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:

  1. $A,B$ symmetric.
  2. $A = \Xi$, $B = \Xi P$ 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.
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    $\begingroup$ For symmetric matrices $\|A\|_2$ is equal to the largest eigenvalue, hence the statement holds for symmetric matrices. $\endgroup$ – daw Dec 1 '14 at 7:57
  • $\begingroup$ @daw Could you explain on why it is true that in the symmetric case "eigenvalues of $A-B, A, B$ are nonnegative implies maximum eigenvalue of $A$ is greater or equal maximum eigenvalue of $B$". The point I don't understand is the connection between eigenvalues of $A-B$ being positive and the spectral radius of $A$ being greater than the spectral radius of $B$. $\endgroup$ – ziutek Dec 1 '14 at 10:53
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    $\begingroup$ @ziutek see my updated answer. $\endgroup$ – Omnomnomnom Dec 1 '14 at 19:08

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 $$ \|B\|_2 = \sup_{\|x\| = 1}x^*Bx \leq \sup_{\|x\| = 1}\left(x^*Bx + x^*(A-B)x\right) = \sup_{\|x\| = 1}x^*Ax = \|A\|_2 $$

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    $\begingroup$ +1 for the first example, but your second doesn't satisfy $A,B\ge0$. $\endgroup$ – Rahul Nov 30 '14 at 16:19
  • $\begingroup$ @Rahul thanks for pointing that out $\endgroup$ – Omnomnomnom Nov 30 '14 at 16:20
  • $\begingroup$ @Omnomnomnom Thanks for explaining the symmetric case. $\endgroup$ – ziutek Dec 2 '14 at 9:16

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