Let \(F\) be a field, let \(V\) be finite-dimensional over \(F\), and let \(T:V\to V\) have irreducible minimal polynomial \(\mu\in F[t]\).
- If \(0\ne v\in V\) and \(V_1\) is the subspace spanned by \(v\) and its images under the positive powers of \(T\), prove that \[ \dim V_1=\deg\mu. \]
- Prove that \(\deg\mu\) divides \(\dim V\).
Solution
Put \[ E\coloneqq F[t]/(\mu). \]
1. The quotient \(E\) is a field and\[ \dim_F E=\deg\mu. \]
The polynomial \(\mu\) is irreducible over the field \(F\), so the ideal \((\mu)\) is maximal in \(F[t]\). Hence \(E\) is a field. If \(d=\deg\mu\), every residue class has a unique representative of degree less than \(d\), so \[ 1,t,\ldots,t^{d-1} \] give an \(F\)-basis of \(E\).
2. The vector space \(V\) carries an \(E\)-vector-space structure defined by\[ [p(t)]\cdot v\coloneqq p(T)v. \]
Since \(\mu\) is the minimal polynomial of \(T\), \[ \mu(T)=0. \] If \(p\equiv q\pmod\mu\), then \(p-q=\mu r\) for some \(r\in F[t]\), so \[ (p(T)-q(T))v = \mu(T)r(T)v =0. \] Thus the action is well-defined. The vector-space axioms follow from the usual polynomial functional calculus for \(T\).
3. If \(0\ne v\in V\), then the \(E\)-subspace \(Ev\) is one-dimensional over \(E\).
The vector \(v\) itself spans \(Ev\) over \(E\). If \(\alpha\in E\) satisfies \[ \alpha v=0 \] and \(\alpha\ne0\), then \(\alpha\) is invertible because \(E\) is a field. Multiplying by \(\alpha^{-1}\) would give \(v=0\), a contradiction. Hence \(v\) is an \(E\)-basis of \(Ev\).
4. The subspace \(V_1\) in part 1 is exactly \(Ev\).
By definition, \[ V_1=F[T]v, \] the \(F\)-span of \(v,Tv,T^2v,\ldots\). The action in step 2 factors \(F[T]\) through the quotient \(E=F[t]/(\mu)\), so its image on \(v\) is precisely \(Ev\).
5. Therefore\[ \boxed{\dim_F V_1=\deg\mu}. \]
By steps <1>3–<1>4, \(V_1\) is one-dimensional over \(E\). Thus step 1 gives \[ \dim_F V_1 = \dim_E V_1\cdot\dim_F E = 1\cdot\deg\mu. \]
6. The space \(V\) is finite-dimensional over \(E\).
Any finite \(F\)-basis of \(V\) is also a finite spanning set over the larger scalar field \(E\), because \(F\) embeds in \(E\). Hence \(V\) has finite dimension over \(E\).
7. One has\[ \boxed{\deg\mu\mid\dim_F V}. \]
Let \[ r\coloneqq\dim_E V. \] By the dimension formula for extension of scalars and step 1, \[ \dim_F V = r\dim_F E = r\deg\mu. \] Thus \(\deg\mu\) divides \(\dim_F V\).
8. Q.E.D.
Step 5 proves part 1, and step 7 proves part 2.