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Let $L$ be a general (possibly non-Hermitian) square matrix and define \begin{equation} A := e^{-i L}. \end{equation}

I am interested in understanding how the singular values of $A$ relate to the properties of $L$.

First, I know that the singular values of a matrix $A$ are the square roots of the eigenvalues of $A^\dagger A$. So in this case, \begin{equation} \sigma_j(A) = \sqrt{\lambda_j(A^\dagger A)} = \sqrt{\lambda_j\big(e^{i L^\dagger} e^{-i L}\big)}. \end{equation}

Second, if $L$ is normal ($[L, L^\dagger] = 0$), then $L$ is unitarily diagonalizable, and the singular values simplify to \begin{equation} \sigma_j(A) = \exp(\operatorname{Im} \lambda_j), \end{equation} where $\lambda_j$ are the eigenvalues of $L$.

My questions are:

  • Is there a similarly simple characterization of the singular values of $e^{-i L}$ for general, non-normal $L$?
  • Are there useful bounds for $\sigma_{\rm min} (e^{-i L})$ in terms of eigenvalues or other properties of $L$?
  • Are there standard references discussing the singular values of matrix exponentials for non-Hermitian matrices?

I would appreciate any references, insights, or known results on this problem.

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  • $\begingroup$ No, there is no simple characterization in the general case. From searching I found this paper, which gives the following bounds: for a square matrix $A$, $$ \sigma_1(e^A)\cdots \sigma_k(e^A) \leq e^{\sigma_1(A)} \cdots e^{\sigma_k(A)}\\ \sigma_1(e^A) + \cdots + \sigma_k(e^A) \leq e^{\sigma_1(A)} + \cdots + e^{\sigma_k(A)}. $$ I'm not sure if these are directly useful, though. I'm not aware of any standard references here $\endgroup$ Commented Oct 19 at 21:10

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No, there is no simple characterization.

We can get a lower bound on $\sigma_\min(e^{-iL})$ using the fact that $\sigma_\max(e^A) \leq e^{\sigma_\max(A)}$: $$ \sigma_\min(\exp(-iL)) = \frac 1{\sigma_\max(\exp(iL))} \geq \frac 1{\exp(\sigma_\max(iL))} = \exp(-\sigma_\max(L)), $$ perhaps that bound is helpful.

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  • $\begingroup$ Thanks @Ben Grossmann. Wouldn't it mean that even if $\sigma_{max}(L)=0$, one would still have $\sigma_{min}(exp(-i L)) = 1$! How can I understand this in simple terms? $\endgroup$ Commented Oct 20 at 13:03

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