Understanding Operator Norm of Matrices

Question

Let $X$ denote the vector space of $n\times n$ complex matrices. To every matrix $A\in X$ one can associate two operator norms:

1. Thinking of $A$ as a map $A\colon \mathbb{C}^n\to \mathbb{C}^n$ or $A\in L(\mathbb{C}^n)$ in short, we define $\|A\|_{L(\mathbb{C}^n)}=\sup_{v\in \mathbb{C}^n, \|v\|=1}{\|Av\|}$.
2. Thinking of $A$ as a map $A\colon X\to X$ or $A\in L(X)$ in short, we define $\|A\|_{L(X)}=\sup_{B\in X, \|B\|_{L(\mathbb{C}^n)}=1}{\|AB\|}$.

It is well-known that $\|A\|_{L(\mathbb{C}^n)}=s_1(A)$ where $s_1$ is the largest singular value of $A$. My questions are:

1. Is there is a (spectral?) characterization of $\|A\|_{L(X)}$? or perhaps a relation between the two above norms?
2. Is there any application of the second norm (in, say functional analysis)? I will explain below why I started thinking about this norm in the first place.
3. I was trying to understand the following paragraph that lead to my questions above. "If $X$ is equipped with the operator norm $\|\cdot\|_{L(\mathbb{C}^n)}$, then its dual space $X^*$ is the space $X$ equipped with the trace norm $\|\cdot\|_1=\text{ sum of singular values}$, and vice versa."

I still don't see why (3) has to be true, but if necessary, and if it turns out to be irrelevant to the other two questions, I can ask it in a separate question.

Any help is appreciated.

Answer 1

I will assume that the norm $\|AB\|$ in the definition of $\|A\|_{L(X)}$ is $\|AB\|_{L(\mathbb C^n)}$.

The two norms are equal: you have, by definition, $$ \|A\|_{L(X)}\leq \|A\|_{L(\mathbb C^n)}, $$ since $\|AB\|_{L(\mathbb C^n)}\leq \|A\|_{L(\mathbb C^n)}\,\|B\|_{L(\mathbb C^n)}=\|A\|_{L(\mathbb C^n)}$.

Conversely, $$ \|A\|_{L(X)}\geq\left\|A\,\frac{A^*}{\|A^*\|}\right\|_{L(\mathbb C^n)} =\frac{\|AA^*\|_{L(\mathbb C^n)}}{\|A\|_{L(\mathbb C^n)}} =\frac{\|A\|^2_{L(\mathbb C^n)}}{\|A\|_{L(\mathbb C^n)}} =\|A\|_{L(\mathbb C^n)}. $$

Regarding the dual $X^*$, it is not hard to see that one can write $$ X^*=\{\text{Tr}(B\,\cdot):\ B\in X\}, $$ giving the identification of every functional in $X^*$ with a matrix $B$. Now let us write $f_B=\text{Tr}(B\cdot)$ and let us look at the norm: $$ \|f_B\|=\sup\{|f_B(A)|:\ \|A\|_{L(\mathbb C^n)}=1\} =\sup\{|\text{Tr}(BA)|:\ \|A\|_{L(\mathbb C^n)}=1\} =\|B\|_1. $$

Here is a proof of the last equality: write $B=UDV$ the singular value decomposition. Using that unitaries preserve the norm, $$ \sup\{|\text{Tr}(BA)|:\ \|A\|_{L(\mathbb C^n)}=1\} =\sup\{|\text{Tr}(DA)|:\ \|A\|_{L(\mathbb C^n)}\} =\sup\{\left|\sum_jD_{jj}A_{jj}\right|:\ \|A\|_{L(\mathbb C^n)}=1\}= \sum_jD_{jj}=\|B\|_1. $$