When considering matrices (or differential operators), $A > B \implies \exp(A) > \exp (B)$?

Question

First, I would like to ask what is the conventionnal definition of $ A > B $ for matrices (or differential operators in my case of interest).

I guess that a natural definition would be that $A-B$ should be positive definite. $$ A > B \iff A - B >0 \iff \langle \psi | (A - B) | \psi \rangle > 0 \ \ \forall \ |\psi \rangle$$ But if there are any equivalent definition, I would be happy to hear.

My question is, from this definition (or an equivalent one), is it true that $$ A > B \implies \exp(A) > \exp (B) $$ and if so, how to prove it ?

Answer 1

Try:

$$ A = \pmatrix{2 & 1\cr 1 & 1\cr}, \ B = \pmatrix{0.99 & 0\cr 0 & 0\cr}$$