3
$\begingroup$

[1] The similar question I post is here.

Let matrices $X$ and $Y$ be positive semidefinite (PSD).

(1) $X \succeq Y$ implies $X^{1/2} \succeq Y^{1/2}$.

(2) $X \succeq Y$ does not imply $X^{2} \succeq Y^{2}$.

I try to prove these two inequalities through Schur Complement but could not succeed. And I utilize reductio ad absurdum which does not work. Could anyone give some hints (maybe these two questions are in the same form so one hint is enough)? Thanks in advance!

[1]: How to prove positive semidefiniteness of two matrices through Schur Complement? .....

$\endgroup$
  • $\begingroup$ For the second you can use an example. Like $X=[[1,1],[1,2]]$ and $Y=[[0,0],[0,1]]$. Then $X-Y=[[1,1],[1,1]]$ which is positive semidefinite, while $X^2=[[2,3],[3,5]]$ and $Y^2=Y$. This gives $X^2-Y^2=[[2,3],[3,4]]$ which is not positive semidefinite. $\endgroup$ – EEE Oct 22 '17 at 13:26
  • $\begingroup$ @EEE Thanks for you help. Good examples for illustration. But I need to prove these cases generally. $\endgroup$ – stander Qiu Oct 22 '17 at 13:37
  • 3
    $\begingroup$ Then you need first to learn that to prove that $A$ does not imply $B$ you only need an example in which $A$ is true and $B$ is false. The example is the proof. $\endgroup$ – EEE Oct 22 '17 at 13:46
  • $\begingroup$ @EEE Oh, sorry. I forget the function of counter-example. You are right, thank you. $\endgroup$ – stander Qiu Oct 22 '17 at 23:39
  • $\begingroup$ @EEE How about the first one? $\endgroup$ – stander Qiu Oct 24 '17 at 0:07
3
+50
$\begingroup$

Proposition 1. Let $r\in(0,1)$. i) If $0<A<B$, then $A^{r}<B^{r}$. ii) If $0\leq A\leq B$, then $A^{r}\leq B^{r}$.

Here is a hilarious (but correct) proof. For i). It is not difficult to prove that

(*) $0<A<B$ implies that $B^{-1}< A^{-1}$. On the other hand, from

$(Eq)\;\int_0^{+\infty}(\dfrac{x^{r+1}}{1+x^2}-\dfrac{x^r}{t+x})dx=\dfrac{\pi}{\sin(r\pi)}(t^r-\cos(r\pi/2))$, we deduce (replace $t$ with $A$)

$A^r=\cos(r\pi/2)I+\dfrac{\sin(r\pi)}{\pi}\int_0^{+\infty}(\dfrac{x^{r+1}}{1+x^2}I-(A+xI)^{-1}x^r)dx$.

Thus $B^r-A^r=\dfrac{\sin(r\pi)}{\pi}\int_0^{+\infty}((A+xI)^{-1}-(B+xI)^{-1})x^r)dx$ is $>0$ according to $(*)$.

For ii). Proceed by continuity.

EDIT 1. Answer to @stander Qiu . If you know an equality between analytic functions, as $(Eq)$, then, clearly, the equality remains valid when you replace $t$ with a diagonal matrix: $f(diag(\lambda_i))=\phi(diag(\lambda_i))$ and even with a diagonalizable matrix: $f(PDP^{-1})=P\phi(D)P^{-1}$. Finally, a general theorem about "matrix function" says that the equality remains valid for any matrix; in other words, it suffices to prove the required equality for diagonalizable matrices.

cf. [1]: Higham, functions of matrices.

Here is a second proof of the required result that does not use integration theory.

Proposition 2. i) If (*) $A>0,0\leq B<A$, then $B^{1/2}<A^{1/2}$. ii) If $0\leq B\leq A$, then $B^{1/2}\leq A^{1/2}$.

Proof. For i). (*)$\implies A^{-1/2}BA^{-1/2}<I\implies \rho((B^{1/2}A^{-1/2})^T(B^{1/2}A^{-1/2}))<1\implies \rho(B^{1/2}A^{-1/2})<1$

$ \implies \rho(A^{-1/4}B^{1/2}A^{-1/4})<1\implies A^{-1/4}B^{1/2}A^{-1/4}<I \implies B^{1/2}<A^{1/2}$.

For ii). Proceed by continuity.

EDIT 2. Answer to @stander Qiu. I don't know any name for $(Eq)$. Perhaps , you think about the Cauchy integral theorem: (cf. [1] p. 8, Frobenius and Poincare for application to matrices).

$f(A)=\int_{\Gamma}f(z)(zI-A)^{-1}dz$ where $A\in M_n(\mathbb{C})$ and $f$ is analytic on and inside a closed contour $\Gamma$ that encloses $spectrum(A)$.

for example, we consider the matrix sign function defined by $sign(z)=z/(z^2)^{1/2}$ when $Re(z)\not= 0$. Then $sign(A)=A(A^2)^{-1/2}$ and using Cauchy on $z^{-1/2}$, we obtain

$sign(A)=\dfrac{2}{\pi}A\int_0^{\infty}(t^2I+A^2)^{-1}dt$. Note that we can also prove this equality, using the $arctan$ function. From this, we can derive another integral formula for $A^r$.

$A^r=\dfrac{\sin(\pi r)}{\pi r}A\int_0^{\infty}(t^{1/r}I+A)^{-1}dt$.

On the other hand, from the equality $\log(x)=\int_0^1(x-1)(t(x-1)+1)^{-1}dt$,

Richter obtains directly $\log(A)=\int_0^1(A-I)(t(A-I)+I)^{-1}dt$.

$\endgroup$
  • $\begingroup$ Amazing.. I have two questions: 1. What is the theorem of that integral? 2. Could you prove the inequality via matrix? I do not mean that your method is wrong, but it is impossible for me to think in this way. lol $\endgroup$ – stander Qiu Oct 25 '17 at 8:06
  • $\begingroup$ Thanks for your answer. A very beautiful answer! I know what matrix function means and how equality preserves, but I want to know what is the name of $(Eq)$ ? $\endgroup$ – stander Qiu Oct 25 '17 at 23:56
  • $\begingroup$ Could you tell me the answer in this room? chat.stackexchange.com/rooms/67697/… @loup blanc $\endgroup$ – stander Qiu Oct 26 '17 at 11:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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