Finite-dimensional vector spaces admit cyclic decompositions, which are direct sum decompositions into invariant subspaces relative to linear operators acting on them. I would now like to use this fact to derive an expression for the characteristic polynomial of a linear operator acting on a finite-dimensional vector space from which it will be immediately apparent that it e.g. satisfies its own characteristic equation.

Suppose that \(T : V \to V\) is a linear operator acting on a finite-dimensional vector space \(V\) over a field \(\mathbb{F}\). Additionally, suppose that the minimal polynomial of \(T\) is given by \(p(x) = \prod\limits_{q(x) \in S} \left ( q(x) \right )^{m_q}\),

letting \(S\) denote a set of irreducible polynomials in \(\mathbb{F}[x]\) and \(m_q\) denote the multiplicity of \(q(x)\) as a factor of \(p(x)\) for all \(q(x) \in S\).

\[\implies V = \bigoplus\limits_{q(x) \in S} \operatorname{ker} \left ( q(T) \right )^{m_q} = \bigoplus\limits_{q(x) \in S} \operatorname{ker} \left ( q(T) \right )^{\operatorname{dim}(V)}\]

[ i.e. \(V\) admits its primary decomposition relative to \(T\), following from a previous post ].

\({\implies}\) For all \(q(x) \in S\):

\[\operatorname{ker} \left ( q(T) \right )^{\operatorname{dim}(V)} = \bigoplus_\limits{ v \in U_q } Z(T; v) \text { for some } U_q \subseteq V\]

[ assuming the fact that finite-dimensional vector spaces admit cyclic decompositions ].

Now, let \(U = \bigcup\limits_{q(x) \in S} U_q\).

\[\implies V = \bigoplus\limits_{v \in U} Z(T; v)\] \[\implies B := \bigcup\limits_{v \in U} \{ v, Tv, \cdots, T^{\left ( \operatorname{dim}(Z(T;v)) - 1 \right) }v \}\]

is a basis for \(V\).

\[\implies A := [T]_B = \bigoplus\limits_{v \in U} C_v\]

is a block-diagonal matrix, given here by a Kronecker sum of companion matrices, letting \(C_v := \left [ T \mid_{Z(T;v)} \right ]_B\) for all \(v\in U\).

\[\implies \phi(x) := \operatorname{det}(xI - A)\] \[= \prod\limits_{v \in U } \operatorname{det}(xI - C_v)\] \[= \prod\limits_{v \in U} p_v(x)\] \[= \prod\limits_{q(x) \in S} \left(q(x)\right)^{\operatorname{dim} \left ( \operatorname{ker} \left ( q(T) \right )^{\operatorname{dim}(V)}\right)}\]

is a expression for the characteristic polynomial of \(T\) of the desired form, letting \(p_v(x)\) denote the relative minimal polynomial of \(v\) for all \(v \in U\).

It is an immediate consequence of the result established in this post that e.g. the minimal polynomial of a linear operator acting on a finite-dimensional vector space divides its characteristic polynomial and hence that it satifies its characteristic equation.

Note that this post is adapted from one of my selected answers on Quora.