Rational canonical form

In one sentence

The rational canonical form does what the Jordan form cannot — it exists over every field — and its main consequence is that similarity never depends on which field you are working in.

Why the exam asks it

Because it answers a question that sounds paradoxical and is a favourite: two real matrices are similar over $\mathbb{C}$; are they similar over $\mathbb{R}$? The instinct is that a bigger field gives more room and therefore more similarities. The rational form shows that instinct is wrong, and the argument is short enough to be examined.

The idea

For a monic polynomial $p(x) = x^d + c_{d-1}x^{d-1} + \cdots + c_0$, its companion matrix is

$$C(p) = \begin{pmatrix} 0 & 0 & \cdots & 0 & -c_0 \\ 1 & 0 & \cdots & 0 & -c_1 \\ 0 & 1 & & 0 & -c_2 \\ \vdots & & \ddots & & \vdots \\ 0 & 0 & \cdots & 1 & -c_{d-1}\end{pmatrix},$$

whose characteristic polynomial and minimal polynomial are both $p$.

Every $A$ over any field $F$ is similar to a block-diagonal matrix of companion matrices $C(d_1), \dots, C(d_k)$, where the monic $d_i$ satisfy $d_1 \mid d_2 \mid \cdots \mid d_k$.

These $d_i$ are the invariant factors, they are unique, and:

• $\chi_A = d_1 d_2 \cdots d_k$;
• $m_A = d_k$, the largest.

No splitting is required anywhere — the construction uses only row and column operations over $F$, via the Smith normal form of $xI - A$ over $F[x]$.

Similarity does not depend on the field

This is the payoff. Suppose $A, B$ have entries in $F$ and are similar over some extension $K \supseteq F$. Their rational canonical forms are computed by operations that never leave $F$, so each has a rational form over $F$; those forms are also rational canonical forms over $K$, where uniqueness applies. So the forms coincide, and $A$ and $B$ are already similar over $F$.

Two real matrices similar over $\mathbb{C}$ are similar over $\mathbb{R}$. Two rational matrices similar over $\mathbb{R}$ are similar over $\mathbb{Q}$. The extension buys nothing.

Contrast this with diagonalisability, which very much does depend on the field: the rotation matrix is diagonalisable over $\mathbb{C}$ and not over $\mathbb{R}$. Similarity to a particular shape can depend on the field; similarity of two given matrices to each other cannot.

Invariant factors and elementary divisors

Factor each invariant factor into powers of irreducibles over $F$; the resulting collection is the list of elementary divisors. Over an algebraically closed field the irreducibles are linear, the elementary divisors are $(x - \lambda)^k$, and their companion matrices are exactly the Jordan blocks. So the Jordan form is the special case of the rational form when everything splits.

Over $\mathbb{R}$ the irreducibles are linear and quadratic, which is why the real canonical form has $2\times2$ rotation-like blocks alongside Jordan blocks.

Reconstructing from data

Given $\chi_A$ and $m_A$, the invariant factors must satisfy $d_k = m_A$, each $d_i \mid d_{i+1}$, and $\prod d_i = \chi_A$. Listing the possibilities is the standard question, and it is the same partition bookkeeping as for Jordan blocks, done with divisibility instead.

For example with $\chi_A = (x-1)^4$ and $m_A = (x-1)^2$: $d_k = (x-1)^2$, and the rest must multiply to $(x-1)^2$ while dividing it — either $(x-1)^2$ alone, giving $d = ((x-1)^2, (x-1)^2)$, or $(x-1), (x-1)$, giving $d = ((x-1), (x-1), (x-1)^2)$. Two possibilities, matching the two Jordan structures exactly.

Where intuition breaks

"Similar over $\mathbb{C}$ but not over $\mathbb{R}$." Impossible. This is the topic's headline, and the option is usually phrased to sound cautious and therefore plausible.

"The rational canonical form needs the eigenvalues to lie in the field." It needs nothing. That is its entire reason for existing.

"The rational form of a diagonalisable matrix is diagonal." Not generally. A diagonal matrix with a repeated eigenvalue has invariant factors that are not linear... more carefully: $\operatorname{diag}(2,2)$ has invariant factors $(x-2), (x-2)$, so its rational form is diagonal here — but $\operatorname{diag}(2,3)$ has a single invariant factor $(x-2)(x-3)$ and its rational form is the companion matrix of that quadratic, which is not diagonal. The rational form is not trying to be diagonal.

"Elementary divisors and invariant factors are the same list." Invariant factors form a divisibility chain; elementary divisors are their irreducible-power pieces. You can recover each from the other, but they are different lists.

"$m_A$ is the product of the invariant factors." It is the largest one; $\chi_A$ is the product.

"A companion matrix is diagonalisable." Only when its polynomial has distinct roots in the field. $C((x-1)^2)$ is $\begin{pmatrix}0&-1\\1&2\end{pmatrix}$, similar to a single Jordan block, not diagonalisable.

The exam's angle

1. If the question compares fields, the answer is almost always "similarity is unchanged". Then check whether it is really asking about diagonalisability, which is field-dependent.
2. Use $\prod d_i = \chi_A$ and $d_k = m_A$ to list possibilities; the divisibility chain does the rest.
3. Translate to Jordan when the field is $\mathbb{C}$ — the bookkeeping is the same and the picture is easier.
4. Remember a companion matrix has $m = \chi$. Any block that is a single companion matrix is cyclic, and that fact alone answers several options.
5. Over $\mathbb{R}$, expect quadratic irreducibles. A real matrix with non-real eigenvalues has $2\times2$ blocks in its real canonical form and no Jordan form at all.

The night before

• Companion matrix of $p$: $\chi = m = p$.
• Rational canonical form exists over every field. Invariant factors $d_1 \mid \cdots \mid d_k$, unique, with $\prod d_i = \chi_A$ and $d_k = m_A$.
• Similarity is independent of the field. Similar over $\mathbb{C}$ $\Rightarrow$ similar over $\mathbb{R}$, for real matrices.
• Diagonalisability is field-dependent. Do not confuse the two.
• Elementary divisors are the irreducible-power factors of the invariant factors. Over an algebraically closed field they give the Jordan blocks.
• Over $\mathbb{R}$ the irreducibles are linear and quadratic, hence real blocks of size $2$.
• $\chi_A = (x-1)^4$, $m_A = (x-1)^2$: two possible invariant-factor lists, matching the two Jordan structures.