In my research, I need a lemma that gives the lower bound of the perturbation of nuclear norm. It is supposed to be
Statement. For any real matrix $M_{n\times n}$ and diagonal matrix $\Lambda_{n\times n}$ with all diagonal entries in $[0,1]$, $$\|M\|_*-\|\Lambda M\Lambda\|_*\geq\sigma_n(M)(n-\|\Lambda\|_F^2),$$ where $\sigma_n(M)$ is the smallest singular value of $M$ and $\|\cdot\|_*$ denotes nuclear norm.
It's easy to prove the statement for the case when $M$ is symmetric and positive semidefinite. Besides, for the general case, I've down simulations with $n\in\{2,3,10,60\}$ and saw results suggesting this statement being valid.
Most existing perturbation theories I found are concerning upper bounds, while I need a lower bound. I tried to prove this statement from scratch with many trials, but only failed so far. I would really appreciate it if anyone could suggest any books, papers or discussions that concern a similar problem or anything helpful?
Remark: An equivalent inequality of the one in the statement is $$\|M\|_*-\|\Lambda M\Lambda\|_*\geq\|\sigma_n(M)I\|_*-\|\Lambda\sigma_n(M)I\Lambda\|_*.$$
I just proved this statement.
Proof. We only need to prove that in case $\sigma_n(M)=1$, $\|M\|_*-\|\Lambda M\Lambda\|_*\geq n-\|\Lambda\|_F^2$. In this case, denote the SVD of $M$ by $Q_1\Sigma Q_2$ then $$\text{LHS}=\text{tr }\Sigma-\Lambda Q_1\Sigma Q_2\Lambda Q_3=\text{tr }\Sigma(I-Q_2\Lambda Q_3 \Lambda Q_1), \text{ for some orthogonal matrix }Q_3.$$ Let $H=Q_2\Lambda Q_3 \Lambda Q_1$, then all diagonal entries of $H$ has the form $\xi^T\Lambda Q_3\Lambda\eta$ with $\|\xi\|_2=\|\eta\|_2=1$, which is bounded above by $\|Q_3\|_{\text{op}}=1$.Thus, all the diagonal entries of $I-H$ is non-negative. This implies that $$\text{LHS}\geq\text{tr }(I-Q_2\Lambda Q_3 \Lambda Q_1)=\text{tr }(I-\Lambda Q_3 \Lambda Q_4^T )=n-\langle\Lambda Q_3,Q_4\Lambda \rangle\geq n-\|\Lambda Q_3\|_F\|Q_4\Lambda\|_F=n-\|\Lambda\|_F^2.$$