Prove that $\|R\|_2 = \|A\|_2^{1/2}$ where $A=R^* R$ is a Cholesky factorization of $A$

Question

Prove that $\|R\|_2 = \|A\|_2^{1/2}$ where $A = R^* R$ is a Cholesky factorization of $A$.

In my book it says that I should use the Singular Value Decomposition.

I have that $\rho(A)=\sqrt{\rho(A^*A)}=\sqrt{\rho(A*A)}=\sqrt{\rho(A^2)}$ $$ \Rightarrow \rho^2(A)=\rho(A^2)$$

So, I have tried:

$R^*R=Q\Sigma V^*$ where $Q\Sigma V^*$, is de SVd of A. Then $\rho(R^*R)=\rho(Q\Sigma V^*)$. With $\rho(.)$ is the bigger eigenvalue.

We conclude that $\rho^2(R)=\rho(A) \Leftrightarrow ||R||=\sqrt{||A||}$.

This last equation comes from the relation between the singular value of A in its norm.

I do not know if this process is well formulated.

Answer 1

Hint: What is the relationship between the eigenvalues of $A$ and the singular values of $R$? Using the fact that $A$ is self-adjoint, what is $\|A\|$ in terms of the eigenvalues of $A$? What is $\|R\|$, in terms of the eigenvalues of $A$?