I was reading a solution of this question
Let $T:V \to V$ and $\|T\|=2\|T^*\|$ over inner product space, then $T=0$
The author solves it when he shows that $tr(T^*T)=0$ and then chooses an orthogonal basis $B$ of $V$ then defines $A:=[T]_B$ and shows that
$$trace(A^*A)=\sum_{i,j}|a_{ij}|=0$$ And therefore $T=0$.
My question is, is there a generalized theorem that given $T,trace (T^*T)$ that can say something about what $T$ looks like?
Off the top of my head, there are two useful facts about $\operatorname{tr}(T^*T)$.
Fact 1: The function $\|T\|_F = \sqrt{\operatorname{tr}(T^*T)}$ defines the "Frobenius norm", also known as the Hilbert-Schmidt norm. Moreover, this norm is induced by the inner product $\langle S,T \rangle = \operatorname{tr}(T^*S)$. This leads to the Cauchy-Schwarz inequality $$ |\operatorname{tr}(T^*S)|^2 \leq \operatorname{tr}(S^*S)\operatorname{tr}(T^*T) $$
Fact 2: In general, we have $$ \operatorname{tr}(T^*T) \geq \sum_{j=1}^n |\lambda_j(T)|^2 $$ and this equality is attained if and only if $T$ is a normal matrix (i.e. $TT^* = T^*T$).