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Homework Help: Positive definite quadratic forms proof

  1. Mar 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Given a real symmetric matrix A, prove that:

    a) A is positive definite if and only if A = (B^T)B for some real invertible matrix B
    b) A is positive semidefinite if and only if there exists a (possibly singular) real matrix Q such that A = (Q^T)Q

    2. Relevant equations
    quadratic form q(x) = a1*x1^2 + ... + an*xn^2
    And possibly the principal axis theorem.

    3. The attempt at a solution
    for part a) I think I know how to show that A is pos def if you assume A = (B^T)B:
    (P^T)AP = D = (P^T)(B^T)BP = ((PB)^T)PB which implies the diagonal entries of the diagonal matrix D are positive since row of (A^T) = column of A.
    I'm not sure how to do it the other way, however. Any hints or advice would be appreciated.

    Cheers,
    W. =)
     
  2. jcsd
  3. Mar 13, 2010 #2

    jbunniii

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    For the "only if" part of (a), do you know a theorem that allows you to express any symmetric matrix in the form

    [tex]A = U^T D U[/tex]

    where [itex]U[/itex] is a unitary matrix (in fact, orthogonal since [itex]A[/itex] is real in your case) and [itex]D[/itex] is a diagonal matrix? If [itex]A[/itex] is positive definite, what can you say about the elements of [itex]D[/itex]?

    Also, I'm not sure I buy your proof of the "if" part. A matrix [itex]A[/itex] is positive definite if for every real vector [itex]x[/itex],

    [tex]x^T A x \geq 0[/tex]

    If [itex]A = B^T B[/itex], then what does that imply about [itex]x^T A x[/itex]?
     
  4. Mar 13, 2010 #3
    I'm not entirely sure this is correct, but I've come up with the following:

    If (X^T)AX > 0 then we have (Y^T)DY > 0 (for X = PY and D = (P^T)AP). This implies that the entries on the diagonal matrix D (the eigenvalues) are all positive. So, write D = (Q^T)Q where Q is the diagonal matrix of the square roots of D's elements. Then
    D = (Q^T)Q = (P^T)AP = (P^1)AP (since P is orthogonal)
    then P(Q^T)Q(P^1) = A
    then (PQ)(Q^T)(P^T) (since (Q^T) = Q)
    then (PQ)(PQ)^T = A
    Take B = (PQ)^T and we have the result.

    As for the other part, I don't think I quite finished my reasoning before....if we can show that the assumption requires that D have all positive entries (take them to be a1 to an) then q(v) = a1*xa^2 + ... + an*xn^2 will necessarily be positive. And since D = (PB)^T(PB) then they must be positive because the diagonal entries of D (which are the only ones that are not 0) are the rows of the transpose of PB multiplied by the columns of PB and these are the same vectors.

    Cheers,
    W. =)
     
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