Typo in spivak's calculus on manifolds?

[Adeimantus]

In the first problem set of chapter 1, problem 1-8(b) deals with angle preserving transformations. In the newest edition of the book the problem is stated

If there is a basis x_1, x_2, ..., x_n and numbers a_1, a_2, ..., a_n such that Tx_i = a_i x_i, then the transformation T is angle preserving if and only if all |a_i| are equal.

In the first edition (the one with the cool blue diagrams that you can actually see), the absolute value bars are omitted on the a_i. However, I'm pretty sure there are counterexamples to both formulations. If the basis happened to be orthogonal, then equal absolute values of the a_i would be enough, but if the basis were oblique, then you would need equality of the a_i. But the author doesn't mention orthogonality until a couple exercises later. Am I missing something?

thanks.

 

[maze]

I found the same problem working through Spivak recently! I think it is a typo, and here's my counterexample:

Consider the matrix:
M=\left(\begin{matrix}1 & -2 \\ 0 & -1 \end{matrix}\right)

This has eigenvector (1,0)' corresponding to eigenvalue 1, and eigenvector (1,1)' corresponding to eigenvalue -1. (this is how I constructed the matrix in the first place actually). Since the eigenvalues are the same in absolute value (+/-1) and the eigenvectors form a basis for R2, according to this problem M should be angle preserving.

On the other hand, compare the angle between (1,0)' and (1,2)', to the angle between M(1,0)'=(1,0)' and M(1,2)'=(-3,-2)'. These angles aren't the same.

 

[Adeimantus]

I found the same problem working through Spivak recently! I think it is a typo, and here's my counterexample:

Consider the matrix:
M=\left(\begin{matrix}1 & -2 \\ 0 & -1 \end{matrix}\right)

This has eigenvector (1,0)' corresponding to eigenvalue 1, and eigenvector (1,1)' corresponding to eigenvalue -1. (this is how I constructed the matrix in the first place actually). Since the eigenvalues are the same in absolute value (+/-1) and the eigenvectors form a basis for R2, according to this problem M should be angle preserving.

On the other hand, compare the angle between (1,0)' and (1,2)', to the angle between M(1,0)'=(1,0)' and M(1,2)'=(-3,-2)'. These angles aren't the same.

Exactly. I considered the same matrix for T for my counterexample. Thanks for confirming. I figured there must be some easy fix to the problem to make it make sense, which is why I called it a "typo" rather than a "mistake". But it must be something deeper than just leaving off the absolute value sign, because if the basis happened to be orthogonal you could have eigenvalues +1 and -1 and still preserve angles.


I think it can be proved that equality of the |a_i| is necessary (but not always sufficient) for T to be angle preserving. If for example |a_j| < |a_i|, the angle between x_i and x_i + x_j would, under the map T, become the angle between x_i and x_i + (a_j / a_i)x_j. Assuming these two angles are equal

angle(x_i, x_i + x_j) = angle(x_i, x_i + (a_j / a_i)x_j).

Applying this relation N times gives

angle(x_i, x_i + x_j) = angle(x_i, x_i + (a_j / a_i)^N x_j) ----> 0 as N ---> infinity. This contradicts the assumption that the x_k 's form a basis.