5

Spivak's Calculus on Manifolds asks the reader to prove this (problem 1-8, pp.4-5):

If there is a basis $x_1, x_2, ..., x_n$ of $\mathbb{R}^n$ and numbers $\lambda_1, \lambda_2, ..., \lambda_n$ such that $T(x_i) = \lambda_i x_i$, $1 \leq i \leq n$, prove that $T$ is angle-preserving iff $\left| \lambda_i \right| = c, 1 \leq i \leq n$.

Here "angle-preserving" means that the linear map $T$ satisfies $$\frac{ \langle x, y \rangle}{\|x\| \|y\|} = \frac{ \langle T(x), T(y) \rangle}{\|T(x)\| \|T(y)\|},$$ and that $T$ is injective.

My first problem with this question is that the claim is false. Taking $n = 2$, $x_1 = (1, 0)$, $x_2 = (1,1)$, $T(x_1) = -x_1$, $T(x_2) = x_2$, and setting $x = x_1$, $y = x_1 + x_2$, the expression in the RHS above evaluates to $0$, while the expression in the LHS evaluates to $\frac{1}{\sqrt{2}}$.

My second, bigger problem is that I'm not really understanding what's going on. An earlier part of the problem had me show that norm-preserving matrices are angle-preserving; this I'm not sure I get. Thus, I'm not sure what true "version" of this statement the author had in mind (was he trying to get a converse?) and I don't know what to do.

Here's my guess:

Looking at some transformations in $\mathbb{R}^2$ (just drawing them), it looks like some of them could be "flip the sign of a basis vector" IF it's an orthogonal basis. However, I can't seem to recover the usual "rotation through an angle $\theta$" transformation this way, so I'm not sure that requiring the basis be orthogonal makes the statement true.

Also, I'm not sure how to take the inner product of vectors which aren't in standard coordinates. Or am I missing something here?

user1551
  • 139,064
Chris
  • 4,865

3 Answers3

8

You are correct. This is one of the few errors in Spivak's Calculus on Manifolds. For this particular exercise, see the following questions:

copper.hat's answer to the first question cited in the above used the same counterexample as yours.

For your second question, I don't think there is any conflict. Spivak simply meant that if $T$ is a norm-preserving map, then it automatically enjoys the property of being angle-preserving. He did not say that the converse is true. In other words, norm preservation is a stronger condition than angle preservation. In fact, $T$ is norm-preserving map if and only if its matrix w.r.t. the standard basis is a real orthogonal matrix $Q$, and $T$ is angle-preserving if and only if its matrix w.r.t. the standard basis is $\lambda Q$ for some scalar $\lambda>0$ and some real orthogonal matrix $Q$ (for a proof, see the aforementioned answer by copper.hat).

Community
  • 1
user1551
  • 139,064
  • Thank you! I did do a bit of searching but none of the language I used had turned up these pages. – Chris Apr 15 '13 at 00:38
1

A possible "true 'version' of this statement the author had in mind" is

Suppose $T(x_i)=\lambda_ix_i$ for some basis $x_1,\dots,x_n$ of $\mathbb R^n$ and numbers $\lambda_1,\dots,\lambda_n.$

  • If $T$ is angle preserving, then every $|\lambda_i|=|\lambda_j|.$
  • If every $\lambda_i=\lambda_j$, then $T$ is angle preserving.

The second part is trivial. A proof of (the contrapositive of) the first part can be found at https://math.stackexchange.com/a/177042.

Community
  • 1
Matt S
  • 143
  • 4
-1

I noticed this too. If you ignore the absolute value on lambda, it's a true statement which can be easily proven.

Alexander Jones
  • 1
  • This is incorrect. The map $T(x,y) = (x,-y)$ has distinct eigenvalues ($1$ and $-1$) and is angle preserving. –  Mar 24 '20 at 04:06