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    Rational canonical form

    FD-XT6HD

    Let \(k\) be a field, let \(V\) be a finite-dimensional \(k\)-vector space, and let \(\phi\colon V\to V\) be a \(k\)-linear map. Make \(V\) a \(k[x]\)-module by letting \(x\) act as \(\phi\). By the structure theorem for finitely generated modules over the principal ideal domain \(k[x]\), there are unique monic nonconstant polynomials \(r_1\divides r_2\divides\cdots\divides r_t\) in \(k[x]\) and a decomposition \(V=V_1\oplus\cdots\oplus V_t\) into \(\phi\)-invariant subspaces with \[ V_i\cong k[x]/(r_i)\quad\text{as } k[x]\text{-modules},\qquad 1\le i\le t. \] The rational canonical form of \(\phi\) is the block diagonal matrix \[ \diag\qty{C_{r_1},C_{r_2},\ldots,C_{r_t}}, \] where \(C_{r_i}\) is the companion matrix of \(r_i\).

    The polynomial \(r_i\) is the minimal polynomial of the restriction \(\phi|_{V_i}\). If \(v_i\in V_i\) corresponds to the class of \(1\) in \(k[x]/(r_i)\) and \(d_i=\deg r_i\), then \(v_i,\phi(v_i),\ldots,\phi^{d_i-1}(v_i)\) is a basis of \(V_i\), and the matrix of \(\phi|_{V_i}\) in this basis is \(C_{r_i}\). Concatenating these bases gives a basis of \(V\) in which \(\phi\) has its rational canonical form.

    Which decomposition of $V$ as a $k[x]\dash$module gives the rational canonical form of $\phi: V \to V$?

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