[proofplan]
The seminormal basis diagonalizes the Jucys-Murphy elements, and the eigenvalue of $X_i$ on the basis vector $v_T$ is the content of the box containing $i$ in $T$. Therefore a vector with prescribed joint eigenvalue $a=(a_1,\dots,a_n)$ can only involve those basis vectors whose full content sequence is equal to $a$. Content separation for standard tableaux says that at most one standard tableau has a given content sequence, so the corresponding simultaneous eigenspace is either zero or the span of one seminormal basis vector.
[/proofplan]
[step:Record the Jucys-Murphy eigenvalues on the seminormal basis]
Let $\operatorname{SYT}(\lambda)$ denote the set of standard Young tableaux of shape $\lambda$. For $T \in \operatorname{SYT}(\lambda)$ and $1 \le i \le n$, let $\operatorname{row}_T(i)$ and $\operatorname{col}_T(i)$ denote the row and column of the box containing $i$ in $T$, and define the content
\begin{align*}
c_i(T) := \operatorname{col}_T(i)-\operatorname{row}_T(i).
\end{align*}
Define the content sequence map
\begin{align*}
c: \operatorname{SYT}(\lambda) &\to \mathbb C^n
\end{align*}
by
\begin{align*}
c(T) := (c_1(T),\dots,c_n(T)).
\end{align*}
Because $S^\lambda$ is a complex Specht module written in seminormal form with basis $(v_T)_{T \in \operatorname{SYT}(\lambda)}$, the hypotheses of [citetheorem:8447] apply. Hence, for every $T \in \operatorname{SYT}(\lambda)$ and every $1 \le i \le n$,
\begin{align*}
X_i v_T = c_i(T)v_T.
\end{align*}
[/step]
[step:Identify the simultaneous eigenspace with the span of basis vectors having content sequence $a$]
Fix $a=(a_1,\dots,a_n) \in \mathbb C^n$. Define
\begin{align*}
A_a := \{T \in \operatorname{SYT}(\lambda) : c_i(T)=a_i \text{ for every } 1 \le i \le n\}.
\end{align*}
We claim that
\begin{align*}
E_a = \operatorname{span}_{\mathbb C}\{v_T : T \in A_a\}.
\end{align*}
First let $v \in E_a$. Since $(v_T)_{T \in \operatorname{SYT}(\lambda)}$ is a basis of $S^\lambda$, there are unique scalars $b_T \in \mathbb C$, indexed by $T \in \operatorname{SYT}(\lambda)$, such that
\begin{align*}
v = \sum_{T \in \operatorname{SYT}(\lambda)} b_T v_T.
\end{align*}
For each $1 \le i \le n$, using the defining condition $X_i v=a_i v$ and the diagonal action from the previous step gives
\begin{align*}
0 = (X_i-a_i)v = \sum_{T \in \operatorname{SYT}(\lambda)} b_T(c_i(T)-a_i)v_T.
\end{align*}
[Linear independence](/page/Linear%20Independence) of the basis $(v_T)_{T \in \operatorname{SYT}(\lambda)}$ implies
\begin{align*}
b_T(c_i(T)-a_i)=0
\end{align*}
for every $T \in \operatorname{SYT}(\lambda)$ and every $1 \le i \le n$. Therefore, if $b_T \ne 0$, then $c_i(T)=a_i$ for all $1 \le i \le n$, so $T \in A_a$. Hence $v$ lies in $\operatorname{span}_{\mathbb C}\{v_T : T \in A_a\}$.
Conversely, if $T \in A_a$, then $c_i(T)=a_i$ for every $1 \le i \le n$, and the diagonal action gives
\begin{align*}
X_i v_T = c_i(T)v_T = a_i v_T
\end{align*}
for every $1 \le i \le n$. Thus $v_T \in E_a$, and by linearity every vector in $\operatorname{span}_{\mathbb C}\{v_T : T \in A_a\}$ lies in $E_a$. This proves the claimed equality.
