# Trace distance between “weighted” Hermitian matrices

The trace norm for a matrix $\mathbf{A}$ (also known as Shatten 1-norm) is defined as follows: $\|\mathbf{A}\|_1=\operatorname{trace}[\sqrt{\mathbf{A}^*\mathbf{A}}]$. It yields a useful distance measure between matrices $\mathbf{A}$ and $\mathbf{B}$, called trace distance: $\|\mathbf{A}-\mathbf{B}\|_1$.

Suppose that $\mathbf{A}$ and $\mathbf{B}$ are Hermitian, and $a,b>0$ are positive scalars. Is the following true:

$$\|a\mathbf{A}-b\mathbf{B}\|_1\leq \max(a,b)\|\mathbf{A}-\mathbf{B}\|_1?$$

No, it cannot be true. Take $A=B$ but $a\ne b$. Then $$\|aA-bB\|_1 = |a-b| \|A\|_1 \not\le \max(a,b) \|A-B\|_1 = 0.$$