Let \(T\) be a linear operator on the finite-dimensional space \(V\), let \(p = p_1^{r_1} \cdots p_k^{r_k}\) be the minimal polynomial for \(T\), and let \(V = W_1 \oplus \cdots \oplus W_k\) be the primary decomposition for \(T\), i.e., \(W_i\) is the null space of \(p_i(T)^{r_i}\). Let \(W\) be any subspace of \(V\) which is invariant under \(T\). Prove that \[ W = (W \cap W_1) \oplus (W \cap W_2) \oplus \cdots \oplus (W \cap W_k). \]
Solution
Let \[ q_i(x)=p_i(x)^{r_i}, \qquad p(x)=q_1(x)\cdots q_k(x). \] The \(q_i\) are pairwise relatively prime. By the polynomial Chinese remainder theorem, there are polynomials \(e_i(x)\) such that \[ e_i\equiv1\pmod{q_i}, \qquad e_i\equiv0\pmod{q_j}\quad(j\ne i), \] and \[ e_1+\cdots+e_k\equiv1\pmod p \] [@HK71]. Put \[ E_i=e_i(T). \] These are the primary projections: \(E_i\) has range \(W_i\), the operators \(E_i\) are pairwise annihilating, and \[ E_1+\cdots+E_k=I. \]
1. Each primary component of a vector in \(W\) still lies in \(W\).
Since \(W\) is invariant under \(T\), it is invariant under every power of \(T\), hence under every polynomial in \(T\). Thus \[ E_i(W)\subseteq W \] for every \(i\). Also \(\operatorname{range}E_i=W_i\), so \[ E_i(W)\subseteq W\cap W_i. \] For any \(w\in W\), \[ w=(E_1+\cdots+E_k)w =E_1w+\cdots+E_kw, \] and each summand lies in \(W\cap W_i\). Hence \[ W\subseteq (W\cap W_1)+\cdots+(W\cap W_k). \] The reverse inclusion is immediate because every \(W\cap W_i\) is contained in \(W\).
2. The sum of the intersections is direct.
The primary decomposition \[ V=W_1\oplus\cdots\oplus W_k \] is direct. Therefore if \[ w_1+\cdots+w_k=0, \qquad w_i\in W\cap W_i, \] then in particular \(w_i\in W_i\), so directness in \(V\) forces every \(w_i=0\). Thus the sum in step 1 is direct, and \[ \boxed{ W=(W\cap W_1)\oplus\cdots\oplus(W\cap W_k).} \]