A <hook partition> $\lambda=(n-k,1^k)$ has cell contents
$$
0,1,\ldots,n-k-1,-1,-2,\ldots,-k.
$$
In a <standard Young tableau>, the unique zero-content cell is $(1,1)$ and contains $1$. The other <eigenvalues> are precisely the remaining contents, in whatever order the tableau supplies. Their product is independent of that order:
$$
\boxed{X_2\cdots X_nw_T=(-1)^k k!(n-k-1)!\,w_T.}
$$
The scalar is nonzero over the <complex numbers>. For the single row or single column, one of the factorials is $0!=1$; the formula still gives $(n-1)!$ or $(-1)^{n-1}(n-1)!$, respectively.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 103 / 3 / c / iii / Solution

The <Gelfand–Tsetlin basis> spans $V^\lambda$, and the scalar computed on its vectors depends only on the shape. Thus
$$
\boxed{X_2\cdots X_nw=\begin{cases}(-1)^k k!(n-k-1)!\,w&\lambda=(n-k,1^k),\\0&\lambda\text{ not a hook},\end{cases}}
$$
for every $w\in V^\lambda$. The product is the sum of the $(n-1)!$ permutations in the <conjugacy class> of an $n$-cycle. Taking <traces> therefore gives $(n-1)!\chi^\lambda((n))$ equal to the displayed scalar times $\dim V^\lambda$.

For a <hook partition>, a <standard Young tableau> is uniquely determined by the choice of its $k$ entries below the top cell, selected from $\{2,\ldots,n\}$. The column and the remaining row are then forced to increase. Hence $\dim V^{(n-k,1^k)}=\binom{n-1}{k}$, and cancellation of the factorials yields
$$
\boxed{\chi^\lambda((n))=\begin{cases}(-1)^k&\lambda=(n-k,1^k),\\0&\lambda\text{ not a hook}.\end{cases}}
$$
This uses the <central character value of a conjugacy-class sum> and tableau counting, without a character rule for removing <rim hooks>.

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 103 / 3 / c / iv / Solution

{title2=$(n-k,1^k)$}

A hook partition has shape $(n-k,1^k)$, $0\leq k\leq n-1$. Equivalently its <Young diagram> has no cell $(2,2)$, so its only cell of content zero is $(1,1)$. Its <standard Young tableaux> are counted by $\binom{n-1}{k}$: choose the $k$ entries below the initial $1$, then the column and remaining row orders are forced.

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