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I have an anti-triangular matrix that is not Hermitian (it is complex and symmetric). I would like to find a method to bound its largest singular value. Might that exist, in general?

Anti-triangular matrices (Skew-triangular (?) matrices and their properties) have vanishing entries below the main anti-diagonal, such as \begin{equation}M=\begin{pmatrix}m_{00}& m_{01}&m_{02} \\ m_{10}& m_{11}&0 \\ m_{20}& 0&0 \end{pmatrix}.\end{equation} The determinant and thus the product of the eigenvalues is easy to determine, with $\mathrm{det}(M)=(-1)^d\prod_{i=0}^d m_{i,d-i}$ only depending on the main anti-diagonal, but the eigenvalues themselves seem more elusive. Does the problem simplify if one is only interested in a single eigenvalue, the one with the largest absolute value?

Other questions (Finding the largest singular value "easily") point to things like the Rayleigh quotient, but I believe that will not help for my non-Hermitian matrices. Also, my matrices are not numerical, so I would prefer to avoid power iteration. If the latter is the only option, it seems to imply that my anti-triangular structure conveys no benefits. Unless, perhaps, there are simplifications to power iteration for anti-triangular matrices? The reason I doubt that is because $M M^\dagger$ is, in general, a full matrix.

Quantum Mechanic
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    singular values are preserved when you multiply by permutation matrices so do that to convert your matrix into a triangular one $T$. Then consider the case of $T$ being nilpotent and see the eigenvalues cannot be used to bound the largest singular value. – user8675309 May 17 '24 at 17:38
  • @user8675309 does that mean the singular values are all given by the absolute values of the components of the main anti-diagonal? I was worried that eigenvalues needed similarity transformations and not just row permutations, but you made me realize that singular values don't have that restriction – Quantum Mechanic May 17 '24 at 17:52
  • No -- reread my second sentence. – user8675309 May 17 '24 at 17:58
  • @user8675309 okay thank you, perhaps you can point out my mistake. If I permute $M$ to $T$ I get a triangular matrix whose diagonal entries are the anti-diagonal of $M$. The eigenvalues of $T$ are those diagonal entries and the singular values of $T$ are the absolute values of those eigenvalues, even if $T$ is nilpotent and the diagonal entries all vanish. Since the singular values of $T$ are the same as those of $M$, for nilpotent $T$ we have all the singular values of $M$ also vanishing, if and only if $M$ has vanishing main anti-diagonal – Quantum Mechanic May 17 '24 at 18:25
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    First prove to yourself that a matrix has all singular values $=0$ iff it is the zero matrix. Then come up with a strictly upper triangular (nilpotent) matrix that isn't the zero matrix. – user8675309 May 17 '24 at 18:27
  • @user8675309 oh I get it, the singular values of $T$ are not the absolute values of $T$'s eigenvalues – Quantum Mechanic May 17 '24 at 18:30

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