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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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    $\begingroup$ Remark: in my corner of functional analysis, $\Vert \cdot \Vert_2$ would normally denote $\ell^2$-norm (i.e. Hilbert-Schmidt norm, a.k.a. Frobenius norm). I'm guessing from context you mean the $\ell^2$-to-$\ell^2$ norm? $\endgroup$ – Yemon Choi May 12 '17 at 1:21
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    $\begingroup$ Unitary real matrices are usually called orthogonal matrices. I was thinking about complex matrices and had not noticed that you were only looking at real ones. $\endgroup$ – Geoff Robinson May 12 '17 at 13:41
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    $\begingroup$ In the case that $U$ is a real orthogonal matrix, any eigenvector ( possibly complex) of $U$ with egenvalue $\alpha$ is also an eigenvalue of $U^{T}$ with eigenvalue ${\bar \alpha},$ and it easily follows that the eigenvalues of the skew symmetric part of $U$ are precisely the imaginary parts of the eigenvalues of $U$. Hence the spectral radius of the skew-symmetric part of $U$ is the maximum of the absolute value of the imaginary parts of the eigenvalues of $U.$ $\endgroup$ – Geoff Robinson May 12 '17 at 13:47
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    $\begingroup$ See my answer below. I don't think there is anything more to say beyond that. $\endgroup$ – Geoff Robinson May 12 '17 at 14:05
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    $\begingroup$ By the way, the title is a bit misleading. You want the norm of the skew-symmetric part of $U,$ and the skew-symmetric part of $U$ is not necessarily unitary itself. $\endgroup$ – Geoff Robinson May 13 '17 at 9:54
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Since I misread the question I will clarify my comments into a formal answer (which is, in the end, quite elementary). Since $U$ is a real orthogonal matrix, it has a basis of eigenvectors when viewed as a complex matrix say $\{v_{1},v_{2}, \ldots,v_{n} \}.$ Also, the eigenvalues of $U$ all lie on the unit circle, and the non-real ones occur in complex conjugate pairs.

Whenever $v$ is an eigenvector of $U$ with eigenvalue $\alpha,$ it is also an eigenvector of $U^{T}$ with eigenvalue ${\bar \alpha}.$ Hence if $v_{j}$ is an eigenvector of $U$ with eigenvector $\alpha_{j},$ then $v_{j}$ is an eigenvector of $\frac{U-U^{T}}{2}$ with eigenvalue $i{\rm Im}(\alpha_{j}).$ Hence the spectral radius of $\frac{U-U^{T}}{2}$ is the maximum element of $\{ |{\rm Im}(\alpha_{j})| : 1 \leq j \leq n \}.$ This can only be $0$ when all eigenvalues of $U$ are $\pm 1,$ and can only be $1$ if $U$ has $i$ as an eigenvalue.

If you prefer, you can write $\alpha_{j} = \exp(i \beta_{j})$ with $0 \leq \beta_{j} < 2 \pi$ and then $\frac{U-U^{T}}{2}$ has spectral radius the maximum element of $\{ |\sin(\beta_{j})|: 1 \leq j \leq n \}$ and $\frac{U - U^{T}}{2}$ has eigenvalues $\{ i\sin(\beta_{j}) : 1 \leq j \leq n \}$ ( even allowing for multiplicities, by a slight abuse of notation).

Since a real skew-symmetric matrix is certainly normal (ie commutes with its transpose), the largest of the absolute values of its eigenvalues is its largest singular value (which is also its operator norm with respect to the Euclidean norm on $\mathbb{R}^{n}).$ Hence this is the maximum element of $\{ |\sin(\beta_{j})|: 1 \leq j \leq n \}.$

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  • $\begingroup$ By studying the spectral radius, you are finding the elementary lower bound for the maximum singular value of the skew-symmetric part. The whole point of the question is to find an upper bound for the maximum singular value of the skew-symmetric part. $\endgroup$ – Astor May 12 '17 at 14:16
  • $\begingroup$ @Astor The spectral radius is the same as the largest singular value, so I don't understand your comment. Furthermore, both Geoff and I are fully aware of the difference between orthogonal matrices, unitary matrices, and (skew-)symmetric matrices $\endgroup$ – Yemon Choi May 12 '17 at 14:28
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    $\begingroup$ I have explicitly given ALL eigenvalues of the skew symmetric part of $U,$ expressed in terms of the eigenvalues of $U$. What else do you want to know? $\endgroup$ – Geoff Robinson May 12 '17 at 14:36
  • $\begingroup$ @YemonChoi It is very likely that I am the ignorant here and if it is the case I apologize. The spectral radius of a matrix is the largest eigenvalue modulus, not the largest singular value. Is there a proof that for all unitary matrices the maximum modulus of its eigenvalues is equal to the maximum singular value? $\endgroup$ – Astor May 12 '17 at 14:37
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    $\begingroup$ It is standard theory that if a real matrix is diagonalizable via a unitary matrix, then its largest singular value is the largest absolute value of its eigenvalues. Any normal matrix may be diagonalized by a unitary matrix. A real matrix $A$ is normal if $A$ commutes with $A^{T}.$ Clearly, any symmetric, any skew-symmetric, and any orthogonal real matrix, is normal. Hence the largest singular value of a real skew-symmetric matrix ( which is also its operator norm), is the largest of teh absolute values of its eigenvalues. $\endgroup$ – Geoff Robinson May 12 '17 at 14:52

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