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.
On
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 $\Phi' = \Phi$.
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 $S_{\alpha_1}\cdots S_{\alpha_k}$): $$ S_{T(\beta)} = TS_\beta T^{-1} = TS_{\alpha_1}\cdots S_{\alpha_k} T^{-1} = TS_{\alpha_1}T^{-1}\cdot TS_{\alpha_2}T^{-1}\cdots TS_{\alpha_k} T^{-1} = S_{T(\alpha_1)}\cdots S_{T(\alpha_k)} = S_{\psi_1}\cdots S_{\psi_k} \in W.$$