Im having trouble understanding how to compute the operator norm of the linear transformation defined by: $$ L\left(x,y\right)=\left(x+3y,y-x\right) $$ I know the operator norm is defined by $$ \left\Vert L\right\Vert =\max_{\left|\left(x,y\right)\right|=1}\left|L\left(x,y\right)\right| $$ So i thought calculating this might help: $$ \left|L\left(x,y\right)\right|^{2}=\left(x+3y\right)^{2}+\left(y-x\right)^{2}=\ldots=2\left(x+y\right)^{2}+8y^{2} $$ But how do i maximize this over the unit sphere?
I wanna say it is maximized when $y=1$ but how do I prove it?
Moreover, is the operator norm of a general linear transformation related somehow to the eigenvalues of its corresponding matrix?
Thanks in advance
Yes, the norm is related to the eigenvalues. More precisely, $$||L|| = \sup \{ \sqrt{\lambda} \ | \ \lambda \in \sigma (A^T A) \}$$ where $\sigma (A^T A)$ denotes the eigenvalues of $A^T A$. In this particular case, it looks like the operator norm is $1+\sqrt{5} \approx 3.2361$