Prove: T is Angle-Preserving iff |ai| are Equal

  • Thread starter Adeimantus
  • Start date
  • Tags
    Ai
In summary, the author shows that if all the angles in a transformation are preserved, then the transformation is angle-preserving. However, if one angle is not preserved, then the transformation is not angle-preserving.
  • #1
Adeimantus
113
1

Homework Statement


1. Problem 1-8(b)

If there is a basis x1,..., xn of Rn and numbers a1,..., an such that Txi = aixi (where T is a linear operator), prove that T is angle-preserving if and only if all the |ai| are equal.



Homework Equations


First a couple of definitions...

Definition (i): If x,y are nonzero vectors in Rn, the angle between them is given by

angle(x,y) = arccos {<x,y>/ ( |x | |y | )}

where < , > denotes the standard inner product and | | the standard norm.

Definition (ii): A linear transformation T is angle-preserving if T is 1-1 and for x,y != 0 we have that angle(Tx, Ty) = angle(x,y).


The Attempt at a Solution


Problem is, I think I see a simple counterexample to the sufficiency part of the proof. Let x1, x2 be a basis for R2 where

x1 = e1 (i.e. standard unit basis vector) and
x2 = e1 + e2

Then suppose Tx1 = x1 and Tx2 = -x2. Now consider the angle between x1 and x1+x2:

angle(x1, x1+x2) = angle(e1, 2e1+e2) ~ 25 degrees, but

angle(Tx1,T(x1+x2)) = angle(x1, x1-x2) = angle(e1, -e2) = 90 degrees.

So even though the absolute values of a1 = 1 and a2 = -1 are equal, it seems T does not preserve angles. What am I misunderstanding about this problem? I am assuming the author means angle-preserving with respect to the standard inner product in which <e1, e2> = 0.
 
Last edited:
Physics news on Phys.org
  • #2
Hi, I know this is a blast from the distant past (14.5 years?) but I was just doing this problem and ran into this exact issue. I, in fact, constructed the identical counter example that you did. When I searched Spivak problem 1-8 and found this, I had to make an account just to say something. IMO, your conclusion is correct, the ##|\lambda|## condition is only for orthogonal bases.

Did you ever get further confirmation on this? I am going to go forward under the assumption that the condition is in general equal lambda except for orthogonal bases, where it expands to equal absolute value of lambda. At any rate, I like this book but Spivak can be a bit fast and loose sometimes, and this seems to be one of those examples where he could stand for a bit more rigorous language at the expense of making the book a bit longer. The book is short so a few more pages would not be too bad, and IMO make the book easier (since I often find myself filling in what he means).

FWIW, I would bet the technical reason behind this outcome is basically that ##0 = -0##. When you equate the cosine subject to the condition that each transformed eigenvector remains co-linear with itself, there are three options (from a 2D perspective). You can leave both fixed (which does not change the relative angle, or any angles), advance both 180 degrees (which does not change the relative angle), or advance one by 180 degrees (which does change the relative angle and corresponds to the lambdas being of opposite sign). In this case, you need to solve:
$$
\begin{align*}
\cos(\theta) &= \cos(\theta+\pi) \\
\cos(\theta) &= \cos(\theta)\cos(\pi) - \sin(\theta)\sin(\pi) \\
\cos(\theta) &= - \cos(\theta)
\end{align*}
$$
This is clearly only possible if ##\cos(\theta) = 0##. In higher dimensions, the above analysis is true for the 2D subspace containing the two eigenvectors.
 
Last edited:
  • #3
Adeimantus said:

Homework Statement


1. Problem 1-8(b)

If there is a basis x1,..., xn of Rn and numbers a1,..., an such that Txi = aixi (where T is a linear operator), prove that T is angle-preserving if and only if all the |ai| are equal.

Homework Equations


First a couple of definitions...

Definition (i): If x,y are nonzero vectors in Rn, the angle between them is given by

angle(x,y) = arccos {<x,y>/ ( |x | |y | )}

where < , > denotes the standard inner product and | | the standard norm.

Definition (ii): A linear transformation T is angle-preserving if T is 1-1 and for x,y != 0 we have that angle(Tx, Ty) = angle(x,y).

The Attempt at a Solution


Problem is, I think I see a simple counterexample to the sufficiency part of the proof. Let x1, x2 be a basis for R2 where

x1 = e1 (i.e. standard unit basis vector) and
x2 = e1 + e2

Then suppose Tx1 = x1 and Tx2 = -x2. Now consider the angle between x1 and x1+x2:

angle(x1, x1+x2) = angle(e1, 2e1+e2) ~ 25 degrees, but

angle(Tx1,T(x1+x2)) = angle(x1, x1-x2) = angle(e1, -e2) = 90 degrees.

So even though the absolute values of a1 = 1 and a2 = -1 are equal, it seems T does not preserve angles. What am I misunderstanding about this problem? I am assuming the author means angle-preserving with respect to the standard inner product in which <e1, e2> = 0.
Maybe you can use that the only matrices that preserve angles are the orthogonal matrices to craft ant
(counter) argument.

 

1. What does it mean for T to be angle-preserving?

Angle-preserving, also known as isometric or congruent, means that the transformation T preserves the size and shape of angles. In other words, if two angles are congruent in the original figure, they will also be congruent in the transformed figure.

2. How can I prove that T is angle-preserving?

To prove that T is angle-preserving, you must show that for any two angles in the original figure that are congruent, their images under T are also congruent. This can be done through various methods, such as using the definition of isometry or showing that the transformation preserves distance and parallelism.

3. What does the notation |ai| stand for?

The notation |ai| represents the absolute value of the scalar factor ai. This scalar factor is used to describe how the original figure is transformed, such as by dilation or reflection.

4. Why are the absolute values of ai important in proving T is angle-preserving?

The absolute values of ai are important because they determine the scale of the transformation. If all the absolute values are equal, it means that the transformation is uniform and preserves angles. If the absolute values are not equal, it means that the transformation may change the size of angles, making it not angle-preserving.

5. Can T be angle-preserving if the absolute values of ai are not equal?

No, for T to be angle-preserving, it is necessary that the absolute values of ai are equal. If they are not equal, it means that the transformation will not preserve angles, as it will change the scale of the figure. In this case, the transformation would be classified as non-uniform or non-isometric.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
5K
Replies
3
Views
2K
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
10
Views
4K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
11K
Back
Top