How to find the maximum of $\|Ax\|$ for all unit vectors $x \in\mathbb R^2$? $A = \left[\begin{matrix} 1  & 0 \\ 3 & -1 \\ \end{matrix} \right]$
How to find the maximum of $\|Ax\|$ for all unit vectors $x \in\mathbb R^2$?
My professor told me the answer is: $\sqrt{\frac{11+ \sqrt{117} }{2}}$
I know this has something to do with the characteristic polynomial of $A$, which is $X^2-11x+1$.

Let $x = (a,b)$, then $||Ax|| = ||(a, 3a - b)|| = \sqrt{a^2 + (3a - b)^2}$.
Consider: $f(a,b) = a^2 + (3a - b)^2 = a^2 + 9a^2 - 6ab + b^2 = 10a^2 - 6ab + b^2$ subject to: $g(a,b) = a^2 + b^2 = 1$. Using Lagrange Multiplier method:
$\nabla f = \lambda\cdot \nabla g$:
$(20a - 6b, 2b - 6a) = (2a\lambda, 2b\lambda)$. So:
$20a - 6b = 2a\lambda$
$2b - 6a = 2b\lambda$.
If $\lambda = 0$, then: $b = 3a$, and $10a = 3b \to a = b = 0$, this can't happen.
So: $\lambda \neq 0$, then if $a = 0$ or $b = 0$, then both $a = b = 0$, and this can't happen. So $a \neq 0$, and also $b \neq 0$. Then:
$\dfrac{10a - 3b}{b - 3a} = \dfrac{a}{b}$
So: $10ab - 3b^2 = ab - 3a^2 \to -a^2 + b^2 = 3ab \to 1 - t^2 = 3t$, with $t = \dfrac{a}{b}$. So:
$t^2 + 3t - 1 = 0 \to t = \dfrac{-3 \pm \sqrt{13}}{2}$. Together with $a^2 + b^2 = 1$, you can solve for $a$, and $b$, and then find $\sqrt{f_{max}}$
