A problem on two bases $\Delta$ and $\Delta'$ of a root system 
$\Phi $ is a root system in Euclid space $E$, $\Delta＝\{\alpha_1, \alpha_2, \dotsc,\alpha_l\}$ is the base of $\Phi$. If $\Delta'＝\{\beta_1, \beta_2, \dotsc, \beta_l\}\subset\Phi$ satisfies that $(\alpha_i, \alpha_j)＝(\beta_i, \beta_j)  $ for all $1\leqslant i, j \leqslant l$, then $\Delta'$ is also a base of $\Phi$.

Apparently $\Delta'$ is a basis of $E$, now all I gotta do is to confirm that every root in $\Phi$ is in the lattice spanned by $\Delta'$ and has coefficients of same sign. Please give me some hint, thanks in advance.
 A: We may assume that the vector space $E$ is spanned by the root system $\Phi$.
Otherwise we can replace the space $E$ by the span of $\Phi$.
The simple system $\Delta$ is then a vector space basis of $E$.
There hence exists a unique vector space endomorphism $w$ of $E$ which maps $\alpha_i$ to $\beta_i$ for all $i = 1, \dotsc, n$.
It follows from the condition
$$
  ( \alpha_i, \alpha_j ) = ( \beta_i, \beta_j )
  \qquad
  \text{for all $i, j = 1, \dotsc, n$}
$$
that the endomorphism $w$ is an isometry.
Is is therefore in particular an isomorphism of vector spaces.
We can now adopt the approach from a very similar question (which is a special case of this question).
We repeat the argument here for the sake of completeness.
It follows that the set $\Delta' = w \Delta$ is a basis for the root system $\Phi' = w \Phi$ in $E$.
It thus sufficies to show that $\Phi' = \Phi$.
We note that
$$
  s_{\beta_i}
  =
  s_{w(\alpha_i)}
  =
  w s_{\alpha_i} w^{-1}
$$
for all $i = 1, \dotsc, l$.
It follows for the Weyl groups $W$ and $W'$ of the root systems $\Phi$ and $\Phi'$ that
$$
  W'
  =
  \langle s_{\beta} \mid \beta \in \Delta' \rangle
  =
  \langle w s_{\alpha} w^{-1} \mid \alpha \in \Delta \rangle
  =
  w \langle s_{\alpha_i} \mid \alpha \in \Delta \rangle w^{-1}
  =
  w W w^{-1} \,.
$$
We now use that $\Delta'$ is contained in $\Phi$.
It follows from this inclusion that
$$
  W'
  =
  \langle s_\beta \mid \beta \in \Delta' \rangle
  \subseteq
  \langle s_\alpha \mid \alpha \in \Phi \rangle
  =
  W \,.
$$
Both $W$ and $W'$ have the same cardinality (since they are conjugated), so $W' = W$.
The root system $\Phi$ is the $W$-orbit of $\Delta$, and the root system $\Phi'$ is the $W'$-orbit of $\Delta'$.
We use the equality $W' = W$ and that $\Delta'$ is contained in $\Phi$ to find that
$$
  \Phi'
  =
  W' \Delta'
  =
  W \Delta'
  \subseteq
  W \Phi
  =
  \Phi \,.
$$
But both root systems $\Phi$ and $\Phi'$ have the same cardinality because $\Phi'$ is the image of $\Phi$ under the isometry $w$.
Thus $\Phi' = \Phi$.
