0
$\begingroup$

The spectral $l^2$ norm of a complex matrix is given by:

$\|A\|= \left( \sum_{k=0}^{N-1} s_k(A)^2 \right)^{1/2}$ where $s_k(A)$ are the singular values of $A$ ordered so as to be non decreasing in $k$.

The weighted $l^2$ norm of real vector $x \in \mathbb{R}^N$ is given by: $\|x\|= \left( \sum_{k=0}^{N-1} w_k x_k^2 \right)^{1/2}$. Here the weights are taken to be such that $w_k \geq 0, \forall k$ and $\sum_{k=0}^{N-1} w_k = 1$. It is well know that this norm arises from inner product given by:

$\langle x, y \rangle = x^T W y$ where $W=\text{diag}(w)$ a diagonal matrix with the weights on the diagonal. One imposes that no weights are zero to avoid degeneracy.

If one defines (as is done in many places) the following matrix norm: $\|A\|= \left( \sum_{k=0}^{N-1} w_k s_k(A)^2 \right)^{1/2}$ (the condition that no weights are zero is no longer needed I don't think) does this arise from some inner product of the set of all complex matrices?

Further to that I'd like to know if it can arise from an inner product such that when restricted to $\mathfrak{su}(N)$ the inner product is always real valued.

$\endgroup$
3
$\begingroup$

I don't think it is even a norm. If you take $w_0 = 1$ and $w_k = 0$ for all $k\geq 1$ then you get $\|A\| = |s_{\min}(A)|$ and it does not satisfy the triangular inequality: $\|e_1e_1^T\| =0$, $\|e_2e_2^T\| =0$, but $\|e_1e_1^T+e_2e_2^T\| \neq 0$. Am I missing something?

$\endgroup$
  • 1
    $\begingroup$ No, your counterexample is fine. The $w_k$ corresponding to the largest singular values had better be positive, so if all $w_k=0$, we have a seminorm, if $w_n > 0$, and the rest are zero, we're ok, and so on. $\endgroup$ – Suvrit Nov 21 '13 at 20:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.