Functional calculation for Hermitian matrices

Question

First, let me recall some useful definitions

We recall that if $A=U \Lambda U^*$ is a Hermitian matrix with $U U^*=U^* U=I$ and $\Lambda=\operatorname{diag}\left(\lambda_1\right)$ and $f: \mathbb{R} \rightarrow \mathbb{R}$ a continuous function then the matrix $f(A)$ is defined by $$ f(A)=U \operatorname{diag}\left(f\left(\lambda_i\right)\right) U^{*} $$

If $f$ is further differentiable, then $$ \operatorname{Tr} f(A+H)=\operatorname{Tr} f(A)+\operatorname{Tr}\left(f^{\prime}(A) H\right)+o(\|H\|), \quad H \in \mathcal{H}_n . $$

My problem Let $f, g: \mathbb{R} \rightarrow \mathbb{R}$ be two continuous continuous functions with compact support. We consider the Hermitian matrices $f\left(\frac{X}{\sqrt{n}}\right)$ and $g\left(\frac{X}{\sqrt{n}}\right)$.

We will fix that $k\in [n]$ and $k \le l$. I want to identify constant $\alpha$ and $\beta$, $\gamma$ such that

\begin{aligned} & \frac{\partial}{\partial X_{k k}}\left\{\operatorname{Tr} f\left(\frac{X}{\sqrt{n}}\right)\right\}=\frac{\alpha}{\sqrt{n}}\left[f^{\prime}\left(\frac{X}{\sqrt{n}}\right)\right]_{k k} \\ & \frac{\partial}{\partial X_{k \ell}}\left\{\operatorname{Tr} f\left(\frac{X}{\sqrt{n}}\right)\right\}=\frac{\beta}{\sqrt{n}}\left[f^{\prime}\left(\frac{X}{\sqrt{n}}\right)\right]_{\ell k}\\ & \frac{\partial}{\partial \bar{X_{k \ell}}}\left\{\operatorname{Tr} f\left(\frac{X}{\sqrt{n}}\right)\right\}=\frac{\gamma}{\sqrt{n}}\left[f^{\prime}\left(\frac{X}{\sqrt{n}}\right)\right]_{k\ell} \end{aligned}

I tried to take values of $H$ and use the above formula but I can't obtain the result. Could you please help me with this problem? Thank you very much for your help.

Answer 1

The $\sqrt{n}$ factors can be absorbed in the $X$ matrices, I can omit them. Consider a Hermitian matrix $X$ and a Hermitian perturbation $\delta X$. Then, for $f:\mathbb{R}\rightarrow\mathbb{R}$ one has $$\operatorname{Tr}f(X+\delta X)=\sum_{i}f(X)_{ii}+\sum_{i}f'(X)_{ii}\delta X_{ii}+\sum_{i<j}\left[f'(X)_{ji}\delta X_{ij}+\overline{f'(X)}_{ji}\overline{\delta X}_{ij}\right]+{\cal O}(\delta X)^2$$ $$\Rightarrow \frac{\partial}{\partial X_{ij}}\operatorname{Tr}f(X)=\begin{cases}f'(X)_{ii}&\text{if}\;\;i=j\\ f'(X)_{ji}&\text{if}\;\;i< j,\end{cases}$$ and also $$\frac{\partial}{\partial \overline{X}_{ij}}\operatorname{Tr}f(X)=\overline{f'(X)}_{ji}=f'(X)_{ij}.$$ So $\alpha,\beta,\gamma$ are all equal to 1.