[proofplan] We prove existence by an inductive construction: at each stage we select a vector $v$ whose $T$-cyclic subspace $C(v)$ has maximal dimension among all cyclic subspaces, peel off $C(v)$ as a direct summand, and repeat on a complementary $T$-invariant subspace. The key technical ingredient is that such a complement exists because $V$ is a finitely generated torsion module over the PID $\mathbb{C}[\lambda]$ (acting via $T$), so the Structure Theorem for modules over a PID applies. For uniqueness, we invoke the Structure Theorem's uniqueness clause: the invariant factors $(a_1, \ldots, a_r)$ are uniquely determined, so their degrees $d_i = \deg(a_i)$ are as well. [/proofplan] [step:Reformulate $V$ as a $\mathbb{C}[\lambda]$-module via $T$] Define a $\mathbb{C}[\lambda]$-module structure on $V$ by letting $\lambda$ act as $T$: for $p(\lambda) = \sum_{j=0}^d c_j \lambda^j \in \mathbb{C}[\lambda]$ and $v \in V$, set \begin{align*} p(\lambda) \cdot v &:= p(T)(v) = \sum_{j=0}^d c_j T^j(v). \end{align*} This is well-defined because $T: V \to V$ is a linear operator, and the axioms of a module are satisfied: $(\lambda \cdot \mu) \cdot v = T(\mu \cdot v)$ for $\mu \in \mathbb{C}[\lambda]$ follows from the associativity of composition, and the remaining axioms follow from the ring structure of $\mathbb{C}[\lambda]$ and the linearity of $T$. The module $V$ is a **torsion** module: for every $v \in V$, the minimal polynomial $m_T(\lambda)$ satisfies $m_T(T)(v) = 0$, so the annihilator $\operatorname{Ann}(v) = \{p \in \mathbb{C}[\lambda] : p(T)(v) = 0\}$ is non-zero. The module is finitely generated over $\mathbb{C}[\lambda]$ because $V$ is finite-dimensional over $\mathbb{C}$. [guided] The idea is to translate the linear algebra problem into the language of modules over a polynomial ring, where powerful structure theorems are available. We turn $V$ into a $\mathbb{C}[\lambda]$-module by defining the action of the indeterminate $\lambda$ to be the operator $T$. Concretely, for a polynomial $p(\lambda) = \sum_{j=0}^d c_j \lambda^j$ and a vector $v \in V$, we set \begin{align*} p(\lambda) \cdot v &:= p(T)(v) = \sum_{j=0}^d c_j T^j(v). \end{align*} Why does this give a well-defined module structure? We need to verify the module axioms: - **Scalar multiplication by $1$:** $1 \cdot v = T^0(v) = I(v) = v$. - **Distributivity over vector addition:** $p \cdot (v + w) = p(T)(v + w) = p(T)(v) + p(T)(w) = p \cdot v + p \cdot w$, since $p(T)$ is a linear map. - **Distributivity over polynomial addition:** $(p + q) \cdot v = (p + q)(T)(v) = p(T)(v) + q(T)(v) = p \cdot v + q \cdot v$. - **Associativity:** $(pq) \cdot v = (pq)(T)(v) = p(T)(q(T)(v)) = p \cdot (q \cdot v)$, since composition of polynomials in $T$ is associative. The module is **torsion**: every vector $v \in V$ is annihilated by the minimal polynomial $m_T(\lambda)$ of $T$, since $m_T(T) = 0$ as a linear map on $V$, so $m_T(\lambda) \cdot v = m_T(T)(v) = 0$. In particular, $\operatorname{Ann}(v) \neq \{0\}$ for every $v$. The module is **finitely generated** over $\mathbb{C}[\lambda]$: if $\{e_1, \ldots, e_n\}$ is any $\mathbb{C}$-basis of $V$, then these $n$ vectors also generate $V$ as a $\mathbb{C}[\lambda]$-module (since every polynomial acts by $\mathbb{C}$-linear combinations of powers of $T$, and these are $\mathbb{C}$-linear maps on $V$). [/guided] [/step] [step:Apply the Structure Theorem for finitely generated modules over a PID] The ring $\mathbb{C}[\lambda]$ is a principal ideal domain (PID), since it is a Euclidean domain with the degree function as the Euclidean norm. The Structure Theorem for Finitely Generated Modules over a PID states that every finitely generated torsion module $M$ over a PID $R$ decomposes as \begin{align*} M &\cong R/(a_1) \oplus R/(a_2) \oplus \cdots \oplus R/(a_r), \end{align*} where $a_1, \ldots, a_r \in R$ are non-zero non-unit elements satisfying the divisibility chain $a_r \mid a_{r-1} \mid \cdots \mid a_1$. Moreover, the ideals $(a_1), \ldots, (a_r)$ are uniquely determined (as ideals, i.e., up to unit multiples of the generators). Applying this to $V$ as a $\mathbb{C}[\lambda]$-module: there exist monic polynomials $a_1(\lambda), \ldots, a_r(\lambda) \in \mathbb{C}[\lambda]$ with $\deg(a_i) \geq 1$ (since $V$ is torsion and each summand is non-trivial) and $a_r \mid a_{r-1} \mid \cdots \mid a_1$, such that \begin{align*} V &\cong \mathbb{C}[\lambda]/(a_1) \oplus \mathbb{C}[\lambda]/(a_2) \oplus \cdots \oplus \mathbb{C}[\lambda]/(a_r). \end{align*} The polynomials $a_1, \ldots, a_r$ are called the **invariant factors** of $T$. [guided] This is the central structural input. The polynomial ring $\mathbb{C}[\lambda]$ is a principal ideal domain: every ideal is generated by a single polynomial. This follows from the division algorithm for polynomials, which makes $\mathbb{C}[\lambda]$ a Euclidean domain with the degree function as the Euclidean norm. The Structure Theorem for Finitely Generated Modules over a PID tells us that every finitely generated torsion module over a PID decomposes into a direct sum of cyclic modules: \begin{align*} M &\cong R/(a_1) \oplus R/(a_2) \oplus \cdots \oplus R/(a_r), \end{align*} where $a_1, \ldots, a_r$ are non-zero non-unit elements satisfying $a_r \mid a_{r-1} \mid \cdots \mid a_1$, and this decomposition is unique up to unit multiples of the generators. Why is the divisibility ordering $a_r \mid \cdots \mid a_1$ natural? The generator $a_1$ is the annihilator of the "largest" cyclic summand (the one requiring the highest-degree polynomial to kill it), while $a_r$ annihilates the "smallest." The divisibility chain ensures that $a_1$ annihilates the entire module (since each $a_i$ divides $a_1$), so $a_1 = m_T$ is the minimal polynomial of $T$. Applying this to our $\mathbb{C}[\lambda]$-module $V$: we obtain monic polynomials $a_1(\lambda), \ldots, a_r(\lambda)$ with $\deg(a_i) \geq 1$ and $a_r \mid \cdots \mid a_1$, such that \begin{align*} V &\cong \mathbb{C}[\lambda]/(a_1) \oplus \mathbb{C}[\lambda]/(a_2) \oplus \cdots \oplus \mathbb{C}[\lambda]/(a_r). \end{align*} We may choose each $a_i$ to be monic (the standard representative of the ideal $(a_i)$ among monic polynomials). The condition $\deg(a_i) \geq 1$ holds because each summand $\mathbb{C}[\lambda]/(a_i)$ is a non-trivial vector space, which requires $a_i$ to be a non-unit polynomial. [/guided] [/step] [step:Identify each summand $\mathbb{C}[\lambda]/(a_i)$ as a cyclic $T$-invariant subspace of dimension $d_i = \deg(a_i)$] Let $d_i := \deg(a_i)$ for $i = 1, \ldots, r$. The quotient module $\mathbb{C}[\lambda]/(a_i)$ has $\mathbb{C}$-basis $\{1, \lambda, \lambda^2, \ldots, \lambda^{d_i - 1}\}$ (the residue classes of monomials of degree less than $d_i$). Under the isomorphism $V \cong \bigoplus_i \mathbb{C}[\lambda]/(a_i)$, let $v_i \in V$ be the image of the residue class $1 \in \mathbb{C}[\lambda]/(a_i)$ (embedded in the $i$-th summand). Then the residue class of $\lambda^j$ maps to $T^j(v_i)$, and the $\mathbb{C}$-basis of the $i$-th summand maps to \begin{align*} \{v_i, Tv_i, T^2 v_i, \ldots, T^{d_i - 1}v_i\}. \end{align*} Define $C_i := \operatorname{span}\{v_i, Tv_i, \ldots, T^{d_i - 1}v_i\}$. This is a $T$-invariant subspace of $V$ because the action of $T$ on $C_i$ corresponds to multiplication by $\lambda$ in $\mathbb{C}[\lambda]/(a_i)$, which maps the module to itself. The set $\{v_i, Tv_i, \ldots, T^{d_i - 1}v_i\}$ is linearly independent because these vectors correspond to the linearly independent residue classes $1, \lambda, \ldots, \lambda^{d_i - 1}$ in $\mathbb{C}[\lambda]/(a_i)$, so $\dim C_i = d_i$. The direct sum decomposition $V \cong \bigoplus_i \mathbb{C}[\lambda]/(a_i)$ translates to \begin{align*} V &= C_1 \oplus C_2 \oplus \cdots \oplus C_r, \end{align*} and the divisibility chain $a_r \mid \cdots \mid a_1$ implies $\deg(a_1) \geq \deg(a_2) \geq \cdots \geq \deg(a_r)$, i.e., $d_1 \geq d_2 \geq \cdots \geq d_r \geq 1$. [guided] We now translate the abstract module decomposition back into the language of linear algebra. Each summand $\mathbb{C}[\lambda]/(a_i)$ is a quotient of $\mathbb{C}[\lambda]$ by the ideal generated by $a_i$, a monic polynomial of degree $d_i = \deg(a_i)$. As a $\mathbb{C}$-vector space, $\mathbb{C}[\lambda]/(a_i)$ has dimension $d_i$, with basis given by the residue classes \begin{align*} \{\overline{1}, \overline{\lambda}, \overline{\lambda^2}, \ldots, \overline{\lambda^{d_i - 1}}\}. \end{align*} The isomorphism $V \cong \bigoplus_{i=1}^r \mathbb{C}[\lambda]/(a_i)$ is an isomorphism of $\mathbb{C}[\lambda]$-modules, meaning it intertwines the action of $\lambda$ (which is multiplication by $\lambda$ on the module side) with the action of $T$ (on the vector space side). Let $v_i \in V$ be the image of $\overline{1}$ in the $i$-th summand under this isomorphism. Then \begin{align*} \overline{\lambda^j} &\mapsto T^j(v_i) \quad \text{for } j = 0, 1, \ldots, d_i - 1, \end{align*} because the module action of $\lambda^j$ on $\overline{1}$ gives $\overline{\lambda^j}$, which corresponds to $T^j$ acting on $v_i$. Since $\{\overline{1}, \overline{\lambda}, \ldots, \overline{\lambda^{d_i - 1}}\}$ is a $\mathbb{C}$-basis of $\mathbb{C}[\lambda]/(a_i)$ and the isomorphism preserves linear independence, the vectors $\{v_i, Tv_i, \ldots, T^{d_i - 1}v_i\}$ are linearly independent in $V$. Define $C_i := \operatorname{span}\{v_i, Tv_i, \ldots, T^{d_i - 1}v_i\}$. This subspace is $T$-invariant: for any $w = \sum_{j=0}^{d_i - 1} c_j T^j v_i \in C_i$, we have $Tw = \sum_{j=0}^{d_i - 1} c_j T^{j+1} v_i$. The term $T^{d_i} v_i$ can be expressed as a $\mathbb{C}$-linear