Proving that cond(ATA) = cond(A)^2

[prhzn]

Homework Statement

Prove that $$\kappa(A^TA) = \kappa(A)^2$$ where $$\kappa(A) = \left\|A\right\|\left\|A^{-1}\right\|$$, or in a more general case, $$\kappa(A) = \left\|A\right\|\left\|A^+\right\|$$, where $$A^+$$ is the pseudoinverse of $$A\in\mathbb{R}^{m\times n}$$

The Attempt at a Solution

My initial guess is to use Cauchy-Schwarz to separate the products of the norms when replacing $$A$$ with $$A^TA$$, giving me

$$\kappa(A^TA) = \left\|A^TA\right\|\left\|(A^TA)^+\right\| \leq \left\|A^T\right\|\left\|A\right\|\left\|A^+\right\|\left\|\left(A^T\right)^+\right\| = \underbrace{\left\|A\right\|\left\|A^+\right\|}_{\kappa(A)}\underbrace{\left\|A^T\right\|\left\|\left(A^T\right)^+\right\|}_{\kappa(A^T)}$$

which gives $$\kappa(A)^2$$ as $$\kappa(A) = \kappa(A^T)$$.

But I'm not sure if I can do it this simple; any inputs?

Thanks in advance

 

[prhzn]

Figured it out - used the definition of the singular values of A, and then prooved that the singular values of ATA are the squared singular values of A.