Proof of the decomposition into generalised eigenspaces

Proof

We split $V$ with polynomials in $T$ that add up to the identity, as in the primary decomposition. Then we compute the dimensions from the characteristic polynomial.

1. Write $p = p_T$, $q_i = (t - \lambda_i)^{m_i}$ and $f_i = \prod_{j \ne i} q_j$, so $p = q_i f_i$ for each $i$. Let $W_i = \ker q_i(T)$, a kernel in $V$.
2. The ring $F[t]$ is a Euclidean domain, hence a principal ideal domain, hence a unique factorisation domain. Each $t - c$ is an irreducible polynomial, since it has degree $1$. Distinct monic polynomials $t - c$ are not associates.
3. By step 2 and the uniqueness of factorisations, each monic divisor of $\prod_j (t - \lambda_j)^{m_j}$ has the form $\prod_j (t - \lambda_j)^{e_j}$ with $0 \le e_j \le m_j$. The same holds for monic divisors of $(t - c)^N$, with a single factor.
4. The ideal generated by $f_1, \dots, f_k$ is principal and nonzero, so it is generated by a monic $d$. Then $d$ divides each $f_i$ and $p$.
5. By step 3, $d = \prod_j (t - \lambda_j)^{e_j}$. Since $d$ divides $f_i$, and $t - \lambda_i$ does not occur in the factorisation of $f_i$, uniqueness gives $e_i = 0$. This holds for each $i$, so $d = 1$.
6. Hence there are $a_1, \dots, a_k \in F[t]$ with $\sum_i a_i f_i = 1$. Let $\pi_i = a_i(T) f_i(T)$. Evaluation at $T$ respects sums and products, so $\sum_i \pi_i = \id_V$, by the definition of the value at $T$.
7. Polynomials in $T$ commute with each other. So each $W_i$ is mapped into itself by every polynomial in $T$. In particular, $W_i$ is $T$-invariant.
8. Let $v \in V$. Then $q_i(T) \pi_i(v) = a_i(T) p(T)(v) = 0$ by the Cayley–Hamilton theorem. So $\pi_i(v) \in W_i$, and $v = \sum_i \pi_i(v)$ lies in $W_1 + \cdots + W_k$.
9. Let $w \in W_j$ and $i \ne j$. Then $q_j$ divides $f_i$, so $\pi_i(w) = 0$. With step 6, $w = \pi_j(w)$.
10. Let $w_j \in W_j$ with $\sum_j w_j = 0$. Applying $\pi_l$ and using step 9 gives $w_l = \pi_l(0) = 0$ for each $l$. With step 8, $V = W_1 \oplus \cdots \oplus W_k$ as a direct sum.
11. We show $W_i = V_{\lambda_i}$, the generalised eigenspace. If $m_i \ge 1$, the definition gives $W_i \subseteq V_{\lambda_i}$ with exponent $m_i$. If $m_i = 0$, then $W_i = \ker \id_V = 0$.
12. Conversely, let $v \in V_{\lambda_i}$, with $(T - \lambda_i \id_V)^N (v) = 0$. Let $j \ne i$ and $u = \pi_j(v)$. By step 7, $(T - \lambda_i \id_V)^N (u) = \pi_j\big((T - \lambda_i \id_V)^N (v)\big) = 0$. Also $q_j(T)(u) = 0$ by step 8.
13. The ideal generated by $(t - \lambda_i)^N$ and $q_j$ has a monic generator $d'$. By step 3, $d'$ divides $q_j$, so $d' = (t - \lambda_j)^e$. It also divides $(t - \lambda_i)^N$, so uniqueness gives $e = 0$ and $d' = 1$.
14. So $1 = b (t - \lambda_i)^N + c q_j$ for some $b, c \in F[t]$. Applying this to $u$ gives $u = 0$ by step 12.
15. Hence $\pi_j(v) = 0$ for $j \ne i$, and $v = \pi_i(v) \in W_i$ by step 8. So $W_i = V_{\lambda_i}$, and $V = V_{\lambda_1} \oplus \cdots \oplus V_{\lambda_k}$.
16. Let $d_i = \dim W_i$, and choose a basis of each $W_i$. Their concatenation $B$ spans $V$ by step 8. It is linearly independent, because a relation splits into vectors $w_i \in W_i$ with sum $0$, so each $w_i = 0$ by step 10.
17. Let $T_i = T|_{W_i}$ be the restriction, and let $A_i$ be its matrix in the chosen basis of $W_i$. By step 7, the matrix of $T$ in $B$ is the block diagonal matrix $\operatorname{diag}(A_1, \dots, A_k)$. Blocks with $d_i = 0$ are omitted, and we set $p_{A_i} = 1$ for them.
18. The matrix $t I - \operatorname{diag}(A_1, \dots, A_k)$ is block diagonal with blocks $t I - A_i$. Using the determinant of a block triangular matrix $k - 1$ times, $p_T = \prod_i p_{A_i}$, by the definition of $p_T$.
19. Let $w \in W_i$ be nonzero with $T(w) = \mu w$. Then $0 = q_i(T)(w) = (\mu - \lambda_i)^{m_i} w$, so $\mu = \lambda_i$. Thus $\lambda_i$ is the only possible eigenvalue of $T_i$.
20. The polynomial $p_{A_i}$ is monic of degree $d_i$ and divides $p_T$ by step 18. By step 3, $p_{A_i} = \prod_j (t - \lambda_j)^{e_{ij}}$.
21. If $e_{ij} > 0$ for some $j \ne i$, then $\lambda_j$ is a root of $p_{A_i}$. By the roots of the characteristic polynomial, it is an eigenvalue of $T_i$. This contradicts step 19, since $\lambda_j \ne \lambda_i$.
22. So $p_{A_i} = (t - \lambda_i)^{d_i}$, by comparing degrees. Then step 18 gives $\prod_i (t - \lambda_i)^{d_i} = p_T = \prod_i (t - \lambda_i)^{m_i}$.
23. The $t - \lambda_i$ are pairwise non-associate irreducible elements by step 2. Uniqueness of factorisation in the unique factorisation domain $F[t]$ gives $d_i = m_i$ for each $i$.
24. So $\dim V_{\lambda_i} = m_i$, which completes the proof.