I'm trying to calculate some norm for a matrix $A = [3, 2; 0,1]$ given the formula $\|A\| = \max_{|v|=1}|Av|$, where $|v|$ is taken to be the Euclidean norm for a vector, i.e. the standard distance measurement in $R^2$. The trouble is there seems to be some ambiguity with which norm I should be taking. In any case I'm a little confused with where to go or what I've done wrong. What I have:
Take a vector $v = [v_1, v_2]^T$ such that $|v| = 1$. Then $Av = [3v_1 + 2v_2; v_2]$. From here, I have $|Av| = \sqrt{(3v_1 + 2v_2)^2 + v_2^2} = \sqrt{9v_1^2 + 12v_1v_2 + 5v_2^2}$. Am I on the right track? If so, where do I go from here? Tips appreciated.
Note that maximizing $|Av|$ is equivalent to maximize $|Av|^2$. Now, using $v_1^2+v_2^2=1$, it reduces to maximize $4v_1^2+12v_1v_2+5$ subject to the restriction $v_1^2+v_2^2=1$. We can solve it using Lagrange multipliers but it gets messy.
Another approach is as follows: note that the matrix $A^tA$ is self adjoint and positive, and so there exists a unitary matrix $V$ such that $V A^tAV^{-1}$ is diagonal: $ \left[ {\begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_2 \\ \end{array} } \right]$ where each $\lambda_i\geq0$,and we may assume that $\lambda_1\geq\lambda_2$. Now,we have
$\|A\|^2 = \sup\{|Av| : |v| = 1\}^2$
$= \sup\{|Av|^2 : |v| = 1\}$
$= \sup\{(A^tAv,v) : |v| = 1\}$
$= \sup\{(V A^tAV^{-1}v, v) : |v| = 1\}$
$= \sup\{\lambda_1v_1^2+\lambda_2 v_2^2: v_1^2+v_2^2=1\}$
$=\lambda_1$
It follows that $\|A\| = \lambda_1$, the largest eigenvalue of $A^tA$. Which in this case is $7+2\sqrt{10}$.