if the matrix such $B-A,A$ is Positive-semidefinite,then $\sqrt{B}-\sqrt{A}$ is Positive-semidefinite

Question

Question:

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

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. — Eric Auld, Nov 29 '13 at 15:31
Question still stands, though. Do you mean just any old square root? — Eric Auld, Nov 29 '13 at 15:32
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. — Julien, Nov 29 '13 at 15:50
At least I have time to link to Pedersen's paper which has a proof in the more general case of $B(H)$. And he mentions that this was first proved by K. Löwner (1934). — Julien, Nov 29 '13 at 15:58
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? — Tom, Nov 29 '13 at 16:00
When $A$ is positive definite and $k=2$, I have given a proof in the thread If $A^2\succ B^2$, then necessarily $A\succ B$. The case where $A$ is p.s.d. follows from a continuity argument. — user1551, Nov 30 '13 at 00:38

Eric Auld · Accepted Answer · 2013-12-01T20:52:47.170

6

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

edited Dec 01 '13 at 20:52

answered Nov 29 '13 at 16:25

Eric Auld

28,127

@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. – Eric Auld Nov 30 '13 at 01:10
There is a counterexample in the thread I linked to :-) Not Pedersen's paper. – Julien Nov 30 '13 at 02:24
How about have $A^{-\frac{1}{4}}B^{\frac{1}{2}}A^{\frac{1}{4}}-I\ge 0$? – math110 Nov 30 '13 at 06:26
@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$(B^{1/2}A^{-1/2})>1$? – Eric Auld Nov 30 '13 at 21:35
@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? – Eric Auld Dec 01 '13 at 13:35
@julien I feel like I should make the answer a community wiki since you helped me so much. – Eric Auld Dec 01 '13 at 13:40
Don't worry about the CW thing. It should be $\geq 1$ and $\geq 0$ everywhere, not $>1$ and $>0$. And for the continuity argument, you could just note that $\epsilon\longmapsto ((A+\epsilon I)x,x)$ is continuous. – Julien Dec 01 '13 at 15:26
@julien Quite right. How does $\epsilon \mapsto \langle (A+\epsilon I)x, x \rangle$ continuous imply that $B_\epsilon ^{1/2}\geq A_\epsilon ^{1/2}$ holds in the limit $\epsilon \to 0$? – Eric Auld Dec 01 '13 at 19:06
I missed a square root. $0\leq ((\sqrt{B_\epsilon}-\sqrt{A_\epsilon})x,x)\rightarrow ((\sqrt{B}-\sqrt{A})x,x)$. Since diagonalizing $A$ in onb yields easily that $\sqrt{A_\epsilon}\longrightarrow \sqrt{A}$. – Julien Dec 01 '13 at 19:23
@EricAuld $\langle A^{-1/2}BA^{-1/2}x,x \rangle \geq 1 $ why is this true? – Babai Aug 11 '16 at 07:08
@EricAuld Why is $<A^{-\frac{1}{2}}B A ^{-\frac{1}{2}}x,x>\geq 1$? – Babai Aug 17 '16 at 16:39
@Babai The general principle is that if $M \geq 0$, then $N^*MN \geq 0$, if $N$ is invertible. In this case $N=A^{-1/2}$, which is self-adjoint. – Eric Auld Aug 17 '16 at 23:49

score 0 · Answer 2 · answered Aug 17 '16 at 18:51

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.

if the matrix such $B-A,A$ is Positive-semidefinite,then $\sqrt{B}-\sqrt{A}$ is Positive-semidefinite

2 Answers2