A subset of roots whose mutual angles agree with those of a simple system I would appreciate help/hints solving the following exercise from Humphreys book "Reflection Groups and Coxeter Groups", page 11, exercise 1.

Let $\Phi$ be a root system of rank $n$ of unit vectors in $V$ and let
  $\Psi\subset\Phi$  be a subset of size $n$. If the mutual angles in
  $\Psi$ agree with those between the roots of some simple system, then
  $\Psi$ is a simple root system.

 A: Adding my own answer using a proposition which appears later in the book. This leaves me curious whether there is a solution using only the material appearing prior to the exercise.
Denote by $\Delta=\{\alpha_1,\dots,\alpha_n\}\subset\Phi$ the simple system with the same angles as $\Psi=\{\psi_1,\dots,\psi_n\}$. $\Phi$ is also normalized so $(\psi_i,\psi_j)=(\alpha_i,\alpha_j)$ for every $1\leq i,j \leq n$. Denote by $W$ the reflection group generated by the reflections $S_\beta$ such that $\beta\in\Phi$.
We need to show that $\Psi$ is a simple root system ($|\Psi|=n$ therefore it is enough to show that $\Psi$ is linearly independent and that every $\beta\in\Phi$ is a linear combination of $\Psi$ in which all the coefficients have the same sign).
First show that $\Psi$ is linearly independent; assume the contrary, so there is some linear combination $ \sum_{i=1}^n c_i\psi_i = 0$ where not all $c_i$ are equal to zero. Therefore
$$ 
0=
(\sum_{i=1}^n c_i\psi_i,\sum_{i=1}^n c_i\psi_i) = 
\sum_{i=1}^n\sum_{j=1}^n c_i c_j(\psi_i,\psi_j) = 
\sum_{i=1}^n\sum_{j=1}^n c_i c_j(\alpha_i,\alpha_j) = 
(\sum_{i=1}^n c_i\alpha_i,\sum_{i=1}^n c_i\alpha_i).
$$
Thus $\sum_{i=1}^n c_i\alpha_i=0$ where not all $c_i$ are equal to zero, in contradiction to $\Delta$ being linearly independent.
To show that every $\beta\in\Phi$ is a linear combination of $\Psi$ in which all the coefficients have the same sign, we will need the following proposition on page $24$ section $1.14$ of the same book where Humphreys shows that all the reflections in $W$ are of the form $S_\beta$ such that $\beta\in\Phi$. That is
$$(*) \; S_\beta\in W \Leftrightarrow \beta\in\Phi $$
Let $T$ be the orthogonal transformation defined by $T(\alpha_i)=\psi_i$ (if $\Phi$ does not span $V$ entirely, then complete $\Delta$ to be a base of $V$ and define that $T$ fixed point wise the additional elements).
Let $\beta$ be a root in $\Phi$ and $\beta=\sum c_i\psi_i$. Applying $T^{-1}$ to both side of the equation we get $T^{-1}(\beta)=\sum c_i T^{-1}(\psi_i)=\sum c_i \alpha_i$. This is a linear combination of $\Delta$, thus proving that $T(\Phi)=\Phi$ shows that all $c_i$ have the same sign completing the proof.
Following $(*)$, it is enough to show that the reflection $S_{T(\beta)}\in W$ in order to show that $T(\beta)\in\Phi$. The following completes the proof (recall that simple reflections generate $W$, thus $S_\beta$ is equal to a composition of some $k$ simple reflection $S_{\alpha_1}\cdots S_{\alpha_k}$):
$$ S_{T(\beta)} = TS_\beta T^{-1} = TS_{\alpha_1}\cdots S_{\alpha_k} T^{-1}  = TS_{\alpha_1}T^{-1}\cdot TS_{\alpha_2}T^{-1}\cdots TS_{\alpha_k} T^{-1} = S_{T(\alpha_1)}\cdots S_{T(\alpha_k)} = S_{\psi_1}\cdots S_{\psi_k} \in W.$$
A: While this question has been inactive for quite some time, I think it is worthwhile to add a solution which only uses material which has been introduced prior to the exercise.
For the sake of completeness I also repeat some of the arguments given in the previous answer.


