Prove that X†AX is positive when A is positive

[Doom of Doom]

I'm supposed to prove this step as part of my proof for existence of Cholesky Decomposition. I can see how to use it in my proof, but I can't seem to be able to prove this lemma:

For any positive (nxn) matrix $$A$$ and any non-singular (nxn) matrix $$X$$, prove that

$$B=X^{\dagger}A X$$

is positive.

____

Let $$X=\left(x_{1}, x_{2}, \ldots, x_{n}\right)$$, where all xi are n-vectors.

I see that
$$b_{i,j}=x_{i}^{\dagger}Ax_{j}$$,
and thus all of the diagonal elements of B are positive (from the definition of a positive matrix).

But where do I go from there?

 

[yyat]

I think you mean http://en.wikipedia.org/wiki/Positive-definite_matrix" .

In that case, you cannot conclude that the diagonal elements of B are positive.
Try to show instead directly that $$v^TBv>0$$ for $$v\neq 0$$ (this is a one-liner), then B is positive definite by definition.

 

[Doom of Doom]

Dur! Yeah, i meant positive definite.

I see it now. It's so easy...

For arbitrary vector $$v$$, let $$z = Xv$$, and thus $$z^{\dagger}=v^{\dagger}X^{\dagger}$$.

So $$v^{\dagger}Bv=v^{\dagger}X^{\dagger}AXv=z^{\dagger}Az>0$$.