Hilbert-Schmidt norm vs Uniform norm 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 $\|A\|_H\leq\|A\|_U$?

 A: 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 $A\in \mathcal M_n$.
A: All norms on a finite dimensional vector space are equivalent.
One strategy to prove it is that we prove that all norms on $\Bbb{R}^{n}$ or $\Bbb{C}^{n}$ are equivalent by using the compactness of the unit sphere by Heine Borel Theorem. And then for the general case, we fix a basis say $\{v_{k}\}$ of $V$ and then define $T(e_{k})=v_{k}$ for the standard ordered basis for $\Bbb{C}^{n}$. Then create a norm $p(x)=||T^{-1}(x)||$ on the given space $V$. Then given any norm say $q$ on $V$. You define a norm $n$ on $\Bbb{C}^{n}$ by setting $n(z)=q(Tz)$ for $z\in\Bbb{C}^{n}$. Now by equivalence with say the euclidean norm on $\Bbb{C}^{n}$ we prove that $p$ and $q$ are equivalent. Thus any norm on $V$ will be equivalent to $p$.
Hilbert Schmidt norm is independent of the choice of an orthonormal basis. This is true for general Hilbert Spaces
Let $\{e_{n}\}$ ,$\{f_{n}\}$  and $\{g_{n}\}$ be orthonormal bases.
Then let $$\sum_{n} ||Ae_{n}||^{2} =\sum_{m,n}\big|\langle Ae_{n},f_{m}\rangle\big|^{2} = \sum_{m,n}\big|\langle e_{n},A^{*}f_{m}\rangle\big|^{2}=\sum_{m}||A^{*}f_{m}||^{2}$$
$$=\sum_{n,m}\big|\langle A^{*}f_{m},g_{n}\rangle\big|^{2} = \sum_{n,m}\big|\langle f_{m},Ag_{n}\rangle\big|^{2}=\sum_{m,n}\big|\langle Ag_{n},f_{m}\rangle\big|^{2}=\sum_{n}||Ag_{n}||^{2}$$
Thus it is independent of the choice of basis.
So to answer your first question, yes, the hilbert schmidt norm for $M_{n}$ is just a norm on a finite dimensional space and hence all norms on it are equivalent to the hilbert schmidt norm.
For the infinite dimensional case, the equivalence may fail. But you are guaranteed that the Hilbert Schmidt norm dominates the operator norm.
Proof that all norms on $\Bbb{K}^{n}$ are equivalent. ($\Bbb{K}$ can mean either $\Bbb{R}$ or $\Bbb{C}$).
Let $n$ be any norm on $\Bbb{K}^{n}$. Let $\{e_{k}\}$ denote the standard basis and let $M=\max(|n(e_{k})|)$ . If $x=\sum_{k=1}^{n}x_{k}e_{k}$ then $n(x)\leq n(\sum_{k=1}^{n}x_{k}e_{k})\leq M\sum_{k=1}^{n}|x_{k}|=Mn||x||$ where $||.||$ denotes the Euclidean norm.
Let $S=\{x\in \Bbb{K}^{n}:||x||=1\}$
Now define $f:S\to \Bbb{R}$ by setting $f(x)=n(x)$ . Then $f$ is continuous (as by the above inequality, f is Lipschitz)
So $f$ is a continuous function on a compact set and hence achieves it's minimum. Note that the minimum has to be a strictly positive number say $C>0$ as $n(x)>0$ if $x\neq 0$
So $n(\frac{x}{||x||})\geq C\,,\forall x\in V$ . So $n(x)\geq C||x||$ . This shows that $n$ is equivalent to the euclidean norm.
