Let \(T\) be a linear operator on the vector space \(V\) over the field \(F\). If \(f\) is a polynomial over \(F\) and \(\alpha\) is in \(V\), let \(f\alpha = f(T)\alpha\). If \(V_1, \ldots, V_k\) are \(T\)-invariant subspaces and \(V = V_1 \oplus \cdots \oplus V_k\), show that \[ fV = fV_1 \oplus \cdots \oplus fV_k. \]
Solution
1. \(fV = fV_1 + \cdots + fV_k\).
1.1. Every \(\alpha \in V\) is \(\alpha = \alpha_1 + \cdots + \alpha_k\) with \(\alpha_i \in V_i\).
\(V = V_1 \oplus \cdots \oplus V_k\).
1.2. \(f\alpha = f(T)\alpha = f(T)\alpha_1 + \cdots + f(T)\alpha_k \in fV_1 + \cdots + fV_k\).
linearity of \(f(T)\).
1.3. Hence \(fV \subseteq fV_1 + \cdots + fV_k\), and the reverse inclusion is clear (each \(fV_i \subseteq fV\)).
2.
2. The sum \(fV_1 + \cdots + fV_k\) is direct.
2.1. Suppose \(f\alpha_1 + \cdots + f\alpha_k = 0\) with \(\alpha_i \in V_i\).
take a relation.
2.2. Then \(f(\alpha_1 + \cdots + \alpha_k) = 0\), so \(\alpha_1 + \cdots + \alpha_k \in \ker f(T)\).
linearity.
2.3. Since each \(V_i\) is \(T\)-invariant, it is \(f(T)\)-invariant, so \(f\alpha_i \in V_i\) for each \(i\).
\(f(T)\) commutes with the projection onto \(V_i\) (since \(V_i\) is \(T\)-invariant).
2.4. Hence \(f\alpha_i = 0\) for each \(i\) (the sum is direct in \(V\)).
\(f\alpha_1 + \cdots + f\alpha_k = 0\) with \(f\alpha_i \in V_i\), and \(V = V_1 \oplus \cdots \oplus V_k\), so each summand is \(0\).
3. Hence \(fV = fV_1 \oplus \cdots \oplus fV_k\).
1 and 2.
4. Q.E.D.
3.