Why is a module with a diagonal operator and distinct diagonal entries cyclic?

[Chris L T521]

Here's this week's problem.

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Problem: Let $V$ be a finite-dimensional $K[X]$-module, and let $\phi$ be the associated operator on $V$. Suppose that $\Delta$ represents $\phi$ with respect to some basis. Prove that if $\Delta$ is a diagonal matrix (no nonzero entries off the diagonal), and the diagonal entries of $\Delta$ are pairwise distinct, then $V$ is a cyclic $K[X]$-module.

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[Chris L T521]

No one answered this week's question. You can find my solution below.

Spoiler

Suppose \begin{equation*}
\Delta =
\left(
\begin{array}{cccc}
\lambda_1 & 0 & \cdots & 0 \\
0 & \lambda_2& \cdots & 0 \\
\vdots& \vdots & \ddots & \vdots\\
0 & 0& \cdots &\lambda_n \\
\end{array}
\right)
\end{equation*}
where $\lambda_i \neq \lambda_j$, for $i \neq j;\ i,j = 1,2,\cdots, n$. Then for $k=1,2,\cdots,n-1$,
\begin{equation*}
\Delta^k = \left(
\begin{array}{cccc}
\lambda^k_1 & 0 & \cdots & 0 \\
0 & \lambda^k_2& \cdots & 0 \\
\vdots& \vdots & \ddots & \vdots\\
0 & 0& \cdots &\lambda^k_n \\
\end{array}
\right)
\end{equation*}

Let $v = (1,1,\cdots,1)^T \in V$, then for $k=1,2,\cdots,n-1$,
$$\phi^k(v) = \Delta^k v = (\lambda^k_1, \lambda^k_2, \cdots, \lambda^k_n)^T.$$

Now we prove that $\{v,\phi(v),\phi^2(v), \cdots, \phi^{n-1}(v)\}$ are linearly independent. Suppose for $k_1,k_2,\cdots,k_n \in K$,
$$k_1 v + k_2\phi(v) + k_3\phi^2(v) + \cdots + k_n\phi^{n-1}(v) = 0.$$

i.e.

\begin{equation*} k_1
\left(
\begin{array}{c}
1 \\
1\\
\vdots \\
1\\
\end{array}
\right)
+k_2 \left(
\begin{array}{c}
\lambda_1 \\
\lambda_2\\
\vdots \\
\lambda_n\\
\end{array}
\right) +\cdots+ k_n \left(
\begin{array}{c}
\lambda^{n-1}_1 \\
\lambda^{n-1}_2\\
\vdots \\
\lambda^{n-1}_n\\
\end{array}
\right) = 0.
\end{equation*}

That is
\begin{equation*}
\left(
\begin{array}{cccc}
1 & \lambda_1 & \cdots & \lambda^{n-1}_1 \\
1 & \lambda_2& \cdots & \lambda^{n-1}_2 \\
\vdots& \vdots & \ddots & \vdots\\
1 & \lambda_n& \cdots &\lambda^{n-1}_n \\
\end{array}
\right)
\left(
\begin{array}{c}
k_1 \\
k_2\\
\vdots \\
k_n\\
\end{array}
\right) = 0.
\end{equation*}

The matrix is a Vandermonde Matrix above, therefore
\begin{equation*}
\det
\left(
\begin{array}{cccc}
1 & \lambda_1 & \cdots & \lambda^{n-1}_1 \\
1 & \lambda_2& \cdots & \lambda^{n-1}_2 \\
\vdots& \vdots & \ddots & \vdots\\
1 & \lambda_n& \cdots &\lambda^{n-1}_n \\
\end{array}
\right) = \prod_{1\leqslant i < j \leqslant n} (\lambda_i - \lambda_j).
\end{equation*}

By the presumption, $\lambda_i \neq \lambda_j$, for $i \neq j$. Hence the determinant above is non-zero. Therefore the matrix is non-singular. Thus the equations admit only zero solution. i.e.
$$k_1 = k_2 = \cdots = k_n = 0.$$
Hence $\{v,\phi(v),\phi^2(v), \cdots, \phi^{n-1}(v)\}$ are linearly independent, and it is a basis for $V$. By definition, $V$ is a cyclic $K[X]$-module.