It is well know that every matrix $A\in M_n(\mathbb{C})$ can be decomposed in a unique way as a sum of its hermitian part $H$ and its skew-hermitian part $K$: $$A=\frac{A+A^*}{2}+\frac{A-A^*}{2}$$

However, there is a relationship between the eigenvalues of $A$, $H$ and $K$ that I´m trying to prove without success:

If the spectrums of $A, H, K$ are $\sigma(A)= \{ \lambda_1,\ldots,\lambda_n \}$, $\sigma(H)= \{ \alpha_1,\ldots,\alpha_n \}$, $\sigma(K)= \{ \beta_1, \ldots, \beta_n \},$ then $$\sum_{i=1}^n\mathopen|\lambda_i\mathclose|^2\leqslant \sum_{i=1}^n\mathopen|\alpha_i\mathclose|^2+ \sum_{i=1}^n\mathopen|\beta_i\mathclose|^2$$ With equality $\text{iff}$ $A$ is normal.


1 Answer 1


Note: My convention for $H$ and $K$ is to take $H = \frac{A + A^*}{2}$ and $K = \frac{A - A^*}{2i}$, so that $A = H + iK$. With this convention, $H$ and $K$ are both Hermitian. Notably, they have real eigenvalues.

First, note that $$ \DeclareMathOperator{\tr}{tr} \sum |\lambda_i|^2 \leq \tr(A^*A) $$ Something along these lines is usually proven together with the spectral theorem for normal matrices following Schur's theorem (see for example Horn and Johnson). Notably, the inequality becomes equality if and only if $A$ is normal.

Now, expand $\tr(A^*A)$ in terms of $H$ and $K$ to get $$ \tr(A^*A) = \tr[(H+iK)^*(H+iK)]= \tr(H^*H)+ i\tr(HK) - i \tr(HK)+\tr(K^*K)=\\ \tr(H^*H)+ \tr(K^*K) $$ Note, however, that both $H$ and $K$ are normal.

Or, a yet more straightforward observation at this point is that since both $H$ and $K$ are Hermitian, we have $$ \tr(H^*H)+ \tr(K^*K) = \tr(H^2)+ \tr(K^2) = \sum_{i} \alpha_i^2 + \sum_i \beta_i^2 $$ where we note that these eigenvalues must be real.

  • $\begingroup$ Is the first inequality troublesome? Let me know. Also, if you have not learned about the Schur decomposition of a matrix, then this proof might not be what was intended. $\endgroup$ Sep 26, 2016 at 12:57
  • $\begingroup$ I knew about the Schur inequality and the Spectral Theorem. Thanks a lot! $\endgroup$
    – Gio V
    Sep 26, 2016 at 15:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.