let the matrix $A,B$ such $B-A,A$ is Positive-semidefinite

show that:

$\sqrt{B}-\sqrt{A}$ is Positive-semidefinite

maybe The general is true?

question 2:

(2)$\sqrt[k]{B}-\sqrt[k]{A}$ is Positive-semidefinite

This problem is very nice,because we are all know this if $$x\ge y\ge 0$$,then we have $$\sqrt{x}\ge \sqrt{y}$$

But in matrix,then this is also true,But I can't prove it.Thank you

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    $\begingroup$ How do you define $\sqrt A$ and $\leq$? $\endgroup$ – Shuchang Nov 29 '13 at 15:28
  • $\begingroup$ By $\sqrt{A}$, do you mean the unique positive square root? But a matrix with positive determinant need not have one... $$\left( \begin{matrix} -1&0\\0&-1 \end{matrix} \right)$$ has no positive square root, though it has a positive determinant. $\endgroup$ – Eric Auld Nov 29 '13 at 15:31
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    $\begingroup$ Imo the correct way to state the question is: is the function $f(z)=z^s$ operator monotone? The answer is yes (but I don't have time now) for $s\in [0,1]$. Interestingly, $f(t)=t^2$ is not. $\endgroup$ – Julien Nov 29 '13 at 15:50
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    $\begingroup$ Do you assume symmetry in the definition of $A$ and $B$? That is, do you implicitly assume that $A$ and $B$ are symmetric given that they are positive semi-definite? $\endgroup$ – Tom Nov 29 '13 at 16:00
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    $\begingroup$ $A,B$ is Positive-semidefinite mean is symmetry $\endgroup$ – math110 Nov 29 '13 at 16:06

Two proofs: (thanks for Julien for helping me fix the first proof)

First proof:

First assume $A,B>0$. Suppose $B \geq A$. $$B-A \geq 0 \\ A^{-1/2}BA^{-1/2} - I \geq 0, $$

since $M\geq 0 \Rightarrow N^{*}MN \geq 0$ for $N$ invertible. This means all eigenvalues of $A^{-1/2}BA^{-1/2}$ are strictly greater than 1. Therefore for all $\|x\|=1$

$$\langle A^{-1/2}BA^{-1/2}x,x \rangle \geq 1 \\ \langle BA^{-1/2}x, A^{-1/2}x \rangle \geq 1 \\ \langle B^{1/2}A^{-1/2}x, B^{1/2}A^{-1/2}x \rangle = \|B^{1/2}A^{-1/2} x\|^2\geq1 \\ \|B^{1/2}A^{-1/2} x\|\geq1$$ so all eigenvalues of $B^{1/2}A^{-1/2}$ are strictly greater than 1 in modulus. But by a similarity transformation, these are also the eigenvalues of $A^{-1/4}B^{1/2}A^{-1/4}$, and those are all positive since $A^{-1/4}B^{1/2}A^{-1/4}$ is positive (being of the form $N^* M N$ for $M>0$ and $N$ invertible). So

$$A^{-1/4}B^{1/2}A^{-1/4}-I \geq0\\B^{1/2}-A^{1/2}\geq0,$$

finishing the case $A,B>0$. To extend to $A,B\geq 0$, let $A_\epsilon:=A+\epsilon I$, and similarly for $B_\epsilon$. Then $B\geq A \Rightarrow B_\epsilon \geq A_\epsilon \Rightarrow (B_\epsilon)^{1/2}\geq (A_{\epsilon})^{1/2} \Rightarrow B^{1/2}\geq A^{1/2}$, if you believe that $\epsilon \mapsto M_\epsilon^{1/2}-N_{\epsilon}^{1/2}$ is right-continuous in the eigenvalues at zero (follows from $\epsilon \mapsto M_\epsilon^{1/2}$ being right-continuous in the eigenvalues).

