First, we have a real matrix $N$ $(m \times m)$, and we don't have much constrains on N. For simplicity, we can normalize $N$ as $$ \frac N{\sqrt{Tr(N^T N)}} $$ Later on we are always using the normalized $N$. Normalized $N$ has a good property: $$ \sum_{i,j} N_{i,j}^2 = 1 $$ or in the form of trace: $$ Tr(N^T N) = 1 $$

We can always separate the matrix $N$ by its symmetric part and its antisymmetric part: $$ N = \frac 1 2 (N+N^T) + \frac 1 2 (N - N^T) = N^s + N^a $$

Then $$ Tr(N^T N) = Tr((N^s)^2 + [N^s,N^a] - (N^a)^2) = 1 $$ $[N^s,N^a]$ is traceless, so that $$ Tr((N^s)^2) - Tr((N^a)^2) =1 $$ We can set $Tr((N^a)^2) $ as $- \omega$, and it is obvious that $0\le \omega \le 1$, so that $Tr((N^s)^2) = 1- \omega$

Then we can set the eigenvalues of $\sqrt{N^T N}$ as $\lambda_0^1, \dots \lambda_0^m$ (the singular values of $N$); the eigenvalues of $N^s$ as $\lambda_s^1, \dots, \lambda_s^m$.

The question is what is the relationship between $\lambda_0^1, \dots, \lambda_0^m$ and $\lambda_s^1, \dots, \lambda_s^m$ and $\omega$?

for example, when $\omega = 0$, $\lambda_0^1, \dots \lambda_0^m$ and $\lambda_s^1 \dots \lambda_s^m$ must be the same, because when $\omega =0$, $N^a$ is a zero matrix.



Assume that $N$ is a real valued matrix. Let $x$ be an eigenvector corresponding to $\lambda_s$, i.e. $N_sx = \lambda_sx$. Note that $N_ax$ is always orthogonal to $x$. Therefore $||Nx||^2 = {\lambda_s}^2 + ||N_ax||^2$. This means that ${\lambda_0^i}^2 \ge {\lambda_s^i}^2 + ||N_ax_i||^2$, where $x_i$ is the corresponding eigenvector.

I don't think interlacing can be established since we don't really have control over $N_a$ beyond the fact that $||N_a||_f = \sqrt{1 - ||N_s||_F^2}$. If the norm of $N_s$ is small then $N_a$ can have significant effect. For example if $||N_ax_2||^2 \ge {\lambda_s^1}^2 + ||N_ax_1||^2 - {\lambda_s^2}^2$, then no interlacing can happen.


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.