case $f(\mathbf{A})=\mathbf{A}$, we have? $$\mathbf{H}^{\mathrm{H}}\mathbf{A}\mathbf{H}\rightarrow N\mathbf{H}^{\mathrm{H}}\mathbf{H}\rightarrow N^2\mathbf{I}.$$

$\begingroup$ I don't see how this would be consistent with MarchenkoPastur $\endgroup$– Carlo BeenakkerCommented Jan 21, 2020 at 9:43

$\begingroup$ Your identity matrix $\mathbf I$ must be $K\times K$ and hence cannot be the limit of anything as $K\to\infty$. $\endgroup$– Iosif PinelisCommented Jan 21, 2020 at 14:19
1 Answer
Let me check this for real matrices and $f(A)={\rm tr}\,A^2$. The $N$ eigenvalues $\mu_n$ of $A$ have in the limit $N\rightarrow\infty$ at fixed $N/K=\lambda\leq 1$ the MarchenkoPastur distribution $$\rho(\mu)=\mathbb{E}\left[\sum_{n=1}^N\delta(\mu\mu_n)\right]=N\frac{\sqrt{\lambda_+\mu/N}\sqrt{\mu/N\lambda_}}{2\pi\lambda\mu},\;\;\lambda_<\mu/N<\lambda_+,$$ with $\lambda_\pm=(1\pm\sqrt\lambda)^2$. The function $f(A)$ tends in this limit to $$f(A)\rightarrow\int_{N\lambda_}^{N\lambda_+}\mu^2\rho(\mu)\,d\mu=N^3(1+\lambda),$$ which differs from the answer $f(N I)=N^3$ conjectured in the OP.