Monotony of the operator exponential $e^A$ I have read somewhere that if $A$ and $B$ are two self-adjoint operators such that $A\geq B$, then we need not have $e^A\geq e^B$. I have tried to find counterexamples using $2\times2$ matrices as well as some online calculators, but without success! I also know that we should avoid commuting $A$ and $B$ for a possible counterexample? 
My query is a simple counterexample and whether there is a difference between the finite and the infinite dimensional cases?
Thanks!
Math.
 A: I am assuming that you define the relation $\geq$ by $\left(  X\geq Y\right)
\Longleftrightarrow\left(  X-Y\text{ is nonnegative semidefinite}\right)  $.
Let $C=\left(
\begin{array}
[c]{cc}
1 & 1\\
1 & 1
\end{array}
\right)  $ and $B=\left(
\begin{array}
[c]{cc}
0 & 0\\
0 & 2
\end{array}
\right)  $. Set $A=B+C$. Then, $A\geq B$, since $A-B=C$ is nonnegative
semidefinite. (Also, $A$, $B$ and $C$ are nonnegative semidefinite.) But
\begin{equation}
e^{A}-e^{B}=\left(
\begin{array}
[c]{cc}
e^{2-\sqrt{2}}\left(  \dfrac{1}{4}\sqrt{2}+\dfrac{1}{2}\right)  -e^{\sqrt
{2}+2}\left(  \dfrac{1}{4}\sqrt{2}-\dfrac{1}{2}\right)  -1 & -e^{\sqrt{2}
+2}\left(  \sqrt{2}+1\right)  \left(  \dfrac{1}{4}\sqrt{2}-\dfrac{1}
{2}\right)  -e^{2-\sqrt{2}}\left(  \sqrt{2}-1\right)  \left(  \dfrac{1}
{4}\sqrt{2}+\dfrac{1}{2}\right)  \\
\dfrac{1}{4}\sqrt{2}e^{\sqrt{2}+2}-\dfrac{1}{4}\sqrt{2}e^{2-\sqrt{2}} &
\dfrac{1}{4}\sqrt{2}e^{\sqrt{2}+2}\left(  \sqrt{2}+1\right)  -e^{2}+\dfrac
{1}{4}\sqrt{2}e^{2-\sqrt{2}}\left(  \sqrt{2}-1\right)
\end{array}
\right)
\end{equation}
has determinant
\begin{align}
-\dfrac{1}{4}\left(  e^{\sqrt{2}}-1\right)  \dfrac{e^{2}\left(  \sqrt
{2}-2\right)  -e^{4}\left(  \sqrt{2}+2\right)  +e^{2}e^{\sqrt{2}}\left(
\sqrt{2}+2\right)  -e^{4}e^{\sqrt{2}}\left(  \sqrt{2}-2\right)  }{e^{\sqrt{2}
}}\approx-8.436 < 0
\end{align}
and thus fails to be nonnegative semidefinite; thus, $e^{A}\geq e^{B}$ does
not hold.
