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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.

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


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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.

    W. =)
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