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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.

Thanks!

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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.

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