Trace-class norm and rank inequality

Question

I am quite new to operators in Hilbert spaces and I have been trying to show that for any linear and bounded operator $T : \mathcal{H} \rightarrow \mathcal{H}$ $$\|T\|_1 \leq \operatorname{rank}(T) \,\|T\|$$ where $\| T \|_1$ is the trace-norm of $T$, $\| T \|$ is its operator norm and $\operatorname{rank}(T)$ is the dimension of $\operatorname{Im}(T)$.

The idea is simple: the trace-norm is the sum of the eigenvalues of $T$, while the right part is the largest eigenvalue multiplied by the number of non-zero eigenvalues and so the inequality holds.

How can one prove this formally?

Answer 1

This can be proven in two steps:

1. $T$ is not trace class

Then $\|T\|_1=\infty$, and so is ${\rm rank}(T)$, because every finite-rank operator is trace class. Also $T\neq 0$, so $\infty\leq\infty\cdot\|T\|=\infty$ which concludes this step.

2. $T$ is trace class

If $T=0$, then $0\leq 0$ and we are done. For $T\neq 0$ we work with the Schmidt representation/singular value decomposition $T=\sum_{j\in J}s_j\langle f_j,\cdot\rangle g_j$ with $J\subseteq\mathbb N$, a unique decreasing null sequence $(s_j)_{j\in J}\subset(0,\infty)$, and two orthonormal systems $(f_j)_{j\in J}$, $(g_j)_{j\in J}$ where, notably, the largest of the $s_j$ equals the operator norm of $T$. Also from this form it is clear that $|J|={\rm rank}(T)$ so $$ \|T\|_1=\sum_{j\in J}s_j\leq\sum_{j\in J}\big(\max_{k\in J}s_k\big)=|J|\big(\max_{k\in J}s_k\big)={\rm rank}(T)\|T\| $$ and we are done.