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Cholesky Decomposition

  1. Mar 6, 2009 #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?
  2. jcsd
  3. Mar 6, 2009 #2
    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: May 4, 2017
  4. Mar 6, 2009 #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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