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A function $f(x)$ is matrix monotone if $f(A)-f(B)$ is positive semidefinite whenever $A-B$ is positive semidefinite for positive semidefinite matrices $A, B$. Is $\sqrt{1+x^2}$ matrix monotone?

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  • $\begingroup$ On what set of matrices ? Have you considered $A=-2$ and $B=1$ ? $\endgroup$
    – user10676
    Mar 6 '14 at 23:49
  • $\begingroup$ Thanks, I should require $A, B$ to be positive semidefinite. $\endgroup$
    – Fischer
    Mar 7 '14 at 2:37
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No. One may be tempted to expect $\sqrt{1+x^2}$ to behave like $x$. However, since $x^2$ is not matrix monotone, it may cause $\sqrt{1+x^2}$ to fail to be matrix monotone. I don't have a nice counterexample at hand, but if you have access to a numerical linear algebra package, you will see that when $$ A=\pmatrix{2.01&1\\ 1&3},\ B=\pmatrix{1\\ &2}, $$ we have $A\succ B\succ 0$ but the eigenvalues of $\sqrt{I+A^2}-\sqrt{I+B^2}$ are $1.83$ and $-8.34\times10^{-3}$.

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I think yes. My general thinking is that the positive definite property is tied to a matrix's determinate being non-zero, and that the square of a matrix will merely amplify the magnitude of the determinate, but not zero it out.

I emphasize that this is strictly a guess.

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  • $\begingroup$ $x^2$ is increasing over $[0, \infty)$, but it is not matrix monotone. $\endgroup$
    – Fischer
    Mar 7 '14 at 2:38

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