Compute the matrix of norms of $A=\begin{bmatrix}3&4\\1&-3\end{bmatrix}$ My work so far
Using the following
$\hspace{30px} L^1\ =\displaystyle \max_{\small 1\le j\le m}(\displaystyle \sum_{i=1}^n |a_{ij}|)\\ \hspace{30px} L^2\ =\sigma_{max}(A)\\ \hspace{30px} L^F\ =\sqrt{\displaystyle \sum_{i} \displaystyle \sum_{j} |a_{ij}|^2}\\ \hspace{30px} L^\infty\ =\displaystyle \max_{\small 1\le i\le n}(\displaystyle \sum_{j=1}^m |a_{ij}|)\\$
Thus,
$L^1=\begin{bmatrix}3&\textbf{4}\\1&\textbf{3}\end{bmatrix}=7\\
L^2=?\\
L^F=\sqrt{3^2+4^2+1^2+(-3)^2}=\sqrt{35}=5.916079783\\
L^\infty=\begin{bmatrix}\textbf{3}&\textbf{4}\\1&-3\end{bmatrix}=7$
However, I'm unsure how to get $L^2$. How would I start off doing this part?
 A: To find the $L^2$ norm, compute the eigenvalues of $A A^T$, then square root and take the largest one. So like
$$
A A^T = \begin{bmatrix}3& 4\\1& -3\end{bmatrix}\cdot \begin{bmatrix}3& 1\\4& -3\end{bmatrix} = \begin{bmatrix}25& -9\\-9& 10\end{bmatrix}\\
$$
$$
\det(\lambda I - A A^T) = \lambda^2-35\lambda + 169
$$Setting this equal to zero and solving gives $\lambda = \frac{35\pm3\sqrt{61}}{2}$, so the $L^2$ norm is $\sqrt{\frac{35+ 3\sqrt{61}}{2}}$.
A: Alternatively, recall that
$$\|A\|_2 = \sup_{\|(x,y)\|_2=1} \|A(x,y)\|_2 = \sup \left\{\left\|\begin{bmatrix}3&4\\1&-3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\right\| : x^2+y^2=1\right\}$$
so subject to $x^2+y^2=1$ we wish to maximize $$f(x,y) = (3x+4y)^2+(x-3y)^2 = 10x^2+25y^2+18xy.$$
Plugging in $y=\sqrt{1-x^2}$ this is $$g(y)=10+15y^2+18y\sqrt{1-y^2}.$$
Solving $g'(y) = 0$ for $y\in[0,1]$ gives $y=\sqrt{\frac12+\frac{5}{2\sqrt{61}}}$ and finally
$$g(y) = \frac{35+ 3\sqrt{61}}{2}$$
so $$\|A\|_2 = \sqrt{g(y)} = \sqrt{\frac{35+ 3\sqrt{61}}{2}}.$$
