Norm of complex symmetric nilpotent matrix

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Let $A$ be a complex square matrix. We know that

  1. $A$ is symmetric but not Hermitian: $A = A^T$ and $A \neq A^*$.

  2. $A$ is nilpotent. In fact, we have $A^2 = 0$.

I need to bound the norm of $A$ in any way possible (lose bounds are acceptable!) Any bound on any norm will do, though I'm most interested in the infinite norm (i.e. the maximum row sum) and the operator norm.

The main problem here is that even though the eigenvalues of $A$ are all zero, the eigenvalues of $A^* A$ are unknown. (Recall the operator norm of $A$ is the square root of the largest eigenvalue of $A^* A$.)

Any help appreciated!

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The answer of user1551 is correct.

Indeed, let $A$ be your complex-valued symmetric, nilpotent matrix. Then, we may define $B=\alpha A$ where $\alpha \in \mathbb{C}$. We can easily check that $B^2 = \alpha^2 A^2 = 0$ and $B^T = \alpha A^T = \alpha A = B$.

If we assume the norm $||A|| = K$, then $||B|| = ||\alpha A|| = |\alpha| ||A|| = |\alpha| K$. Because we have not stated anything about $\alpha$, we can assume it to be a large number, and thus, the norm is unbounded.

Example:

$A = \begin{pmatrix} -i\alpha & \alpha \\ \alpha & i \alpha \end{pmatrix}$, $\alpha \in \mathbb{R}$.

Then, $A$ is a complex-valued symmetric, nilpotent matrix, but you can easily check that

$A^* A = \begin{pmatrix} 2 \alpha^2 & -2 i \alpha^2 \\ 2 i \alpha^2 & 2 \alpha^2 \end{pmatrix}$ , which has eigenvalues $0$ and $4\alpha^2$.

Again, $\alpha$ can be a large number.

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Answering my own question. I found this paper useful. In particular, part (2) of Theorem 3.8 says that for a complex symmetric matrix, $\lambda$ is a singular value of $A$ if and only if the problem $$ A x = \lambda \bar{x} $$ has nonzero solution $x$. In my particular case, I was able to use this to bound the operator norm of $A$ (i.e. the largest $\lambda$ that satisfies this equation). In the end, I did not use the fact that $A$ is nilpotent, so I suspect that this result may be strengthened with more work.