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If you have a Hermitian matrix $C$ that you can rewrite using Cholesky decomposition, how can you use this to show that $C$ is also positive definite? $C$ is positive definite if $x^\top C x > 0$ and $x$ is a vector.

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From Wikipedia:

If A can be written as LL* for some invertible L, lower triangular or otherwise, then A is Hermitian and positive definite.

$A=LL^*\implies x^*Ax=(L^*x)^*(L^*x)\ge 0$

Since $L$ is invertible, $L^*x\ne 0$ unless $x=0$, so $x^*Ax>0\ \forall\ x\ne 0$

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I know that's what wikipedia says, but how do you prove that to be the case? –  user1855952 Mar 7 '13 at 10:43
    
@user1855952: Sounds good now? –  Bravo Mar 7 '13 at 10:47
    
I see. I'm just a little bit confused over one step where you do xAx = (Lx)*(Lx). So xAx = x*(LL*)x which you could split up as (xL)*(Lx) but that's different from what you have. –  user1855952 Mar 7 '13 at 10:51

If $C = LL^{\ast}$ with $L$ lower triangular, then $$ \langle Cx, x \rangle = \langle LL^{\ast} x, x \rangle = \langle L^{\ast} x, L^{\ast} x \rangle \geq 0$$ Thus $C$ is positive semi-definite. Note that, as stated, you cannot show that $C$ is positive definite unless you also know that $L$ is invertible.

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How do you go from <LLx, x> to <Lx, L*x> ? And then I still don't see how that is necessarily >= 0 –  user1855952 Mar 7 '13 at 10:45
    
The very definition of $A^{\ast}$ is that $\langle Ax, x \rangle = \langle x, A^{\ast} x \rangle$. –  Christopher A. Wong Mar 7 '13 at 19:01

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