Using Canonical Forms
The characteristic polynomial is the product of the invariant factors, i.e. \begin{align*} \chi_A(x) = \prod_{j=1}^n f_j(x) .\end{align*}
The minimal polynomial of \(A\) is the invariant factor of highest degree, i.e. \begin{align*} \min_A(x) = f_n(x) .\end{align*}
For a linear operator on a vector space of nonzero finite dimension, TFAE:
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The minimal polynomial is equal to the characteristic polynomial.
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The list of invariant factors has length one.
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The Rational Canonical Form has a single block.
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The operator has a matrix similar to a companion matrix.
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There exists a cyclic vector \(\mathbf{v}\) such that \(\mathop{\mathrm{span}}_k\left\{{T^j \mathbf{v} {~\mathrel{\Big\vert}~}j = 1, 2, \cdots}\right\} = V.\)
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\(T\) has \(\dim V\) distinct eigenvalues
Minimal / Characteristic Polynomials
A trick for finding characteristic polynomials: \begin{align*} \chi_A(t) &= \sum_{k=0}^n (-1)^k \operatorname{tr}\qty{\bigwedge\nolimits^k A} t^{n-k} \\ &= t^n - \operatorname{tr}\qty{A} t^{n-1} + \operatorname{tr}\qty{\bigwedge\nolimits^2 A}t^{n-2} - \cdots \pm \operatorname{tr}\qty{\bigwedge\nolimits^{n-1} A} t \mp \operatorname{det}(A) ,\end{align*} using that
\begin{align*} {\bigwedge\nolimits^0 A} \coloneqq 1 \\ {\bigwedge\nolimits^1 A} \coloneqq A \\ \operatorname{tr}\qty{\bigwedge\nolimits^n A} = \operatorname{det}(A) .\end{align*}
Moreover, the intermediate traces are easy to compute by hand: \begin{align*} \operatorname{tr}\qty{\bigwedge\nolimits^\ell A} = \sum \operatorname{det}\qty{M^{\ell}} ,\end{align*} where the sum is taken over all \(\ell\times\ell\) principal minors: determinants of the \(n \choose \ell\) principal matrices which are obtained by choosing \(\ell\) diagonal entries to keep and and deleting the rows and columns for every entry not chosen. Equivalently, one can select \(n-\ell\) diagonal entries and delete the corresponding row/column for each.
To factor this polynomial, the rational roots test can be useful: for \(f(t) = a_nt^n + \cdots + a_1 t + a_0\), rational roots are of the form \(p/q\) where \(p \divides a_n\) and \(q\divides a_0\). Note that this simplifies greatly for \(f\) monic! Once you have a root, apply polynomial long division to get a smaller problem, and hopefully this continues to work until it’s factored.
Fix some notation: \begin{align*} \min_A(x): \quad & \text{The minimal polynomial of } A \\ \chi_A(x): \quad & \text{The characteristic polynomial of } A .\end{align*}
The minimal polynomial of a linear map \(T\) is the unique monic polynomial \(\min_T(x)\) of minimal degree such that \(\min_T(T) = 0\).
The characteristic polynomial of \(A\) is given by \begin{align*} \chi_A(x) = \operatorname{det}(A - xI))= \operatorname{det}(SNF(A - xI)) .\end{align*}
If \(A\) is upper triangular, then \(\operatorname{det}(A) = \prod_{i} a_{ii}\)
The minimal polynomial divides the characteristic polynomial, and in particular \(\chi_A(A) = 0\).
By minimality, \(\min_A\) divides \(\chi_A\). Every \(\lambda_i\) is a root of \(\min_A(x)\): Let \((\mathbf{v}_i, \lambda_i)\) be a nontrivial eigenpair. Then by linearity, \begin{align*} \min_A(\lambda_i)\mathbf{v}_i = \min_A(A)\mathbf{v}_i = \mathbf{0} ,\end{align*} which forces \(\min_A(\lambda_i) = 0\).
Finding Minimal Polynomials
Let \(m(x)\) denote the minimal polynomial \(A\).
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Find the characteristic polynomial \(\chi(x)\); this annihilates \(A\) by Cayley-Hamilton. Then \(m(x) \divides \chi(x)\), so just test the finitely many products of irreducible factors.
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Pick any \(\mathbf{v}\) and compute \(T\mathbf{v}, T^2\mathbf{v}, \cdots T^k\mathbf{v}\) until a linear dependence is introduced. Write this as \(p(T) = 0\); then \(\min_A(x) \divides p(x)\).