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I was asked to prove the following:

Let $A \in M_{n}(\mathbb{C})$, then there is an Hermitian matrix $H$ and an skew-Hermitian matrix $K$ such that $A=H+K$.

If $\sigma(A) = \{\lambda_1, \lambda_2, \cdots, \lambda_n\}$, $\sigma(H) = \{\alpha_1, \alpha_2, \cdots, \alpha_n\}$ and $\sigma(K) = \{\beta_1, \beta_2, \cdots, \beta_n\}$ are the sets with all eigenvalues of $A,H$ and $K$ respectively, show that:

$ \overset{n}{\underset{i=1}{\sum}}\mid \lambda_{i} \mid ^{2} \leq \overset{n}{\underset{i=1}{\sum}}\mid \alpha_{i} \mid ^{2} + \overset{n}{\underset{i=1}{\sum}}\mid \beta_{i} \mid ^{2}. $

I know that by using Schur's Theorem $A \sim T$ where $T$ is a upper triangular matrix such that the eigenvalues of A are in its principal diagonal,

$T = \left( \begin{array}{rcccl} \lambda_1 & t_{12} & t_{13} & \cdots & t_{1n}\\ 0 & \lambda_2 & t_{23} & \cdots & t_{2n}\\ 0 & 0 & \lambda_3 & \cdots & t_{2n}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & \lambda_n\\ \end{array} \right).$

So $\overset{n}{\underset{i=1}{\sum}}\mid \lambda_{i} \mid ^{2} \leq ||T||_F^2 = \overset{n}{\underset{i=1}{\sum}}\mid \lambda_{i} \mid ^{2} + \overset{n}{\underset{\underset{i<j \leq n}{i=1}}{\sum}}\mid t_{ij} \mid ^{2}$.

Where $||\cdot||_F$ is the Frobenius' norm, i.e., $||M||_F = \sqrt {Trace(M^*M)}$, where if $M := [m_{ij}]$ then $M^* := [ \ \overline{m_{ji}} \ ]$.

By the other hand, $H,K$ are normal matrices so by applying The Spectral Theorem I obtain that

$||H||_F^2 = \overset{n}{\underset{i=1}{\sum}}\mid \alpha_{i} \mid ^{2}$

$||K||_F^2 = \overset{n}{\underset{i=1}{\sum}}\mid \beta_{i} \mid ^{2}$

But I don't know how to connect them in order to demonstrate this inequality.

Thanks for your help.

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1 Answer 1

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Remember that $A=H+K$, so:

$||A||_F^2 = ||H+K||_F^2 = Tr\left((H+K)^*(H+K)\right)$ $= Tr\left(H^2-KH+HK-K^2)\right)$ $= Tr(H^2)-Tr(KH)+Tr(HK)+Tr(-K^2)$ $= Tr(H^*H)-Tr(KH)+Tr(KH)+Tr(K^*K)$ $= Tr(H^*H)+Tr(K^*K) = ||H||_F^2 + ||K||_F^2.$

Therefore:

$ \overset{n}{\underset{i=1}{\sum}}\mid \lambda_{i} \mid ^{2} \leq ||A||^2 = ||H||_F^2 + ||K||_F^2 = \overset{n}{\underset{i=1}{\sum}}\mid \alpha_{i} \mid ^{2} + \overset{n}{\underset{i=1}{\sum}}\mid \beta_{i} \mid ^{2}.$

So

$ \overset{n}{\underset{i=1}{\sum}}\mid \lambda_{i} \mid ^{2} \leq \overset{n}{\underset{i=1}{\sum}}\mid \alpha_{i} \mid ^{2} + \overset{n}{\underset{i=1}{\sum}}\mid \beta_{i} \mid ^{2}. $

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