I have a question about the weighted norm inequality: The weighted norm of a vector $x\in R^{M\times N}$ is defined by: $\left \| X\right \|_{w,*} = \sum_{_{i}}\left |w_{i}\sigma _{i}\left ( X \right ) \right |$ where $w_{i}$ denotes entries of the weighted matrix and $\sigma_{i}\left ( X \right )=\sqrt{\lambda _{i}\left ( X^{T}X \right )}$ stands for the singular values of X.
Lemma: $\forall A, B \in R^{M\times N} $ satisfies $ A^{T}B=0$, we have: $\left \| A+B \right \|_{w,*}\geqslant \left \| A \right \|_{w,*}$
I have tried to prove the above lemma, but not success Please help me do it. Thank you very much.
The orthogonality condition $A^TB=0$ implies that $$\sigma_i^2(A+B)=\lambda_i((A+B)^T(A+B))=\lambda_i(A^TA+B^TB).$$ Since the matrices $A^TA$ and $B^TB$ are symmetric positive semi-definite, it follows from the Weyl theorem that $$\lambda_i(A^TA)\leq\lambda_i(A^TA+B^TB) \quad \Rightarrow \quad \sigma_i(A)\leq\sigma_i(A+B).$$ Now just plug these inequalities to your norm.