$f(t) = \min\{1,t\}$ not operator monotone

Question

I want to show that the function on $\mathbb{R}^+$ $f(t) = \min \{1,t\}$ is not operator monotone on the complex $2\times 2$ matrices. My plan is to find matrices $A$ and $B$, $B\geq A$, such that the spectrum of $A$ is in $[0,\infty]$ and the spectrum of $B$ contains elements greater than 1. Then applying $f$ should 'make $B$ smaller' such that $f(B) \not \geq f(A)$. However, I struggle to find a concrete example. Can someone help me finding such $A$ and $B$?

Answer 1

You are on the right track. Actually any pair of matrices such that

1. $A$ has spectrum on $[0, 1]$,
2. $B$ has an eigenvalue on $(1, \infty)$, and
3. $B - A$ is of rank $1$
4. $A$ and $B$ do not commute

will work as an example.

Indeed, $f(t) \leq t$ together with 2. imply that $B \neq f(B) \leq B$. Since also $f(A) = A$, $$ f(B) - f(A) = f(B) - A \leq B - A. $$ Since the RHS is of rank $1$, and the second inequality is not equality, the only way the LHS can be positive is that $f(B) - f(A) = c(B - A)$ for some $0 \leq c < 1$, or $$ A = \frac{1}{1 - c} \left(f(B) - c B\right). $$ But this implies that $A$ commutes with $B$, contradiction.

As a concrete example one may for instance take $$ A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},\;\; B = A + \begin{bmatrix} \frac12 & \frac12 \\ \frac12 & \frac12 \end{bmatrix} = \begin{bmatrix} \frac32 & \frac12 \\ \frac12 & \frac12 \end{bmatrix}. $$ Again $f(A) = A$. One may calculate $f(B)$ using the definitions directly but the task might be slightly easier using the following observation: if $f$ and $g$ agree on the spectrum of $B$, then $f(B) = g(B)$. One may check that the eigenvalues of $B$ are $1 \pm \frac{1}{\sqrt{2}}$, so it suffices to find a polynomial $p$ with $$ p\left(1 - \frac{1}{\sqrt{2}}\right) = 1 - \frac{1}{\sqrt{2}}\; \text{ and } p\left(1 + \frac{1}{\sqrt{2}}\right) = 1; $$ then $f(B) = p(B)$. Degree $1$ example $p(x) = \frac{1}{2} x + \frac{1}{2} - \frac{1}{2 \sqrt{2}}$ will do, and thus $$ f(B) = p(B) = \frac{1}{2} \begin{bmatrix} \frac32 & \frac12 \\ \frac12 & \frac12 \end{bmatrix} + \left(\frac{1}{2} - \frac{1}{2 \sqrt{2}} \right) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ = \begin{bmatrix} \frac54 - \frac{1}{2 \sqrt{2}} & \frac{1}{4} \\ \frac{1}{4} & \frac34 - \frac{1}{2 \sqrt{2}} \end{bmatrix}. $$ Finally, $$ f(B) - f(A) = \begin{bmatrix} \frac14 - \frac{1}{2 \sqrt{2}} & \frac{1}{4} \\ \frac{1}{4} & \frac34 - \frac{1}{2 \sqrt{2}} \end{bmatrix} \not\geq 0, $$ as the determinant of the matrix is $\frac{1}{4} \left(1 - \sqrt{2}\right) < 0$.