Proving Positivity of Quadratic Forms Using Eigendecomposition

[GenAlg]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

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

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

I attempted to work directly on $$h(x) = X^t(AA^t)X$$ as well by trying to complete the square for 2x2, 3x3, ... matrices $$AA^t$$ (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?

 

[Mathnerdmo]

I don't know most of the methods you used in your solution attempt, however...

Rewrite $$h(x)$$ in terms of $$Y = A^tX$$

 

[VKint]

Hint: Try rewriting $$X^t (AA^t) X$$ as an inner product of something with itself.

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