Question about Angle-Preserving Operators This an exercise out of Spivak's "Calculus on Manifolds".

Edit:  There was a typo in the exercise as is noted below in the answers.  The statement has been edited to reflect this.

Given $x,y\in\mathbb{R}^{n}$, the angle between $x$ and $y$ is defined by

$$\angle(x,y) = \arccos\left(\frac{\langle x,y \rangle}{|x|\cdot |y|}\right),$$ where $\langle x,y \rangle$ denotes the standard Euclidean inner product.

A linear operator $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is said to be angle-preserving if
$\angle(T(x),T(y)) = \angle(x,y)$ for every $x,y\in\mathbb{R}^{n}$.

The exercise as stated:

Let $\{x_{1},\dots, x_{n}\}$ be a basis for $\mathbb{R}^{n}$.  Then suppose that $\lambda_{1}, \dots, \lambda_{n}\in \mathbb{R}$ are such that $Tx_{j} = \lambda_{j}x_{j}$ for each $j = 1,\dots, n$.

Then $T$ is angle-preserving only if (not if and only if!)$|\lambda_{i}| = |\lambda_{j}|$ for every $1\leq i\leq j\leq n$.

I'm having problems with the $(\Rightarrow)$ direction.

My best attempt (which seems to lead nowhere) is to suppose that $|\lambda_{j}|\neq |\lambda_{k}|$.  Then by assumption,
\begin{align*}
\angle(Tx_{j},Tx_{k}) &amp; = \arccos\left(\frac{\langle Tx_{j},Tx_{k} \rangle}{|Tx_{j}|\cdot |Tx_{k}|}\right)\\
&amp; = \arccos\left(\frac{\langle \lambda_{j}{x_{j}},\lambda_{k}{x_{k}} \rangle}{|\lambda_{j}{x_{j}}|\cdot |\lambda_{k}{x_{k}}|}\right)\\
&amp; = \arccos\left(\frac{\lambda_{j}\lambda_{k}\langle {x_{j}},{x_{k}} \rangle}{|\lambda_{j}|\cdot|\lambda_{k}|\cdot|{x_{j}}|\cdot |{x_{k}}|}\right)\\
&amp; = \arccos\left(\text{sign}(\lambda_{j})\text{sign}(\lambda_{k})\frac{\langle {x_{j}},{x_{k}} \rangle }{|{x_{j}}|\cdot |{x_{k}}|}\right)\\
\end{align*}
may also be calculated as
\begin{align*}
\angle(Tx_{j},Tx_{k}) &amp; =  \angle(x_{j},x_{k})\\
&amp; = \arccos\left(\frac{\langle x_{j},x_{k} \rangle}{|x_{j}|\cdot |x_{k}|}\right).
\end{align*}

Then since $\arccos$ is injective, I believe I can make the jump that $\text{sign}(\lambda_{j})\text{sign}(\lambda_{k}) = 1$, which does not resemble the conclusion that I should arrive at.

Note:  I wasn't sure what tag to put this under, so anyone who knows better please feel free to adjust.

Thanks for any help you can give.
If necessary, rescale each $x_i$ so they all have unit length and relabel the indices so that $|\lambda_1|\neq |\lambda_2|$.

Now, consider the vectors $v_1 = x_1 + x_2$ and $v_2 = x_1-x_2$.

First, I claim that $v_1$ and $v_2$ are orthogonal, for \begin{align*} \langle v_1, v_2\rangle &amp;= \langle x_1+x_2,\; x_1-x_2\rangle \\\ &amp;= \langle x_1, x_1\rangle + \langle x_1, x_2\rangle - \langle x_2,x_1\rangle - \langle x_2, x_2\rangle \\\ &amp;= |x_1|^2-|x_2|^2 \\\ &amp;=0\end{align*} where the last equality follows since both $x_1$ and $x_2$ are unit length.

Next, I claim that $Tv_1$ and $Tv_2$ are not orthogonal.  For, \begin{align*} \langle Tv_1,  Tv_2\rangle &amp;= \langle \lambda_1 x_1 + \lambda_2 x_2,\; \lambda_1 x_1 - \lambda_2 x_2\rangle \\\ &amp;= \lambda_1^2 \langle x_1, x_1\rangle +\lambda_1 \lambda_2 \langle x_1, x_2\rangle - \lambda_2\lambda_1 \langle x_2,x_1\rangle - \lambda_2^2 \langle x_2, x_2\rangle \\\ &amp;= \lambda_1^2 |x_1|^2 - \lambda_2^2 |x_2|^2 \\\ &amp;= \lambda_1^2 - \lambda_2^2\end{align*}

and this last line is $0$ iff $|\lambda_1| = |\lambda_2|$.
New bug in above arguments! If $AA^T=\sigma ^2I$ we cannot say $A=\sigma I$. Counterexample:  take a diagonal matrix with $+\sigma$ or $-\sigma$ on diagonal, and $0$ elsewhere.

The very final answer to these circle of ideas is 
Which I have proved with ALL details, and optimal care to such issues as orientation.
For 1 => 2 see ALL Orthogonality preserving linear maps from $\mathbb R^n$ to $\mathbb R^n$?
Please challenge.