I am currently working on the following problem:
Let $A \in \mathbb{R}^{n \times n}$ be a symmetric and positiv-definite with Cholesky decomposition $A = LL^T$. Define $(x,y)_A:= y^TAx$
For $C \in \mathbb{R}^{n \times n}$ show that $||C||_{A}$ can be computed using the largest eigenvalue of the matrix $L^{-1}C^TACL^{-T}$
I believe this should follow directly from the definition of the matrix norm, but I'm struggling to see how. I would greatly appreciate any hints.
By definition, \begin{align} \|C\|_A^2 &=\max\{\|Cx\|_A^2:\ \|x\|_A=1\}\\[0.2cm] &=\max\{(Cx,Cx)_A:\ \|x\|_A=1\}\\[0.2cm] &=\max\{x^TC^TACx:\ x^TAx=1\}\\[0.2cm] \end{align}
The equality $x^TAx=1$ can be written $x^TLL^Tx=1$. If we let $z=L^Tx$, then the condition $x^TAx=1$ becomes $z^Tz=1$. Hence, using that $x=L^{-T}z$, we get $$ \|C\|_A^2=\max\{zL^{-1}C^TACL^{-T}z:\ z^Tz=1\}. $$ This is the usual operator norm of the matrix $L^{-1}C^TACL^{-T}$, so $\|C\|_A$ is the square root of the largest eigenvalue of $L^{-1}C^TACL^{-T}$.