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.
Best Answer
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 $\{ |\sin(\beta_{j})|: 1 \leq j \leq n \}.$