Let $M_n$ denote the set of all $n\times n$ matrices over complex and define the norm $\|\cdot\|_H$ on $M_n$, called Hilbert-Schmidt norm, by
$$\|A\|_H=\sqrt{\sum_{1\leq i, j\leq n}|a_{ij}|^2}$$
where $A=(a_{ij})$. On the other hand, $M_n$ has the topology induced by the uniform norm
$$\|A\|_U=\text{max}\{\|Ax\|:x\in\mathbb{C}^n ,\|x\|=1\}$$
where $\|\cdot\|$ is the Euclidean norm. I would like to know whether the Hilbert-Schmidt norm topology is weaker (or stronger) than the uniform norm topology or whether they are equivalent.
My attempt. I head these facts
$$\|A\|_U^2=\max\{\lambda_i:\text{$\lambda_i\in\mathbb{R}$ is an eigenvalue of $A^*A$}\}$$
and
$$\|A\|_H^2=\text{trace}(A^*A).$$
If they are true then we have
$$\|A\|_U^2\leq\sum_i\lambda_i=\text{trace}(A^*A)=\|A\|_H^2.$$
My question.
- Is the Hilbert-Schmidt norm independent of the choice of an orthonormal basis for $\mathbb{C}^n$?
- Are these facts true? How do I show them?
- Can we prove the other direction $\|A\|_H\leq\|A\|_U$?
Best Answer
All the norms of finite-dimensional spaces are equivalent and hence they produce the same topology.
Say $\|\cdot\|$ is an abritrary norm in $\mathcal M_n$, then the function $$ \|\cdot\| :\mathcal M_n\to\mathbb R, $$ is continuous, i.e., for every $A_0\in\mathcal M_n$ and $\varepsilon>0$, there exists a $\delta>0$, such that $$ \|A-A_0\|_H<\delta \qquad\Longrightarrow\qquad \|A-A_0\|<\varepsilon. $$ The set $\mathcal K=\{A\in\mathcal M_n:\|A\|_H=1\}$ is compact and hence $\|\cdot\|$ attains minimum and value on $K$, i.e. there exist $c_2\ge c_1>0$, such that $$ c_1\le \|A\|\le c_2 $$ for all $A\in \mathcal K$. Thus $$ c_1\|A\|_H\le \|A\|\le c_2\|A\|_H $$ for all $A\in \mathcal M_n$.