Second Proof (from Lax's book Linear Algebra and its Applications): This relies on the fact that if $A(t)$ is a matrix-valued function whose derivative is everywhere positive definite, then $t_2 > t_1 \Rightarrow A(t_2) > A(t_1)$ (unstrict inequality corresponds to positive semidefiniteness). Also it uses the lemma that if $A$ is positive definite and $AB+BA$ is positive definite, then $B$ is positive definite. And it uses that positive semidefinite matrices are a convex subset of $\mathbb{C}^{n^2}$ (all these proofs can be found in Lax's book in the section on matrix inequalities).

Let $B\geq A \geq 0$. Define a function $M(t) = A + t(B-A)$. It is positive on $[0,1]$ by convexity (as a matter of fact, it is positive for all $t\geq0$, I believe). Further, it has a positive semidefinite matrix as its derivative. Also $\sqrt{M(t)}$ is positive semidefinite. Define $R(t) = \sqrt{M(t)}$. Then $R^2 = M$, so

$$R\dot{R} + \dot{R} R = M'(t) \geq 0.$$

$R$ is positive and $R\dot{R} + \dot{R} R$ is positive, so $\dot{R}$ is positive. Therefore $R$ is a non-decreasing function of $t$ and $R(1) \geq R(0)$. QED

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  • $\begingroup$ @julien Very nice! Why is $f(s)=s^2$ not a monotone operator, as you mentioned? Maybe I'll just look at the paper you linked. $\endgroup$ – Eric Auld Nov 30 '13 at 1:10
  • $\begingroup$ There is a counterexample in the thread I linked to :-) Not Pedersen's paper. $\endgroup$ – Julien Nov 30 '13 at 2:24
  • $\begingroup$ How about have $A^{-\frac{1}{4}}B^{\frac{1}{2}}A^{\frac{1}{4}}-I\ge 0$? $\endgroup$ – math110 Nov 30 '13 at 6:26
  • $\begingroup$ @julien I think there is a gap in my logic...without thinking, I assumed that the square root of $A^{-1/2}BA^{-1/2}$ was $A^{-1/4}B^{1/2}A^{-1/4}$...I'm interested how you would show the passage from line 2 to line 3 in more detail. Why does Sp$(A^{-1/2}BA^{-1/2})>1 \Rightarrow $ Sp$(B^{1/2}A^{-1/2})>1$? $\endgroup$ – Eric Auld Nov 30 '13 at 21:35
  • $\begingroup$ @julien Very helpful. I've changed it, let me know what you think. One remaining improvement would be to show that $\epsilon \mapsto M_\epsilon^{1/2}-N_{\epsilon}^{1/2}$ is continuous in the eigenvalues. Is is clear to you why that is? $\endgroup$ – Eric Auld Dec 1 '13 at 13:35

Another proof (short and simple) from "Linear Algebra and Linear Models" by R. B. Bapat.

Lemma Let $A$ and $B$ be $n\times n$ symmetric matrices such that $A$ is positive definite and $AB+BA$ is positive semidefinite, then Y is positive semidefinite.

Proof of $B\geq A \implies B^{\frac{1}{2}}\geq A^{\frac{1}{2}}$

First consider the case, when $A$ and $B$ are positive definite.

Let $X=(B^{\frac{1}{2}}+ A^{\frac{1}{2}})$ and $ Y=(B^{\frac{1}{2}}- A^{\frac{1}{2}})$,

then $XY+YX=2(B-A)$

Now, $(B-A)$ is positive semidefinite implies (given) $\implies 2(B-A)$ is positive semidefinite. Also $X=(B^{\frac{1}{2}}+ A^{\frac{1}{2}})$ is positive definite as positive linear combination of positive definite matrices is positive definite.

Hence by the lemma, $Y=(B^{\frac{1}{2}}- A^{\frac{1}{2}})$ is positive semidefinite. Therefore, $B^{\frac{1}{2}}\geq A^{\frac{1}{2}}$

The case, when $A$ and $B$ are positive semidefinite matrices can be dealt as the other answer.

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