The spectral norm of $A$ is bigger than that of $B$, then $A-B$ is positive semidefinite

Question

Given are $A$ and $B$ symmetric $n\times n$ matrices.

If $A - B$ is positive semi-definite, then the spectral norm of $A$ is bigger than that of $B$.

The question is that:

Given then the spectral norm of $A$ is bigger than that of $B$, can we derive $A - B$ is positive semi-definite?

Many thanks.

Answer 1

No.

Let $A = \left[\begin{array}{cc}3 & 0 \\ 0 & 1\end{array}\right]$, $B = \left[\begin{array}{cc}1 & 0 \\ 0 & 2\end{array}\right]$. Then $\|A\|_2 = 3$, $\|B\|_2 = 2$. Thus $\|A\|_2 > \|B\|_2$. But $C = A - B = \left[\begin{array}{cc}2 & 0 \\ 0 & -1\end{array}\right]$ is not semi-positive definite.