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Let $U \in \mathbb{R}^{n \times n}$ be a unitary matrix, $U$ can be nonsymmetric, its eigenvalues can be complex numbers and all have modulus $1$.

Is there an upper bound for the maximum singular value of its skew symmetric part (which is not necessarily unitary) depending on its eigenvalues?

i.e.: Is there an $f$ such that $\left\|\frac{U - U^T}{2} \right\|_2 = \sigma_\text{max}\left(\frac{U - U^T}{2}\right) \le f\left(\lambda_i\left(U\right)\right)$ ?

More details:

Observe that if $U=I$ (eigenvalues are real) $\Rightarrow \left\|\frac{U - U^T}{2} \right\|_2 = \sigma_\text{max}\left(\frac{U - U^T}{2}\right) = 0$, and if $U$ is skew-symmetric (eigenvalues purely imaginary) $\Rightarrow\left\|\frac{U - U^T}{2} \right\|_2 = \sigma_\text{max}\left(\frac{U - U^T}{2}\right) = 1$. Therefore there is a relationship between the norm $\left\|\frac{U - U^T}{2} \right\|_2 = \sigma_\text{max}\left(\frac{U - U^T}{2}\right)$ and the argument of the eigenvalues of $U$, i.e. $f\left(\lambda_i\left(U\right)\right) = f\left(\text{arg}(\lambda_i\left(U\right))\right)$.

Further notes: in my work $U$ is the unitary factor of the polar decomposition of an M-matrix, but this may be irrelevant.

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    $\begingroup$ Your question is confusing: you say "unitary" (usually used in complex matrices) but consider $U^T$ which makes me think that you consider $U$ real. If $U$ is real, the only possible eigenvalues are $1$ and $-1$. $\endgroup$ – Martin Argerami May 12 '17 at 1:40
  • $\begingroup$ math.stackexchange.com/questions/1717713/… $\endgroup$ – dantopa May 12 '17 at 2:18
  • $\begingroup$ @MartinArgerami Thanks for your interest! In my case $U$ is real, but unitary does not imply symmetric. When $U$ is nonsymmetric, its eigenvalues are complex numbers (regardless of the matrix being real) and they lie at the unit circle in the complex plane. I am looking for a relationship between the maximum singular value of the skew-symmetric part of $U$ (not necessarily unitary) and the eigenvalues of $U$. $\endgroup$ – Astor May 12 '17 at 9:34
  • $\begingroup$ @dantopa Thanks for the link! But I am looking for a relationship between the maximum singular value of the skew-symmetric part of $U$ (not necessarily unitary) and the eigenvalues of $U$ here. How do I use the fact of the eigenvalues having modulus 1? $\endgroup$ – Astor May 12 '17 at 9:36
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    $\begingroup$ @MartinArgerami Thanks for the clarification, our areas of math seem to be disjoint then. Do you think you could navigate through the awful, awful $\|\cdot\|_2$ notation to address the actual question? $\endgroup$ – Astor May 12 '17 at 14:15
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A unitary is normal, so it is diagonalizable by unitary conjugation. So we can think of $U $ diagonal, with the eigenvalues as the diagonal entries. The imaginary part is then the diagonal matrix of the imaginary parts of the eigenvalues. So $$ \left\|\frac {U-U^*}2\right\|=\max\{|\text {Im}\,\lambda_j|:\ j=1,\ldots,n\}. $$

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  • $\begingroup$ Are you assuming that the maximum singular value is equal to the spectral radius of $\frac{U - U^T}{2}$? Why? $\endgroup$ – Astor May 12 '17 at 14:23
  • $\begingroup$ Also, I wrote $U \in \mathbb{R}^{n \times n}$, so I probably should have said "orthogonal" not unitary. My apologies. $\endgroup$ – Astor May 12 '17 at 14:24
  • $\begingroup$ To the spectral radius of $\frac {U-U^*}{2i} $, yes. Because it is selfadjoint. But you don't even need that. In a diagonal matrix, you can calculate the singular values explicitly. $\endgroup$ – Martin Argerami May 12 '17 at 14:27

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