Positive coroot pairings of a dominant integral weight

For every simple root ${\alpha }_i$ one has $⟨\lambda ,{\alpha }_i^{\vee }⟩\in {\mathbb{Z}}_{\ge 0}$ by dominance. Applying $s_i$ to $2\rho ={\sum }_{\alpha \in {\mathrm{\Phi }}^{+}}\alpha$ and using that $s_i$ permutes the positive roots other than ${\alpha }_i$ while $s_i{\alpha }_i=-{\alpha }_i$ gives $2s_i\rho =2\rho -2{\alpha }_i$, whereas $s_i\rho =\rho -⟨\rho ,{\alpha }_i^{\vee }⟩{\alpha }_i$. Comparing the two expressions gives $⟨\rho ,{\alpha }_i^{\vee }⟩=1$, so $⟨\lambda +\rho ,{\alpha }_i^{\vee }⟩\in {\mathbb{Z}}_{>0}$.

Let $\beta \in {\mathrm{\Phi }}^{+}$ and write $\beta ={\sum }_i{n}_i{\alpha }_i$ with $n_i\in {\mathbb{Z}}_{\ge 0}$, not all zero. Put $c_i:=⟨{\omega }_i,{\beta }^{\vee }⟩=\frac{2({\omega }_i,\beta )}{(\beta ,\beta )}$. Then $c_i\in \mathbb{Z}$ because ${\omega }_i\in P$ and ${\beta }^{\vee }$ is the coroot of the root $\beta$, and $c_i\ge 0$ because $({\omega }_i,\beta )={\sum }_j{n}_j({\omega }_i,{\alpha }_j)=\frac{1}{2}{n}_i({\alpha }_i,{\alpha }_i)\ge 0$ and $(\beta ,\beta )>0$. Since the ${\omega }_j$ are a basis of $E$ and both sides have the same pairing with every ${\omega }_j$, this gives the identity ${\beta }^{\vee }={\sum }_i{c}_i{\alpha }_i^{\vee }$: every positive coroot is a nonnegative integral combination of the simple coroots, and for $\beta >0$ at least one $c_i$ is positive.

Combining the two previous steps, $⟨\lambda +\rho ,{\beta }^{\vee }⟩={\sum }_i{c}_i⟨\lambda +\rho ,{\alpha }_i^{\vee }⟩$ is a nonnegative integer combination in which at least one coefficient is positive and every paired simple coroot contributes at least $1$; hence it lies in ${\mathbb{Z}}_{>0}$. For a negative root $-\beta$ one has $⟨\lambda +\rho ,(-\beta {)}^{\vee }⟩=-⟨\lambda +\rho ,{\beta }^{\vee }⟩<0$, so $\lambda +\rho$ pairs nontrivially with the coroot of every root and is therefore regular: it lies on no root hyperplane.

A regular vector lies in some open chamber, and $W$ acts simply transitively on open chambers, so its stabilizer is trivial. If $w\circ \lambda =w^{\mathrm{\prime }}\circ \lambda$, then $w(\lambda +\rho )=w^{\mathrm{\prime }}(\lambda +\rho )$ and hence $(w^{\mathrm{\prime }-1}w)(\lambda +\rho )=\lambda +\rho$; triviality of the stabilizer gives $w^{\mathrm{\prime }-1}w=1$, that is $w=w^{\mathrm{\prime }}$.

For $w\in W$ and $\beta \in \mathrm{\Phi }$ one has $(w\beta {)}^{\vee }=w({\beta }^{\vee })$ because $w$ preserves the form, and pairing with $w(\lambda +\rho )$ against $w({\beta }^{\vee })$ equals pairing against ${\beta }^{\vee }$ because $w$ is an isometry. Hence $⟨w(\lambda +\rho ),(w\beta {)}^{\vee }⟩=⟨\lambda +\rho ,{\beta }^{\vee }⟩$, and since $w\circ \lambda +\rho =w(\lambda +\rho )$ the second displayed identity is the same statement; the nonnegative-integral-combination statement at a simple coroot, transported along $W$, is the "$W$-invariance" used throughout.