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Question:

The Three-norm on $R^n$ is defined as:

$$||x||_3=(|x_1|^3+\cdots+|x_n|^3)^{1/3}$$

The natural matrix norm it induces on $R^{n \times n}$ is

$$||A||_3 = \max\{||Ax||_3 : ||x||_3=1\}$$

For $y \in R$, let

$$A_y = \left(\begin{matrix} 1 & y \\ 0 & 1 \end{matrix}\right)$$

1) Give a table showing $||A_y||_3$ for $y = 1, \ldots, 9$.

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Since it's about numerical method, you have to write a program. What have you done? –  Davide Giraudo Apr 28 '12 at 18:53
    
Working with Maple, writing a script. In the last 2 days, i have tried several things. But i cannot find much information on the 3 norm. the 1 2 and infinity are simple. Honestly, i have nothing done and i have no idea where to start. Virtually given up. –  Special--k Apr 28 '12 at 18:58
    
It's an optimization problem with constraints... Hint: You can replace the constraint $\|x\|_3=1$ by $\|x\|_3\leq 1$ which makes it a convex constraint... –  Dirk Apr 28 '12 at 20:02
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1 Answer

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.

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