Rational Canonical Form

\(\not\implies\): In general, \(\min_A \divides \chi_A\), so suppose they’re not equal. Set \(n\coloneqq\deg \chi_A\), then if \(n' \coloneqq\deg \min_A < n\), using that \(\min_A(A) = 0\) this exhibits a linear dependence in \(\left\{{v, Av, \cdots, A^{n'} v}\right\}\) for any \(v\). In particular, since \(n>n'\), any set \(\left\{{v, Av,\cdots,A^nv}\right\}\) has a linear dependence.

\(\implies\): Apply the structure theorem to write \(V\cong \bigoplus_{i=1}^m k[x]/\left\langle{p_i}\right\rangle\). Since \(\chi_A(x) = \prod p_i(x)\) and \(\min_A(x) = p_m(x)\), this forces \(m=1\) – one way to see this is that \(\dim_k V = \sum_{i=1}^m \dim_k k[x]/\left\langle{p_i}\right\rangle\), where \(\deg \chi_A = \dim_k V\) and \(\deg \min_A = \dim k[x]/\left\langle{p_m}\right\rangle\). For these to be equal, this forces \(\dim_k k[x]/\left\langle{p_i}\right\rangle = 0\) for \(1\leq i \leq m-1\), making \(V\) a cyclic \(k[x]{\hbox{-}}\)module. So \(V = k[x]\curvearrowright\mathbf{v}\) for some \(\mathbf{v}\in V\), which is the desired cyclic vector, and \begin{align*} V = \left\{{f(x).v {~\mathrel{\Big\vert}~}f\in k[x]}\right\} = \mathop{\mathrm{span}}_k\left\{{A^k v {~\mathrel{\Big\vert}~}k\geq 0}\right\} .\end{align*} By Cayley-Hamilton, \(\chi_A(A) = 0\) and so \(A^n\) is a linear combinations of \(A^k\) for \(0\leq k \leq n-1\), so \(V= \mathop{\mathrm{span}}_k \left\{{A^k v {~\mathrel{\Big\vert}~}0\leq k \leq n-1}\right\}\).

• Let \(k[x] \curvearrowright V\) by \(p(x) \curvearrowright\mathbf{v} \coloneqq p(T)(\mathbf{v})\), making \(V\) into a finitely generated torsion \(k[x]{\hbox{-}}\)module.

  • Note that \(k[x]{\hbox{-}}\)submodules are exactly \(T{\hbox{-}}\)invariant subspaces.

• \(k\) a field implies \(k[x]\) a PID, so apply structure theorem to obtain an invariant factor decomposition \begin{align*} V \cong k[x] / \left\langle{\chi_T(x)}\right\rangle \cong \bigoplus_{i=1}^m k[x] / \left\langle{ p_i(x) }\right\rangle && p_1(x) \divides p_2(x) \divides \cdots p_m(x) .\end{align*}

• Since each factor is submodule, each corresponds to a \(T{\hbox{-}}\)invariant subspace \(V_i\) where \(p_i\) is the minimal polynomial of \(T\) restricted to \(V_i\).

  • The largest invariant factor \(p_m\) is the minimal polynomial of \(T\), their product is the characteristic polynomial. This follows because \(p_m(x)\curvearrowright V = 0\), since \(p_i\divides p_m\) for all \(i\), forcing \(\min_A \divides p_m\) by minimality.

• Write \(V \cong \bigoplus_{i=1}^m V_i\) as a \(k[x]{\hbox{-}}\)module, where \(V_i \coloneqq k[x] / \left\langle{ p_i(x) }\right\rangle\), then \(T\) is a block matrix \(\bigoplus_{i=1}^m T_i\) where \(T_i\) is the restriction of \(T\) to \(V_i\): \begin{align*} \left(\begin{array}{ccccc}T_{1} & 0 & 0 & \cdots & 0 \\ 0 & T_{2} & 0 & \cdots & 0 \\ \vdots & & \ddots & & \vdots \\ 0 & \cdots & & & T_{n}\end{array}\right) .\end{align*}

• It suffices to determine the form of a single \(M_i\), so without loss of generality suppose \(m=1\) so \(V = V_1 = k[x] / \left\langle{ p(x) }\right\rangle\) is a cyclic \(k[x]{\hbox{-}}\)module with \(\deg p(x) = n\).

• \(\chi_M(x) = \min_M(x) \iff\) there exists a cyclic vector \(\mathbf{v}\), so the set \(\left\{{\mathbf{v}_i}\right\}_{i=0}^{n-1} \coloneqq\left\{{ \mathbf{v}, T\mathbf{v}, T^2\mathbf{v}, \cdots, T^{n-1}\mathbf{v} }\right\}\) is a basis for \(V_1\).

  • If there is any linear independence, this gives a polynomial relation \(\sum_{i=1}^{n'} a_iT^i\mathbf{v} = 0\) for some \(n'<n\), but then \(q(x) \coloneqq\sum_{i=1}^{n'} a_i x^i\) is a polynomial annihilating \(T\), contradicting the minimality of \(p(x)\).
  • So this yields \(n\) linearly independent vectors in \(k^n\), so it’s a basis.

• What is \(M_i\) in this basis? Check where basis elements are mapped to by \(T\), noting that \begin{align*} p(T) = \sum_{i=1}^{n}a_i T^i\mathbf{v} = T^n + a_{n-1} T^{n-1}\mathbf{v} + a_{n-2} T^{n-2} + \cdots + a_1 T\mathbf{v} + a_0 \mathbf{v} = 0 ,\end{align*} using the minimal polynomial we can write

  • \(T\mathbf{v}_0 = \mathbf{v}_1\)
  • \(T\mathbf{v}_2 = T^2 \mathbf{v}_0\)
  • \(T\mathbf{v}_3 = T^3 \mathbf{v}_0\)
  • \(\cdots\)
  • \(T\mathbf{v}_{n-2} = T^{n-1}\mathbf{v}\)
  • \(T\mathbf{v}_{n-1} = T^n\mathbf{v} = -a_{n-1}T^{n-1}\mathbf{v} - \cdots - a_1 T\mathbf{v} - a_0 \mathbf{v}\)

• So we have \begin{align*} M_1 = \begin{bmatrix} 0 & & & & -a_0 \\ 1 & 0 & & & -a_1 \\ & 1 & 0& & -a_2 \\ & & \ddots & 0 & \vdots \\ & & & 1 & -a_{n-1} \end{bmatrix} .\end{align*}