Determine the $\|A\|_2$ vector-induced norm This seems to be a very easy exercise, but as I never had previous contact with vector-induced norm of a matrix, I am stuck.


Given the matrix $$A = \begin{bmatrix} 2 & 0\\ 0 & 1\end{bmatrix}$$ compute its vector-induced $\| \cdot \|_2$ norm, defined as $$\|A\|_2=\sup_{v\neq 0}\frac{\|Av\|_2}{\|v\|_2}$$


My attempt so far was as follows. Define a vector $v=(v_1,v_2)$ so that I end up with:
$$\|A\|_2=\sup_{v\neq 0}\frac{\|Av\|_2}{\|v\|_2}=\sup_{v\neq 0}\sqrt{\frac{4v_{1}^{2}+v_{2}^{2}}{v_{1}^{2}+v_{2}^{2}}}$$
It seems reasonable to choose a vector $v$ such that $\|v\|_2=1$, then this would result in a problem where I want to find:
$$\|A\|_2=\sup_{\|v\|_2=1}\sqrt{4v_{1}^{2}+v_{2}^{2}}$$
Which kind of seems like an optimization problem where the constraint is:
$$\sqrt{v_{1}^{2}+v_{2}^{2}}=1$$
And this is where I get stuck! Is this a correct way to solve this? If yes, could you help me finish this line of thought? Would Lagrange Multiplier do the job?
I really would like to solve it "the hard way" before using other properties/definitions such as:
$$\|A\|_2=\sigma_{\max}(A)$$
Thanks in advance!
 A: Here's one way to find it for a particular matrix such as the one in your example. Based on the work you've already done, observe that
$$\|Av\|_2^2=4v_1^2+v_2^2\le4v_1^2+4v_2^2=4\|v\|_2^2 \quad \implies \quad \|A\|_2\le2.$$
Now if you can find a specific vector $v$ such that this value of $2$ is attained, i.e. such that $\|Av\|=2$ for a vector $v$ with $\|v\|_2=1$, then you will have demonstrated that $\|A\|_2=2$. And finding such a vector has a lot to do with eigenvectors….
A: Alternative:
You want to solve $$\max 4v_1^2 + v_2^2$$
subject to $$v_1^2+v_2^2=1$$
Let $y_i = v_i^2$, we have
$$\max 4y_1+y_2$$
subject to $$y_1+y_2=1$$
$$y_i \ge 0 ,\forall i \in \{1,2\}$$
which is just a linear programming problem which has optimal solution at the corners of the polyhedral. 
We want $y_1$ to be large. Hence the optimal $4y_1+y_2=4(1)+0=4$.
Taking square root gives us $2$ as the optimal solution 
A: Given a matrix $A$ the singular values of $A$ are just the square roots of the eigenvalues of the matrix $A^*A$.  The largest singular value is the operator norm which is the quantity you want to compute.
