Rational Canonical Form

Invariant factors, companion blocks, and a field-independent canonical form.

The idea

Rational canonical form. Every linear operator $T$ on a finite-dimensional vector space over a field $k$ is similar to exactly one block-diagonal matrix $\operatorname{diag}\bigl(C(a_1), \ldots, C(a_m)\bigr), \qquad a_1 \mid a_2 \mid \cdots \mid a_m,$ where the $a_i$ are monic polynomials — the invariant factors of $T$ — and $C(a)$ is the companion matrix of $a$. Two operators are similar if and only if their lists of invariant factors agree.

Why ask for this rather than a diagonalization? Because eigenvalues live wherever the characteristic polynomial happens to have roots, and that may be nowhere in $k$: a rotation of the real plane has no real eigenvalue, so no diagonal and no triangular form over $\mathbb{R}$. The rational canonical form demands no roots at all. It exists over every field and is computed entirely inside $k$, which is what rational refers to here.

The companion matrix of a monic $p(x) = x^{d} + b_{d-1}x^{d-1} + \cdots + b_{0}$ is the $d \times d$ matrix carrying $1 s on the subdiagonal and the entries $-b_{0}, \ldots, -b_{d-1}$ down its last column; its characteristic polynomial is $p$ itself.

Where the form comes from: let $x$ act on $V$ as $T$, making $V$ a module over $k[x]$, a principal ideal domain. The classification of finitely generated modules over such a ring splits $V$ into cyclic pieces $k[x]/(a_i)$ with $a_1 \mid \cdots \mid a_m$, and writing $T$ in the natural basis of one cyclic piece produces exactly a companion block.

Ways to work on it

• Walkthrough. Invariant factors, companion blocks, and the minimal versus characteristic polynomial.
• Practice. Read off companion-matrix data from an invariant factor.
• Hardest. Build invariant factors from elementary divisors and find the minimal polynomial.

Not sure where to start? Take the ten-question placement test.