Let $A$ be a complex square matrix. We know that
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$A$ is symmetric but not Hermitian: $A = A^T$ and $A \neq A^*$.
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$A$ 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!
Best Answer
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.