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Call a linear map $T: \mathbb{R}^n \to \mathbb{R}^n$ angle preserving in case $T$ is injective and the angle between $Tx$ and $Ty$ is equal to the angle between $x$ and $y$ for all $x, y \in \mathbb{R}^n - \{ 0 \}$, where the angle between $x$ and $y$ is $$\arccos \frac{x \cdot y}{||x||||y||}.$$ Suppose that $\{ x_i: 1 \le i \le n \}$ is a basis of $\mathbb{R}^n$ and also that $Tx_i = \lambda_i x_i$ for some $\lambda_i$. I'm trying to prove that $T$ is angle preserving if and only if the $|\lambda_i|$ are equal, but the "only if" direction is proving a pain. I'd guess it boils down to choosing $x$ and $y$ cleverly in the angle preservation condition, but no luck so far. Can anyone offer some guidance? Thanks.

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    $\begingroup$ The standard proofs of this theorem (see for instance cs.bsu.edu/homepages/fischer/math445/angles.pdf) state that $\lambda_i \gt 0$, and it seems to be a necessary condition. $\endgroup$
    – lmsteffan
    Apr 8, 2013 at 13:53
  • $\begingroup$ @lmsteffan The OP's conditions include injectivity, which preclude $\lambda_i=0$. $\endgroup$
    – rschwieb
    Apr 8, 2013 at 14:35
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    $\begingroup$ @rschwieb yes, my remark was more about the non-negativity of the $\lambda_i$. If we allow $\lambda < 0$ then I think we may have an inversion of the angle. $\endgroup$
    – lmsteffan
    Apr 8, 2013 at 14:42
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    $\begingroup$ @lmsteffan Reflections like $[x,y]\mapsto [x,-y]$ or even $[x,y]\mapsto [-x,-y]$ are certainly angle preserving, so I don't think the OP wants to eliminate those... $\endgroup$
    – rschwieb
    Apr 8, 2013 at 15:36

2 Answers 2

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Here is the geometric argument which should translate into an algebraic one.

Let $x_i$ and $x_j$ be two of the basis elements corresponding to eigenvalues $\lambda_i\neq \lambda_j$. It follows that the two vectors are linearly independent, and can form the sides of a nondegenerate triangle.

The triangle with vector sides $x_i,x_j, x_i-x_j$ is transformed by $T$ into the triangle $T(x_i),T(x_j)$, $T(x_i-x_j)$, which must share all the same angles as the first triangle. As such, they are similar triangles, and the corresponding side-lengths are proportional.

Since $\|T(x_i)\|=|\lambda_i|\|x_i\|$, we can see that the constant of proportionality is $|\lambda_i|$, so $\|T(x_j)\|=|\lambda_i|\|x_j\|$ as well. But then the equality $\|T(x_j)\|=|\lambda_i|\|x_j\|=|\lambda_j|\|x_j\|$ implies that $|\lambda_i|=|\lambda_j|$.

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  • $\begingroup$ How are you getting $||T(x_j) ||=|{\lambda}_i| \text{ } ||x_j||$ ? $\endgroup$
    – user422112
    Jul 23, 2018 at 2:31
  • $\begingroup$ @ThatIs One of the properties of the norm is that $\|\lambda x\|=|\lambda|\|x\|$, $\endgroup$
    – rschwieb
    Jul 23, 2018 at 13:12
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Think of the case $\mathbb{R}^2$ with basis $\{e_1=(1,0),e_2=(0,1)\}$, and say $T(e_1)=e_1$ and $T(e_2)=2e_2$. Then the angle between $(1,0)$ and $(1,1)$ will not be preserved (it will be greater than $45$ degrees). This is really all you need for the general case since if you argue by contradiction you just have to assume that two of the $\lambda_i$s are different.

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