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X^t(AA^t)X ≥ 0 for all x∈R^n?

  1. Mar 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that the quadratic form [tex]h(x) = X^t(AA^t)X[/tex] is greater than or equal to 0 for all [tex]x[/tex] in [tex]R^n[/tex].

    2. Relevant equations



    3. The attempt at a solution

    Since [tex](AA^t)^t = (A^t)^t A^t = AA^t[/tex], [tex]AA^t[/tex] can (according to the spectral theorem) be diagonalized in an orthonormal eigenvector basis. Assuming [tex]X = TX'[/tex] to be the relation between the bases, it follows that (since T is orthogonal):

    [tex]h(x) = X^t(AA^t)X = (TX')^t(AA^t)(TX') = (X'^t T^t) (AA^t)(TX') = X'^t (T^{-1} AA^t T)X' = X'^t D X' = \mu_1 x_1'^2 + \mu_2 x_2'^2 + ... \mu_n x_n'^2[/tex]

    The problem is thus equivalent to showing that [tex]\mu_1 \ge 0, \mu_2 \ge 0, ... , \mu_n \ge 0[/tex]. I haven't gotten anywhere with that approach.

    I attempted to work directly on [tex]h(x) = X^t(AA^t)X[/tex] as well by trying to complete the square for 2x2, 3x3, ... matrices [tex]AA^t[/tex] (i.e. by using induction), and the proof would, if some relevant transformation is valid and all eigenvalues are ≥0, follow from Sylvester's law of inertia. I didn't get anywhere with that approach either.

    Ideas?
     
  2. jcsd
  3. Mar 4, 2010 #2
    I don't know most of the methods you used in your solution attempt, however...

    Rewrite [tex]h(x)[/tex] in terms of [tex]Y = A^tX[/tex]
     
  4. Mar 4, 2010 #3
    Hint: Try rewriting [tex] X^t (AA^t) X [/tex] as an inner product of something with itself.

    Alternatively, if you want to explicitly prove that all eigenvalues are positive, consider an eigenvector [tex] y [/tex], with [tex] AA^t y = \lambda y [/tex]. What's [tex] (AA^t y) \cdot y [/tex]? Now apply a basic result about the transpose to get what you want.
     
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