Is it true that $A \geq B$ implies $\|A\|_2 \geq \|B\|_2$ for $A,B \geq 0$? All matrices are real and not necessarily symmetric. Denote by $A \geq B$ the condition that $(A-B)$ has eigenvalues with non-negative real parts. Denote by $\| \cdot \|_2$ the $L_2$ matrix norm.
Is it true that $A \geq B$ implies $\|A\|_2 \geq \|B\|_2$ for $A,B \geq 0$?

Edit: Now I see it doesn't hold in general, I would also be grateful if someone could provide additional conditions on $A$ and $B$ for which $\|A\|_2 \geq \|B\|_2$ would hold.
I am particularly interested in whether the statement can still be rescued the following cases:


*

*$A,B$ symmetric.

*$A = \Xi$, $B = \Xi P$ where $P$ is a stochastic matrix (rows sum to 1) and $\Xi$ a diagonal matrix with the principal left eigenvector of $P$ (i.e. stationary distribution) on the diagonal.


Not necessarily.  For example, we can take
$$
A = \pmatrix{1&0\\0&1}, \quad B = \pmatrix{1&1\\0&1}
$$
the only eigenvalue of $A-B$ is $0$, so $A \geq B$.  However, $\|A\|_2 = 1 < \|B\|_2$.

Regarding your update: when $A,B$ are symmetric, this amounts to showing that if $A,B,$ and $A-B$ are positive semidefinite, then $\|A\|_2 \geq \|B\|_2$.
Note that
$$
\|B\|_2 = \sup_{\|x\| = 1}x^*Bx \leq 
\sup_{\|x\| = 1}\left(x^*Bx + x^*(A-B)x\right) = \sup_{\|x\| = 1}x^*Ax = \|A\|_2
$$
