# Exercise 6.A.17 in "Linear Algebra Done Right 3rd Edition" by Sheldon Axler. I am worried if my solution is ok.

I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.

6.A.17 Prove or disprove: there is an inner product on $$\mathbb{R}^2$$ such that the associated norm is given by $$||(x,y)||=\max\{x,y\}$$ for all $$(x,y)\in\mathbb{R}^2$$.

I solved this exercise but I am worried if my solution is ok because this exercise appears to be unnaturally too easy.

My solution is the following:

If there is an inner product on $$\mathbb{R}^2$$ such that the associated norm is given by $$||(x,y)||=\max\{x,y\}$$ for all $$(x,y)\in\mathbb{R}^2$$, then $$0<||(-1,-1)||=\max\{-1,-1\}=-1$$.

• Yep, that's a good counterexample. May 10, 2022 at 2:41
• This is correct, assuming no transcription errors from the question. A trickier (and IMHO more worthwhile) problem is to show that $\|(x, y)\| = \max \{|x|, |y|\}$ cannot be generated by an inner product (and it wouldn't surprise me if this is what Axler intended). May 10, 2022 at 2:41
• I would bet serious money that @TheoBendit is right, and that it's a misprint. If you want to understand linear algebra, you should definitely be able to answer the "corrected" version. May 10, 2022 at 3:21
• The missing absolute values in Exercises 17 and 18 in Section 6.A of Linear Algebra Done Right are typos that will be corrected in the next edition of the book. May 10, 2022 at 14:19
• @tchappy ha I am working on the fourth edition of Linear Algebra Done Right now. It is still at least a year away from completion. May 11, 2022 at 19:15

As covered in the comments, you need absolute values to get what is called the $$\infty$$-norm: $$\|(x, y)\|_{\infty} = \max\{|x|, |y|\}.$$ This is the limit as $$p \to \infty$$ of the $$p$$-norm $$\|(x, y)\|_p = (|x|^p + |y|^p)^{1/p}$$ which generalizes the Euclidean norm ($$p = 2$$).

The rule for when a norm comes from an inner product is called the Parallelogram law which says that if $$\| \cdot \|$$ comes from an inner product, then for all $$u, v$$ $$\|u + v\|^2 + \|u - v\|^2 = 2(\|u\|^2 + \|v\|^2).$$ Moreover, if this identity holds for all $$u$$ and $$v$$, the inner product is recovered "by polarization" as $$\langle u, v \rangle = \frac{\|u + v\|^2 - \|u - v\|^2}{4}.$$

So continuing with Axler's exercise, show that the $$\infty$$-norm (with the absolute values) does not satisfy the Parallelogram law. Slightly more challenging: show that the Parallelogram law holds for all $$u, v$$ for the $$p$$-norm if and only $$p = 2$$. I recommend thinking about an actual parallelogram.

• Thank you very much for your perfect answer. May 10, 2022 at 4:30
• By chance, the next exercise (6.A.18) is "Suppose $p>0$. Prove that there is an inner product on $\mathbb{R}^2$ such that the associated norm is given by $||(x,y)||=(x^p+y^p)^{1/p}$ for all $(x,y)\in\mathbb{R}^2$ if and only if $p=2$." Are you a prophet? May 10, 2022 at 4:40
• @tchappyha It's only a norm if $p \ge 1$ (otherwise it doesn't satisfy the triangle inequality). Also you need absolute values again. This family of norms (including $p = \infty$) is very common and if you're discussing which norms come from an inner product, you're all but required to include such an exercise or example. May 10, 2022 at 4:53

Thank you very much Theo Bendit.

Prove or disprove: there is an inner product on $$\mathbb{R}^2$$ such that the associated norm is given by $$||(x,y)||=\max\{|x|,|y|\}$$ for all $$(x,y)\in\mathbb{R}^2$$.

My solution:

Assume that $$\langle(x,y),(x,y)\rangle=\max\{|x|,|y|\}^2$$.
Then, $$\langle(1,0),(1,0)\rangle=\langle(0,1),(0,1)\rangle=1$$.
Then, $$1=\langle(1,1),(1,1)\rangle=\langle (1,0)+(0,1),(1,0)+(0,1)\rangle=\langle(1,0),(1,0)\rangle+2\langle(1,0),(0,1)\rangle +\langle(0,1),(0,1)\rangle=1+2\langle(1,0),(0,1)\rangle +1=2+2\langle(1,0),(0,1)\rangle$$.
So, $$\langle(1,0),(0,1)\rangle=-\frac{1}{2}$$.
Then, $$2=\langle(1,2),(1,2)\rangle=\langle (1,0)+(0,2),(1,0)+(0,2)\rangle=\langle(1,0),(1,0)\rangle+4\langle(1,0),(0,1)\rangle +4\langle(0,1),(0,1)\rangle=1-2+4=3$$.