$\lVert A^TA\rVert_2$=$\lVert A\rVert_2^2$ and $\kappa_2(A^TA)=\kappa_2(A)^2$

Question

I have a question from James Demmel's book. Chapter 2 Question 2.10

Let A be n $\times$ m matrix with n$\geq$m, show that $\lVert A^TA\rVert_2$=$\lVert A\rVert_2^2$ and $\kappa_2(A^TA)=\kappa_2(A)^2$

and Let M be n $\times$ n and positive definite and L be its Cholesky factor so that $M = LL^T$. Show that $\lVert M \rVert_2$ = $\lVert L\rVert_2^2$ and $\kappa_2(M)=\kappa_2(L)^2$.

What I try is not correct and I cannot reach the proof.

Answer 1

$\|Ax\|^2 = \langle Ax, Ax \rangle = \langle x, A^TAx \rangle \le \|x\| \|A^T A x\| \le \|A^TA\|\|x\|^2$. Taking the $\sup $ over $\|x\| \le 1$ shows that $\|A\|^2 \le \|A^TA\|$.

$\langle u, A^TA v\rangle = \langle A u, A v\rangle \le \|A\|^2 \|u\| \|v\|$. Taking the $\sup$ over $\|u\| \le 1, \|v\|\le 1$ gives $\|A^TA\| \le \|A\|^2$.

$\kappa_2(A^TA) = \|A^T A\| \|(A^T A)^{-1} \| = \|A^T A\| \| A^{-1} A^{-T} \| = \|A\|^2 \|A^{-1}\|^2 = \kappa_2^2(A)$.

The other statement follows immediately from the above.