A problem on norm preserving and angle preserving and their relations. I want to solve the following problem and finding some difficulties:-

I have done the part (a) easily.
My problem is in part (b) and (c).
In part (b) after calculation I have achieved that $\lambda_i \lambda_j=|\lambda_i| |\lambda_j|$.
From this we can only say that they are of same sign.
I can not understand how can their absolute values are equal.
In part (c) if I assume that (b) is true then we can say that all tha eigen values are same absolute values.Am I correct?
Can someone help me please?
 A: I assume the $x_i$s are an orthonormal basis.
Now it is sufficient if we can prove $(b)$ on a two-dimensional subspace of $\mathbb{R}^n$, because if $(b)$ is true on say $span \{x_1, x_2\}$ then $\lvert \lambda_1 \rvert = \lvert \lambda_2 \rvert$ and by the same argument has to be true on $span \{x_2, x_3\}$ i.e. $\lvert \lambda_2 \rvert = \lvert \lambda_3 \rvert$. But through this we have also proved that $\lvert \lambda_1 \rvert = \lvert \lambda_3 \rvert$. This carries through all $n$ base vectors.
First we note that $arccos$ is injective on its domain $[-1,1]$. Thus the notion of equal angles may just as well be stated without the $arccos$ acting on the argument.
Let $x=x_1 + x_2$ and $y=x_1$. Then we have that:
$\frac{\langle Tx, Ty \rangle}{\|T x\| \|T y\|} = \frac{\lambda_1^2}{\sqrt{\lambda_1^2 + \lambda_2^2} \cdot \lvert \lambda_1 \rvert }$
Equating this with $\frac{\langle x, y \rangle}{\|x\| \|y\|} = \frac{1}{\sqrt{2}}$ gives:
$\lvert \lambda_1 \rvert = \frac{1}{\sqrt{2}} \sqrt{\lambda_1^2 + \lambda_2^2} \implies 
2 \lambda_1^2 = \lambda_1^2 + \lambda_2^2 \implies \lambda_1^2 = \lambda_2^2 \implies \lvert \lambda_1 \rvert = \lvert \lambda_2 \rvert$ q.e.d
Thus we have proved that if $T$ is angle preserving, then $\lvert \lambda_i \rvert = \lvert \lambda_j \rvert$ for all $i, j$.
To prove the other direction, you have to choose general vectors $x, y$ in our two-dimensional subspace. But if you write it down, you will see that, since $\lambda_1^2 = \lambda_2^2$, all $\lambda$s cancel out and the angle remains the same.
$\implies (b)$ is true.
As of $(c)$, I don't really know what answer is expected... Given $(b)$ you have already fully characterised angle-preserving linear transforms from $\mathbb{R}^n$ to $\mathbb{R}^n$.
A: If a linear map preserves right angles then it preserves all angles and we can then deduce that all it has to be a multiple of an orthogonal matrix, ie. $AA^T=1$. See this:
ALL Orthogonality preserving linear maps from $\mathbb R^n$ to $\mathbb R^n$?
