I am trying to solve the following problem:
For $A$ positive definite, let the Cholesky factorization be $A = R R^T$. Prove that $$\|R\|_2 \le \sqrt{\|A\|_2}$$ and that the vector $2$-norm of $R_1 R_1^T$ will never be greater than $\|A\|_2$ where $R_i$ is the $i$-th column of $R$.
Here, the matrix $2$-norm is defined as
$$\|A\|_2 := \max_{\|x\|_2 = 1} \|Ax\|_2$$
For the first bit, expanding $R^T = U_R \Sigma_R V_R^T$ via a singular value decomposition gives that in fact $\|R\|_2 = \sqrt{\| A \|_2}$, as $\Sigma_R^2 = \Sigma_A$. But I am unsure how to prove that $2$-norm of $R_1 R_1^T$ will never be greater than $\|A\|_2$. Could anyone help with this?
Hint: There's no need to invoke the SVD here. Note that $$ A = RR^T = \sum_{i=1}^n R_iR_i^T $$ is a sum of positive semidefinite matrices. If $A,B$ are positive semidefinite, then there is an inequality relating the eigenvalues of $A$ and $B$ to the eigenvalues of $A + B$ that I suspect you have somewhere in your notes. If you do not, then you can deduce this inequality using the fact that for symmetric matrices $M$, $$ \lambda_{\max}(M) = \max_{\|x\| = 1} x^TMx. $$
To elaborate, we have the following.
$$ \lambda_{\max}(A + B) = \max_{\|x\| = 1} x^T(A + B)x = \max_{\|x\| = 1}(x^TAx + x^TBx) \\ \geq \max_{\|x\| = 1} (x^TAx + 0) = \lambda_{\max}(A). $$ From there, set $A = R_1R_1^T$ and $B = \sum_{i=2}^n R_iR_i^T$.