Let's take the chain of permutation subgroups $S_1=\{1\}\subseteq S_2\subseteq .....\subseteq S_n$,where $S_{i-1}$ sits inside $S_i$ canonically as the stabilizer of $\{i\}$.Then for this inductive chain of subgroups we define Gelfand-Tsetlin algbera to be the subalgebra of $\mathbb C[S_n]$ generated by $Z_1\cup Z_2\cup.....\cup Z_n$, where $Z_i=$ center of the group algebra $\mathbb C[S_i]$.It can be shown that this special sub-algebra can be generated by so-called Young-Jucys-Murphy (YJM) elements,which are defined as follows,
$X_i:=(1\ i)+(2\ i)+.......+(i-1\ i)$;by definition $X_1=0$.
In other words,our Gelfand_Tsetlin algebra $GZ_n=\langle X_1,X_2,......,X_n\rangle$.
Now,take a finite dimensional irreducible complex representation $V$ of $S_n$ and restrict the representation to $S_{n-1}$,then restrict to $S_{n-2}$ and so on until we get down to irreducible $S_1=\{1\}$ modules i.e. one dimensional subspaces. Then up to a scalar we can choose a basis for this one-dimensional subspaces. These basis are known as Gelfand-Tsetlin basis or $GZ$-basis for $V.$
We also know that, YJM elements and hence any element of $GZ$ algebra act diagonally on the $GZ$-basis. Now I have following two questions.
$(i)$ If $v$ is an eigenvector of every element of $GZ_n$, then how can we conclude that $v$ belongs to the $GZ$-basis of $V$?
$(ii)$ Let $u,v$ be two $GZ$-vectors. If $u$ and $v$ have same eigenvalues for every element of $GZ_n$,then can we say $u=v$?
Any help would be appreciated. Thanks in advance.