[guided]
The point of this step is to translate a statement about simultaneous eigenspaces into a statement about which seminormal basis vectors are allowed to appear. Fix $a=(a_1,\dots,a_n) \in \mathbb C^n$, and define
\begin{align*}
A_a := \{T \in \operatorname{SYT}(\lambda) : c_i(T)=a_i \text{ for every } 1 \le i \le n\}.
\end{align*}
This is the set of standard tableaux whose Jucys-Murphy eigenvalue sequence is exactly $a$.
We prove that
\begin{align*}
E_a = \operatorname{span}_{\mathbb C}\{v_T : T \in A_a\}.
\end{align*}
Let $v \in E_a$. Since the vectors $(v_T)_{T \in \operatorname{SYT}(\lambda)}$ form a basis of $S^\lambda$, the vector $v$ has a unique expansion
\begin{align*}
v = \sum_{T \in \operatorname{SYT}(\lambda)} b_T v_T
\end{align*}
with coefficients $b_T \in \mathbb C$.
Now use the simultaneous eigenvector condition. For each $1 \le i \le n$, the condition $v \in E_a$ says $X_i v=a_i v$, hence
\begin{align*}
0 = (X_i-a_i)v.
\end{align*}
Substituting the basis expansion of $v$ and using the seminormal eigenvalue formula $X_i v_T=c_i(T)v_T$, we obtain
\begin{align*}
0 = \sum_{T \in \operatorname{SYT}(\lambda)} b_T(c_i(T)-a_i)v_T.
\end{align*}
Because the $v_T$ are linearly independent, every coefficient in this basis expansion is zero:
\begin{align*}
b_T(c_i(T)-a_i)=0
\end{align*}
for every $T \in \operatorname{SYT}(\lambda)$ and every $1 \le i \le n$.
This coefficient equation is the key filtering mechanism. If $b_T \ne 0$, then the factor $c_i(T)-a_i$ must be zero for every $i$, so $c_i(T)=a_i$ for all $1 \le i \le n$. Equivalently, $T \in A_a$. Therefore no basis vector outside $A_a$ can occur with a nonzero coefficient in a vector of $E_a$, and so
\begin{align*}
E_a \subseteq \operatorname{span}_{\mathbb C}\{v_T : T \in A_a\}.
\end{align*}
For the reverse inclusion, take $T \in A_a$. By definition of $A_a$, we have $c_i(T)=a_i$ for every $1 \le i \le n$. Applying the seminormal eigenvalue formula gives
\begin{align*}
X_i v_T = c_i(T)v_T = a_i v_T
\end{align*}
for every $1 \le i \le n$. Thus $v_T \in E_a$. Since $E_a$ is a complex vector subspace of $S^\lambda$, every complex linear combination of such $v_T$ also lies in $E_a$. Hence
\begin{align*}
\operatorname{span}_{\mathbb C}\{v_T : T \in A_a\} \subseteq E_a.
\end{align*}
Combining the two inclusions proves the equality.
[/guided]
[/step]
[step:Use content separation to show that at most one basis vector can occur]
We show that $A_a$ has at most one element. Suppose $S,T \in A_a$. Then, for every $1 \le i \le n$,
\begin{align*}
c_i(S)=a_i=c_i(T).
\end{align*}
Thus $S$ and $T$ are standard tableaux of the same size and have identical content values for every entry. By [citetheorem:8445], content separation gives $S=T$. Therefore $|A_a| \le 1$.
From the previous step,
\begin{align*}
E_a = \operatorname{span}_{\mathbb C}\{v_T : T \in A_a\}.
\end{align*}
If $A_a=\varnothing$, then this span is $0$. If $A_a=\{T\}$ for a unique $T \in \operatorname{SYT}(\lambda)$, then
\begin{align*}
E_a = \mathbb C v_T,
\end{align*}
which is one-dimensional because $v_T$ is a nonzero basis vector. Hence $E_a$ is either $0$ or one-dimensional, as required.
[/step]
