# Irreducible characteristic polynomial and invariant subspaces

Let $$K$$ be a field and let $$V$$ be a finite-dimensional vector space over $$K$$. Let $$\alpha$$ be an endomorphism of $$V$$, with irreducible characteristic polynomial. I'm trying to show that there is no proper non-zero subspace $$W \subset V$$ with $$\alpha(W) \subset W$$.

So far, I've consider $$V$$ as a $$K[X]$$-module by $$K[X] \times V \to V : (\sum \lambda_i X^i,v) \mapsto \sum \lambda_i \alpha^i(v)$$ And then noticed that the $$\alpha$$-invariant subspaces of $$V$$ are exactly the $$K[X]$$-submodules of $$V$$. Furthermore, since $$K[X]$$ is a PID and $$V$$ finite dimensional, we have by the structure theorem $$V \cong K[X]/p_1^{e_1} \oplus \ldots \oplus K[X]/p_k^{e_k},$$ with $$p_1,\ldots,p_k \in K[X]$$ irreducible. Also since the characteristic polynomial of $$\alpha$$, $$p_\alpha$$ is irreducible, it is equal to the minimum polynomial, and we have that $$K[X]/(p_\alpha)$$ is a field, which is expect we need to use somewhere.

Any hints on how to proceed?

• Note that, by construction, $p_1^{e_1}\cdots p_k^{e_k}$ is the characteristic polynomial of $\alpha$. Use this fact to see that $V$ is in fact a field. Can you say anything stronger now about the $K[X]$-submodule of $V$? Mar 12, 2019 at 11:17
• Thanks. I'm wondering why $p_1^{e_1} \cdots p_k^{e_k}$ is the characteristic polynomial of $\alpha$? Mar 12, 2019 at 11:24
• Check out Lang's Algebra, Theorem 2.1, XIV S2. p.557, and Theorem 3.5 below in the same reference. The complete argument should appear there, sorry I don't have time to try and write it down Mar 12, 2019 at 12:03
• Sure, thanks for the reference, I will check that out. Mar 12, 2019 at 12:12

This is very simple. Suppose $$W$$ is an invariant subspace for$$~\alpha$$, then one can restrict $$\alpha$$ to$$~W$$ to obtain an endomorphism $$\alpha|_W$$ of$$~W$$. Then the characteristic polynomial of $$\alpha|_W$$ divides the characteristic polynomial of $$\alpha$$, contradicting (if $$0\subset W\subset V$$) the assumption that the latter is irreducible.
The fact that the characteristic polynomial of the restriction to an invariant subspace$$~W$$ divides that of the full endomorphism is immediate from the definition of the characteristic polynomial, using a basis of $$V$$ that extends a basis of$$~W$$ to get a matrix for the endomorphism: this matrix is block upper triangular, with a matrix for the restricted endomorphism as upper left diagonal block.