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Cholesky Decomposition

  • #1
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 [tex]A[/tex] and any non-singular (nxn) matrix [tex]X[/tex], prove that

[tex]B=X^{\dagger}A X[/tex]

is positive.

____

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

I see that
[tex] b_{i,j}=x_{i}^{\dagger}Ax_{j}[/tex],
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?
 

Answers and Replies

  • #2
316
0
I think you mean http://en.wikipedia.org/wiki/Positive-definite_matrix" [Broken].

In that case, you cannot conclude that the diagonal elements of B are positive.
Try to show instead directly that [tex]v^TBv>0[/tex] for [tex]v\neq 0[/tex] (this is a one-liner), then B is positive definite by definition.
 
Last edited by a moderator:
  • #3
Dur! Yeah, i meant positive definite.

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

For arbitrary vector [tex]v[/tex], let [tex]z = Xv[/tex], and thus [tex]z^{\dagger}=v^{\dagger}X^{\dagger}[/tex].

So [tex]v^{\dagger}Bv=v^{\dagger}X^{\dagger}AXv=z^{\dagger}Az>0[/tex].
 

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