Matrix $3$-norm 
The $3$-norm on $\mathbb R^n$ is defined as
$$\|x\|_3 := \sqrt[3]{|x_1|^3+\dots+|x_n|^3}$$
The natural matrix norm it induces on $\mathbb R^{n \times n}$ is
$$\|A\|_3 = \max_{\|x\|_3=1} \|Ax\|_3$$
For $y \in \mathbb R$, let
$$A_y = \left(\begin{matrix} 1 & y \\ 0 & 1 \end{matrix}\right)$$
Give a table showing $\|A_y\|_3$ for $y = 1, \dots, 9$.

 A: Well, I've gone through it. I would say the reason this is given to you in $\mathbb R^2$ is for you to see some actual pictures. First, I like $x,y$ for the coordinates, let us call the matrix $A_w$ for $w = 1, 2, \ldots,9.$ The "superellipse" given by $$ |x|^3 + |y|^3 = 1,  $$ appearance discussed HERE, 
 can be parametrized in the first quadrant by $$ x = (\cos t)^{2/3},  \; \;  y = (\sin t)^{2/3},  \; \; 0 \leq t \leq \pi/2.   $$ In the second quadrant,   $$ x = - |\cos t|^{2/3},  \; \;  y = (\sin t)^{2/3},  \; \; \pi / 2 \leq t \leq \pi.   $$ If you have (or write) a function that is traditionally called "signum," where signum of a real number is $1$ if the number is positive, $-1$ if the number is negative, and $0$ if the number is itself $0,$ you can write the parametrization for the entire superellipse. Anyway, the matrix $A_w$ takes such a column vector with entries $x,y$ to $(x+wy,y).$ You can simply have the computer tell you the value of the 3-norm at these points $(x+wy,y)$ for a fairly fine division of $t.$ Once you have the values of $t$ where the 3-norm is largest, restrict to that region and subdivide the $t$ values 10 times smaller. The "sheared" superellipse is $$ x = (\cos t)^{2/3} + w (\sin t)^{2/3},  \; \;  y = (\sin t)^{2/3},  \; \; 0 \leq t \leq \pi/2.   $$ The linear transformation you were given is called a "shear" from physics traditions.
Meanwhile, it is of course true that this can be done with Lagrange multipliers, but the calculation is not elegant and I do not think that is what the instructor wants. Program this, output some pictures, keep subdividing $t$ to get better accuracy, learn something gritty and hands-on.  
A: We have
$$A_y x = \begin{pmatrix} x_1 + y x_2 \\ x_2 \end{pmatrix} .$$
You want to maximize
$$(x_1 + yx_2)^3 + x_2^3$$
with the constraint
$$x_1^3 + x_2^3 = 1.$$
In other words, you want to find the saddle point of the Lagrangian
$$L(x,\lambda) = (x_1 + yx_2)^3 + x_2^3 + \lambda (x_1^3 + x_2^3 - 1).$$
This is very simple to do in any CAS or even by hand.
