Under which conditions is $\|AB\|_2 = \|A\|_2\|B|\|_2$

Question

I know that $\|AB\|_2 \leq \|A\|_2\|B\|_2$. However, let suppose that I have a matrix A, that is symmetric and positive definite. So, there exist a factorization such that $A = LL^t$ (Cholesky). Suppose $\|A\|_2 = 4$. I am interested in knowing the value of $\|L^t\|_2$. My first guess is that it should be 2, because $\|L^t\|_2$ = $\|L\|_2$, using the fact that $\|LL^t\|_2 = \|A\|_2$ and asuming that $\|L\|_2$$\|L^t\|_2 = \|LL^t\|_2$. However I don't know when does this hold, or how to prove this in this case.

Answer 1

As Omnomnomnom already mentioned in his comment $\|LL^T\|_2 = \|L\|_2^2$ is true for any matrix $L\in\mathbb{R}^{n\times n}$:

When calculating the 2-norm \begin{align*} \|L\|_2 = \max\{\|Lx\|_2 : x\in\mathbb{R}^n, \|x\|_2 = 1\} \end{align*} of $L$ we can apply any norm-preserving change of coordinates $U, V$ in the image space and the pre-image space of $L$, respectively ($\|Lx\|_2 = \|U^T Lx\|_2 = \|U^T L V y\|_2$ with $x= Vy$, $\|x\|_2 = \|Vy\|_2 = \|y\|_2$).

Let $L = U\Sigma V^T$ be the singular value decomposition of $L$. (That means $\Sigma$ is a diagonal matrix $\Sigma=\operatorname{diag}(\sigma_1,\ldots,\sigma_r,0,\ldots,0)$ with $\sigma_{\rm max}:=\sigma_1\geq \ldots \geq \sigma_r > 0$ where $r\in\mathbb{N}$ is the rank of $L$ and $U,V$ are orthonormal and therefore norm-preserving.)

The norm-preservation of $U,V$ leads to \begin{align*} \|L\|_2 = \|\Sigma\|_2 = \sigma_{\rm max} \end{align*} which implies $\|LL^T\|_2 = \|L\|_2^2$: \begin{align*} \|LL^T\|_2 = \|U\Sigma V^T V \Sigma U^T\|_2 = \|\Sigma^2\|_2 = \sigma_{\rm max}^2 = \|L\|_2^2. \end{align*}