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  "body": "\n## Proof\n\nWe 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.\n\n1. 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|02YFME3RGW]] in $V$.\n2. The ring $F[t]$ is a [[Euclidean domain|0DQG39K2NH]], hence a [[principal ideal domain|0QKVE1GD9A]], hence a [[unique factorisation domain|055HT85T3S]]. Each $t - c$ is an [[irreducible polynomial|00X2A9G9W3]], since it has degree $1$. Distinct monic polynomials $t - c$ are not associates.\n3. 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.\n4. The [[ideal generated|030TZ0DQWT]] 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 $p$.\n5. 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 $d = 1$.\n6. 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 $T$|0BNEXSKNRV]].\n7. 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|0EP59ESQ94]].\n8. Let $v \\in V$. Then $q_i(T) \\pi_i(v) = a_i(T) p(T)(v) = 0$ by the [[Cayley–Hamilton theorem|0QEM68Y0DE]]. So $\\pi_i(v) \\in W_i$, and $v = \\sum_i \\pi_i(v)$ lies in $W_1 + \\cdots + W_k$.\n9. Let $w \\in W_j$ and $i \\ne j$. Then $q_j$ divides $f_i$, so $\\pi_i(w) = 0$. With step 6, $w = \\pi_j(w)$.\n10. 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|041JTZCK9Q]].\n11. We show $W_i = V_{\\lambda_i}$, the [[generalised eigenspace|07PNDWA7CE]]. If $m_i \\ge 1$, the definition gives $W_i \\subseteq V_{\\lambda_i}$ with exponent $m_i$. If $m_i = 0$, then $W_i = \\ker \\id_V = 0$.\n12. 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.\n13. 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 $d' = 1$.\n14. 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.\n15. 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 $V = V_{\\lambda_1} \\oplus \\cdots \\oplus V_{\\lambda_k}$.\n16. Let $d_i = \\dim W_i$, and choose a [[basis|0VTQTPQGAV]] 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.\n17. Let $T_i = T|_{W_i}$ be the [[restriction|0PFPWKBDTJ]], and let $A_i$ be its [[matrix|0YYMY95FWS]] in the chosen basis of $W_i$. By step 7, the matrix of $T$ in $B$ is the [[block diagonal matrix|0M0SNKZ482]] $\\operatorname{diag}(A_1, \\dots, A_k)$. Blocks with $d_i = 0$ are omitted, and we set $p_{A_i} = 1$ for them.\n18. 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|0J9TVTNB4K]] $k - 1$ times, $p_T = \\prod_i p_{A_i}$, by the [[definition of $p_T$|0BZ81AJNWF]].\n19. 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 $T_i$.\n20. The polynomial $p_{A_i}$ is monic of degree $d_i$ and divides $p_T$ by step 18. By step 3, $p_{A_i} = \\prod_j (t - \\lambda_j)^{e_{ij}}$.\n21. If $e_{ij} \u003e 0$ for some $j \\ne i$, then $\\lambda_j$ is a root of $p_{A_i}$. By [[the roots of the characteristic polynomial|053QVHF6N7]], it is an eigenvalue of $T_i$. This contradicts step 19, since $\\lambda_j \\ne \\lambda_i$.\n22. So $p_{A_i} = (t - \\lambda_i)^{d_i}$, by comparing degrees. Then step 18 gives $\\prod_i (t - \\lambda_i)^{d_i} = p_T = \\prod_i (t - \\lambda_i)^{m_i}$.\n23. The $t - \\lambda_i$ are pairwise non-associate irreducible elements by step 2. Uniqueness of factorisation in the [[unique factorisation domain|06YMXJMXYS]] $F[t]$ gives $d_i = m_i$ for each $i$.\n24. So $\\dim V_{\\lambda_i} = m_i$, which completes the proof.\n\n## Idea\n\nThe polynomials $f_i$ have no common factor, so a combination of them equals $1$. Plugging in $T$ gives maps $\\pi_i$ that add up to the identity, and Cayley–Hamilton shows that $\\pi_i$ lands in $\\ker (T - \\lambda_i)^{m_i}$. For the dimensions, $T$ is block diagonal with one block for each eigenvalue. Each block has characteristic polynomial a power of $t - \\lambda_i$, and unique factorisation matches the exponents.\n",
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