Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Just to avoid confusion, a function is called matrix monotone in an interval $[a, b]$ if $A - B \geq 0$ implies $f(A) - f(B) \geq 0$ for any Hermitian Matrices $A, B$ (we can restrict to finite dimensions) with spectrum in $[a, b]$. ($\geq 0$ means that the matrix is positive semi-definite)

I am currently interested in the function $f(t) = 1 - ( 1 - t^{1/2} )^2$ and the interval $[0, 1]$. The function is monotonically increasing and concave in this interval, so there is reason to hope that it is also operator monotone.

However, I do not know how to prove this and was wondering if there exists a general strategy to prove operator monotonicity of functions. Any ideas?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Operator monotonicity is rather tricky. Counterexamples will generally involve non-commuting matrices. On the other hand, you can assume one of the matrices is diagonal. Another clue is that often the extreme cases are the ones to look at. So I'll try $ B = \pmatrix{b & 0\cr 0 & 0\cr}$, $A = \pmatrix{a & c\cr c & d\cr}$, where $b = \dfrac{c^2+a^2-a}{c^2+a-1}$ and $d = 1 - \dfrac{c^2}{1-a}$ and $0 \le a \le 1 - c^2$. After some experimentation, I found that with, say, $a=4/5$ and $c=1/5$,

$$ A = \pmatrix{4/5 & 1/5\cr 1/5 & 4/5\cr},\ B = \pmatrix{3/4 & 0\cr 0 & 0\cr}$$ $$ f(A) = \pmatrix{\frac{1}{5}+\sqrt{\frac{3}{5}} & \frac{4}{5} - \sqrt{\frac35}\cr \frac{4}{5} - \sqrt{\frac35} & \frac{1}{5}+\sqrt{\frac{3}{5}}\cr},\ f(B) = \pmatrix{\sqrt{3}-\frac34 & 0\cr 0 & 0\cr}$$ and $f(A)_{11} < f(B)_{11}$. So this function is not operator monotone.

share|improve this answer
    
Thanks @Robert. This is very helpful and solves my immediate problem. Still, is there a generic strategy to answer the question whether a function is operator monotone in the positive? –  Marco Tomamichel Jun 15 '12 at 4:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.