Lemma:
  Let $V$ be a finite dimensional euclidian vector space.
  If $v_1, \dotsc, v_n \in V$ are linearly independent, then for $w_1, \dotsc, w_n \in W$ with $(v_i, v_j) = (w_i, w_j)$ for all $i, j$, there exists an orthogonal transformation $T \colon V \to V$ with $Tv_i = w_i$ for every $i$.
Proof:
  Let $v_{n+1}, \dotsc, v_m$ be an orthonormal basis of $\langle v_1, \dotsc, v_n \rangle^\perp$ and let $w_{n+1}, \dotsc, w_m \in \langle w_1, \dotsc, w_n \rangle^\perp$ be linearly independent.
  Then $v_1, \dotsc, v_m$ is a basis of $V$, so there exists a unique linear map $T \colon V \to V$ with $T v_i = w_i$ for every $i$.
  By assumption and construction we have that
  $$
    (T v_i, T v_j)
  = (w_i, w_j)
  = (v_i, v_j)
  \qquad
  \text{for all $i, j$},
$$
  and therefore $(T v, T w) = (v, w)$ for all $v, w \in V$ by bilinearity.

Suppose that $\Psi = \{ \beta_1, \dotsc, \beta_n \}$ and let $\Delta = \{ \alpha_1, \dotsc, \alpha_n \} \subseteq \Phi$ be a simple system such that for all $i, j$ the angle between $\beta_i$ and $\beta_j$ agrees with the angle between $\alpha_i$ and $\alpha_j$.
As $\Phi$ consists of unit vectors it follows that $(\alpha_i, \alpha_j) = (\beta_i, \beta_j)$ for all $i, j$.
By the above lemma there exists an orthogonal transformation $T \colon V \to V$ with $T \alpha_i = \beta_i$ for all $i$.
Since $\Delta$ is a simple system of the root system $\Phi$, it follows that $T \Delta = \Psi$ is a simple system for the root system $T \Phi =: \Phi'$.
We will show that $\Phi' = \Phi$.
Let
$$
  W  := \langle s_\alpha \mid \alpha \in \Phi \rangle
  \quad\text{and}\quad
  W' := \langle s_\alpha \mid \alpha \in \Phi' \rangle
$$
be the finite reflection groups associated to the root systems $\Phi$ and $\Phi'$.
By Theorem 1.5 both $W$ and $W'$ are generated by corresponding simple reflections, i.e.
$$
  W  = \langle s_\alpha \mid \alpha \in \Delta \rangle
  \quad\text{and}\quad
  W' = \langle s_\alpha \mid \alpha \in \Psi \rangle.
$$
Since $\Psi \subseteq \Phi$ it follows that
$$
            W'
  =         \langle s_\alpha \mid \alpha \in \Psi \rangle
  \subseteq \langle s_\alpha \mid \alpha \in \Phi \rangle
  =         W.
$$
On the other hand we have that
\begin{align*}
    W'
  &=  \langle s_\alpha \mid \alpha \in \Psi \rangle
   =  \langle s_\alpha \mid \alpha \in T \Delta \rangle
   =  \langle s_{T \alpha} \mid \alpha \in \Delta \rangle
\\
  &=  \langle T s_\alpha T^{-1} \mid \alpha \in \Delta \rangle
   =  T \langle s_\alpha \mid \alpha \in \Delta \rangle T^{-1}
   =  T W T^{-1},
\end{align*}
so $W'$ and $W$ have the same cardinality.
Together with $W' \subseteq W$ we find that $W = W'$.
By using Corollary 1.5 it now follows that
$$
            \Phi'
  =         W' \Psi
  =         W \Psi
  \subseteq W \Phi
  =         \Phi.
$$
From $\Phi' = T \Phi$ we also know that $\Phi'$ and $\Phi$ have the same cardinality, so $\Phi' = \Phi$.
