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as $A$ is hermitian positive definite so all eigen values are real and strictly positive. we can diagonalize $A=PDP^{-1}$ for some invertible $P$ and we can chose $B=P\sqrt{D}P^{-1}$ and hence $a$ is true?

$BB'$ is symmetric that is true, I am not sure about positive definiteness, let $X$ be any eigen vector corresponding to the eigen value $\lambda$ then $BB^TX=\lambda X\Rightarrow XBB^TX=\lambda XX^T\Rightarrow X(BB^T-\lambda I)X^T=0$.well please help.

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For (a), what can you say about the eigenvalues of a positive definite Hermitian matrix? Can you take their square roots? For (b), note that $\langle x , B^*Bx \rangle = \|Bx\|^2$. $B$ is non-singular, what does that tell you about $\|Bx\|$? –  copper.hat Dec 17 '12 at 18:03
    
I do not get how $<x,B^* Bx>=< B^*x,Bx> =||Bx||^2$ and I can not getting any clue as $B$ is any arbitrary matrix. –  Bunuelian Trick Dec 17 '12 at 18:17
    
$B$ is not arbitrary, it is non-singular. A basic (defining, actually) property of adjoints is $\langle x, A^* y \rangle = \langle A x, y \rangle$. –  copper.hat Dec 17 '12 at 18:19
    
well $BB^T$ need not be positive definite always i guess, I got one counter example of a symmetric matrix which is not positive definite. –  Bunuelian Trick Dec 17 '12 at 18:40
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$\langle x , B^*Bx \rangle = \langle Bx , Bx \rangle = \|Bx\|^2$, and since $B$ is invertible, you have $\|Bx\|=0$ iff $x=0$. What does that say about $B^*B$. –  copper.hat Dec 17 '12 at 19:49
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(a) You wrote $A=PDP^{-1}$ for some invertible $P$ and chose $B=P\sqrt{D}P^{-1}$. This is not good enough. We need $B$ to be positive definite, but your $B$ is not necessarily Hermitian. But you are close. Note that Hermitian matrices can be diagonalized in a very special way (which is described in Wikipedia).

(b) You need to check that $x^TB^TBx>0$ for every real vector $x$ (that's what "positive definite" means). Note that $x^TB^TBx=(Bx)^T(Bx)$.

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I have checked that if $B$ is invertible then the assertion is true, so both the statement $a,b$ are true? right? –  Bunuelian Trick Jun 17 '13 at 14:25
    
@TaxiDriver Yes, both statements are true, but I don't get the "so" in your comment. That $B$ is invertible in (b) has nothing to do with statement (a). How did you conclude that "so statement a is true"? –  user1551 Jun 17 '13 at 15:06
    
No No, "so" was just a use to make sure whether both are true or not –  Bunuelian Trick Jun 17 '13 at 15:09
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