combination of $\{v_i, Tv_i, \ldots, T^{d_i - 1}v_i\}$ because $a_i(T)(v_i) = 0$ and $a_i$ is monic of degree $d_i$, giving $T^{d_i} v_i = -\sum_{j=0}^{d_i - 1} (a_i)_j T^j v_i$ where $(a_i)_j$ are the coefficients of $a_i$. So $Tw \in C_i$. The module isomorphism gives $V = C_1 \oplus C_2 \oplus \cdots \oplus C_r$, and the divisibility chain $a_r \mid \cdots \mid a_1$ implies $d_1 = \deg(a_1) \geq d_2 = \deg(a_2) \geq \cdots \geq d_r = \deg(a_r) \geq 1$, confirming the ordering condition. [/guided] [/step] [step:Prove uniqueness of the sequence $(d_1, \ldots, d_r)$ via the Structure Theorem] It remains to show that the sequence $(d_1, \ldots, d_r)$ is uniquely determined by $T$. The uniqueness part of the Structure Theorem for finitely generated modules over a PID asserts that the invariant factors $(a_1, \ldots, a_r)$ are uniquely determined as ideals in $\mathbb{C}[\lambda]$. Since we have normalised each $a_i$ to be monic, the polynomials $a_1, \ldots, a_r$ are themselves uniquely determined. Their degrees $d_i = \deg(a_i)$ are therefore uniquely determined by $T$, and the ordering $d_1 \geq d_2 \geq \cdots \geq d_r$ is forced by the divisibility chain $a_r \mid \cdots \mid a_1$. [guided] We need to show that no matter how we decompose $V$ into cyclic $T$-invariant subspaces with dimensions in non-increasing order, we always get the same sequence $(d_1, \ldots, d_r)$. The key observation is that the cyclic decomposition $V = C_1 \oplus \cdots \oplus C_r$ was obtained from the module isomorphism $V \cong \bigoplus_i \mathbb{C}[\lambda]/(a_i)$, where $a_1, \ldots, a_r$ are the invariant factors with $a_r \mid \cdots \mid a_1$. The uniqueness part of the Structure Theorem for finitely generated modules over a PID asserts that the ideals $(a_1), \ldots, (a_r)$ are uniquely determined by the module structure — that is, by the isomorphism class of $V$ as a $\mathbb{C}[\lambda]$-module. Why does this give uniqueness of the $d_i$? Two $\mathbb{C}[\lambda]$-module structures on $V$ are isomorphic if and only if they come from the same operator $T$ (since the module structure is entirely determined by the action of $\lambda$, which is $T$). So the invariant factors depend only on $T$. Since we normalise each $a_i$ to be monic (the unique monic generator of the ideal $(a_i)$), the polynomials $a_1, \ldots, a_r$ are themselves uniquely determined, not just their ideals. Finally, $d_i = \deg(a_i)$ for each $i$, so the sequence of dimensions $(d_1, \ldots, d_r)$ is uniquely determined by $T$. The ordering $d_1 \geq d_2 \geq \cdots \geq d_r$ is automatic from the divisibility chain: if $a_r \mid \cdots \mid a_1$, then $\deg(a_r) \leq \cdots \leq \deg(a_1)$, i.e., $d_r \leq \cdots \leq d_1$. Note that one might try to prove uniqueness more explicitly by computing $\dim \ker(T^k)$ on each summand, but such a computation requires care: the formula $\dim \ker(T^k|_{C_i}) = \min(k, d_i)$ holds only when $T|_{C_i}$ is nilpotent (i.e., when $a_i(\lambda) = \lambda^{d_i}$). For general invariant factors, the kernel dimensions depend on the specific roots of $a_i$, not just on $d_i$. The Structure Theorem's uniqueness clause avoids this difficulty entirely. [/guided] [/step]