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.
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.