hi, i'm having trouble seeing where to even begin...
how can i show (||A||_2)^2 = p(A^T * A) for every matrix A in R^(nxn) where p is spectral radius? (A^T is transpose)
any help much appreciated. thanks!
What do you mean by $\displaystyle \|A\|_2$? I would normally take it to mean the Hilbert–Schmidt norm, namely $\displaystyle \|A\|_2^2 = \textstyle\sum_{i,j}|a_{ij}|^2$. But I don't think that can be the case here. For example, if A is the identity 2×2 matrix then $\displaystyle \|A\|_2^2 = 2$, but $\displaystyle \rho(A^{\textsc t}A) = 1$.
Hi, Thank you for your reply.
I am referring to the operator matrix norm (p-norm with p=2).
I have been told that maximising |Av| is the same as maximising v'A'Av and in some way this will help me determine my answer, however, I do not see why this it is the case that they are the same or how it can help me. Help much appreciated.
Okay, the result is certainly true for the operator norm $\displaystyle \|A\|$. The first thing you need is that the spectral radius of a positive definite matrix is equal to the norm. (Basically the reason for that is that the matrix is diagonalisable.) The matrix $\displaystyle A^{\textsc t}A$ is positive definite, so you actually only need to prove that $\displaystyle \|A^{\textsc t}A\| = \|A\|^2$.
If $\displaystyle \|v\|\leqslant1$ then $\displaystyle \|Av\|^2 = |(Av)^{\textsc t}(Av)| = |v^{\textsc t}(A^{\textsc t}A)v| \leqslant \|v\|\|A^{\textsc t}A\|\|v\|\leqslant \|A^{\textsc t}A\|$. Take the sup over all such v to see that $\displaystyle \|A\|^2\leqslant\|A^{\textsc t}A\|$.
For the reverse inequality, $\displaystyle \|A^{\textsc t}A\|\leqslant \|A^{\textsc t}\|\|A\| = \|A\|^2$.