Once a weight vector of a representation $\psi$ is known, we have to find other weights by applying the operations $\psi(Y_j)$ or $\psi(X_j)$; but at the time being we don't know if this gives us another weight vector or null. The Weyl group is some sort of a symmetry group of the weights and thus given any weight - we don't need to know a corresponding weight vector - the Weyl group gives us other weights, which, by the way, have also the same multiplicity. We start by another more geometric approach to weights: The subalgebra ${\cal H}$ generated by $H_1$ and $H_2$ is a maximal commutative subalgebra of $\sla(3,\C)$, it's called the Cartan subalgebra of $\sla(3,\C)$. The set $N$ of all $g\in\SU(3)$ that leave ${\cal H}$ invariant under the adjoint representation $\Ad$, i.e. $\Ad(g)({\cal H})\sbe{\cal H}$ - this is called the normalizer of ${\cal H}$ and it's obviously a subgroup of $\SU(3)$ (cf. section). Let us investigate this subgroup more closely: for any $g\in N$ the matrices $gH_jg^{-1}$ are in ${\cal H}$, hence they must be diagonal, i.e. the standard basis vectors $e_1,e_2,e_3$ are eigen-vectors. Since on the other hand the eigen-vectors of $gH_jg^{-1}$ are $ge_1,ge_2$ and $ge_3$, we infer that $ge_1,ge_2$ and $ge_3$ must be a permutation of the standard basis $e_1,e_2$ and $e_3$ up to factors of modulus $1$, i.e.: there are $\theta_1,\theta_2,\theta_3\in\R$ and a permutation $\pi\in S_3$ such that
$$
ge_j=e^{i\theta_j}e_{\pi(j)}
\quad\mbox{and}\quad
\sign(\pi)e^{i\sum\theta_j}=1~.
$$
Conversely, any such $g$ lies in $N$. A weight $\l$ was defined as a pair of complex numbers: the eigen-values of $\psi(H_1)$ and $\psi(H_2)$ for some common eigen-vector $x$, but in fact for every $H\in{\cal H}$ we have $H=a_1H_1+a_2H_2$ and $\l(H)\colon=a_1\l(H_1)+a_2\l(H_2)$ is the eigen-value of $H$ with eigen-vector $x$. Thus we may conveniently think of a weight as a linear functional $\l$ on the space ${\cal H}$. Moreover, endowing ${\cal H}$ with the euclidean product:
$$
\la H,G\ra\colon=\tr(HG^*)
$$
we know that any linear functional $\l$ can be written as $\l(H)=\la H,G\ra$ for a unique $G\in{\cal H}$. Thus we've got our new definition: a weight of a representation $\psi$ of $\sla(3,\C)$ in some finite dimensional vector-space $E$ is a vector $\l\in{\cal H}$ such that there exists some $x\in E\sm\{0\}$ such that
$$
\forall H\in {\cal H}:\quad
\psi(H)x=\la H,\l\ra x
$$
Hence weights are particular vectors in the Cartan algebra ${\cal H}$ of $\sla(3,\C)$!
1. If $\l_1=diag\{x,y,z\}$, then $x+y+z=0$, $x-y=1$ and $y-z=0$, i.e. $y=-1/3$, $z=-1/3$ and $x=2/3$, i.e. $\l_1$ is the electric charge operator $Q$; analogously we get: $\l_2=diag\{1/3,1/3,-2/3\}$ which is the hypercharge operator $Y$. 3. $\la H_1+H_2,\l_1+\l_2\ra=1-0+0-1=0$.
For $g\in N$ and $H,G\in{\cal H}$ we have:
$$
\la\Ad(g)H,\Ad(g)G\ra
=\tr(gHg^{-1}(gGg^{-1})^*)
=\tr(gHG^*g^{-1})
=\tr(HG^*)
=\la H,G\ra
$$
and thus $\Ad(g)$, $g\in N$, acts isometrically on $({\cal H},\la.,.\ra)$. Also, roots are weights of the adjoint representation; hence, in our new interpretation, a root is an element $r\in{\cal H}$ such that for some root vector $X\in\sla(3,\C)$:
$$
\forall H\in{\cal H}:\quad
\ad(H)X=\la r,H\ra X~.
$$
Since $\l_1,\l_2$ is the dual basis to $H_1,H_2$, we must have: $r_1=2\l_1-\l_2=diag\{1,-1,0\}=H_1$ and $r_2=-\l_1+2\l_2=diag\{0,1,-1\}=H_2$.
The root of $X_1$ is $(2,-1)$, which is in our new interpretation: $2\l_1-\l_2=H_1$; similarly the root of $X_2$ is $(-1,2)$, which now becomes $-\l_1+2\l_2=H_2$; the root of $Y_1$ is $(-2,1)$, which now becomes $-2\l_1+\l_2=-H_1$ and finally the root of $Y_2$ is $(1,-2)$, which now becomes $\l_1-2\l_2=-H_2$.
$\proof$
First we show, that for all $g\in N$ the vector $\Psi(g)x$ is a weight vector with weight $\Ad(g)\l$: for all $H\in{\cal H}$ we have $g^{-1}Hg\in{\cal H}$ and $\psi(g^{-1}Hg)=\Psi(g)^{-1}\psi(H)\Psi(g)$ (cf. 
