I have a matrix $A$ and I wish to say that $\|Ax\|_2 \leq c\|x\|_2$ for all $x$, with some $0 \leq c \leq 1$. $A$ is asymmetric, and I have access to the eigenvalues, but not the singular values of $A$. In fact, after checking, it is not the case that $\sqrt{\sigma_{\max}(A^TA)} =\max_{i}|\lambda_{i}(A)|$.
In general, it is known that we can only guarantee $\sqrt{\sigma_{\max}(A^TA)} \geq \max_{i}|\lambda_{i}(A)|$ for asymmetric matrices, with a counterexample that shows equality is not guaranteed being counterexample
$$
A = \left[\begin{matrix} 1 & 1+\epsilon \\ 1 & 1 \end{matrix}\right].
$$
However, it seems that, since any $x$ can be written as a linear combination of an orthogonal basis, then using the eigenvalue decomposition $A = U D U^T$, then writing $x = Uv$,
$$
\|Ax\|_2 = \|UDU^TUv\|_2 = \|Dv\|_2 \leq \max_{i}|D_{ii}| \|v\|_2.
$$
Therefore I am still tempted to say that this contraction holds with $C = \max_{i}|\lambda_{i}(A)|$.
What's wrong with my logic?
Best Answer
You can't unitarily diagonalize unless $A$ is normal. It doesn't have to be symmetric but it does have to be normal. Your counterexample was not normal.