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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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    $\langle x , B^*Bx \rangle = \langle Bx , Bx \rangle = \|Bx\|^2$, and since $B$ is invertible, you have $\|Bx\|=0$ iff $x=0$. What does that say about $B^*B$.2012-12-17

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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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    No No, "so" was $j$ust a use to make sure whether both are true or $n$ot2013-06-17