# When does a matrix have a cyclic decomposition and when a rational form?

I'm reading Hoffman's "Linear Algebra" and this question comes to my mind: suppose $V$ is a finite-dimensional vector space on a field $F$, and $T$ a linear operator on $V$, when does a matrix has Cyclic Decomposition, and when does it has a Rational Form?

The Cyclic Decomposition Theorem (Hoffman, Thm 7.3) says:

Let $T$ be a linear operator on a finite-dimensional vector space $V$ and let $W_0$ be a proper $T$-admissible subspace of $V$. There exist non-zero vectors $a_1, \dots , a_r$ in $V$ with respective $T$-annihilators $p_1, \dots , p_r$ such that $V = W_0 \oplus Z(\alpha_1; T) \oplus \dots \oplus Z (\alpha_r ; T)$ ;

So the Cyclic Decomposition exists iff there is a $T$-admissible proper subspace $W_0 \subsetneq V$ to start from. Then what is the the condition for $W_0$'s existence?

A Rational Form is actually the Cyclic Decomposition with $W_0=\{0\}$. So does this mean the condition for the existence of a rational form is that $T$'s character values are all in field $F$?

For example a $n\times n$ matrix $T$ on real number field $\mathbb R$ has a rational form, so long as all the characteristic values of $T$ are all real numbers?

The cyclic decomposition theorem actually says that a cyclic decomposition always exists, since you can take $W_0=\{0\}$. Mentioning $W_0$ in that version of the cyclic decomposition theorem is just done in order to allow a proof by induction on the (remaining) dimension: by constructing an appropriate new cyclic factor so that its sum with the old $W_0$ is still $T$-admissible. (That $W_0$ is required to be a proper subspace is to avoid the case $r=0$, but it would have been better to just allow that case.)

Cyclic decompositions lead to "rational forms"; special matrices similar to the original one over the the original field (no field extension). These forms exist whether or not the eigenvalues (characteristic values) live over the base field, since blocks of the matrix are companion matrices of the annihilators of the cyclic factors which could be any polynomials, not just those of the form $X-\lambda$. Cyclic decompositions are not unique however, not even up to some coarse notion of equivalence; for instance the number of cyclic factors can vary among decompositions. There is a notion of Rational Canonical Form, in which the additional condition is imposed that the annihilators of the cyclic factors each divide the next (or in another flavour each are a multiple of the next); this does not yet make the cyclic decomposition unique, but it does make the sequence of annihilators unique, and therefore the associated rational form.

The rational canonical form corresponds to a decomposition in as few cyclic factors as possible. For instance if the whole space is cyclic it will just be a single companion matrix; moreover this is often the case, for instance if $T$ is diagonalisable without repeated eigenvalues, or more generally if the minimal polynomial equals the characteristic one. In general its decomposition contains one cyclic direct factor whose annihilator is as large as possible, namely equal to the minimal polynomial, while a complementary $T$-stable subspace is (if nonzero) similarly decomposed recursively (although this is not the method by which the decomposition is found). Since no decomposition of cyclic factors is attempted corresponding to factoring of minimal polynomials, the rational canonical form does not change when extending scalars to a larger field. As a consequence the entries of the rational canonical form live in the smallest field for which any matrix of $T$ can be found.

There is however a different decomposition that does reflect factoring of the minimal polynomial over the field at hand, which is the primary decomposition. It is not necessarily a decomposition into cyclic factors, but has the advantage over decompositions into cyclic factors that the decomposition itself is canonical: it does not depend on any choices (but it does depend on the field). As a consequence the primary decomposition is compatible with any $T$-invariant subspace $W\subseteq V$: it is always true that $W$ is direct sum of subspaces of the primary factors of$~V$, namely of its intersections with those factors, which intersections form the primary factors of$~W$. (Nothing of this kind holds for decompositions into cyclic factors corresponding to the rational canonical form).

A decomposition into a maximal number of cyclic factors (which are therefore as small as possible) can be obtained be decomposing each primary factor of$~V$ into cyclic factors (this decomposition is not unique, thought in this case all possible decompositions are isomorphic). The result is called the primary rational canonical form. The number of its cyclic factors may increase as one extends scalars to a larger field (if this leads to the minimal polynomial factoring into smaller irreducible factors).

• I thought I got it... mmm... is it true that with field extension, the minimal polynomial might factor out more, hence the annihilators might change and so does the rational form, as it's built by the blocks of companion matrix of the annihilators? Apr 17, 2015 at 1:20
• Thanks for this nice answer. Could you give some more details on your claim that the cyclic decomposition corresponding to the rational canonical form and possibly give an example? Jun 24, 2016 at 9:53
• How can I see the statement "the primary decomposition is compatible with any T -invariant subspace" is true? It seems it is identical to Hoffman and Kunze 6.8.10. Jun 12, 2023 at 8:06
• @UalibekNurgulan: I think there are basically two ways (but it was some time ago that I wrote this answer, so maybe one of them is not OK, in which case use the other ;-). (1) compatibility follows from uniqueness of the primary decomposition: the intersections of the submodule $W$ with the primary factors of $V$ are primary factors of $W$, necessarily the primary factors of $W$. (2) The decomposition theorem also gives that the projection onto each factor is a polynomial in $T$; that same polynomial in the restriction of $T$ to $W$ gives the projection for the latter, showing compatibility Jun 12, 2023 at 13:33
• @UalibekNurgulan: I'll try to say this without using long formulae, which are always hard to digest. The primary factor for any irreducible polynomial $P$ is by definition the kernel of a sufficiently high power of $P[T]$ (higher powers don't make the kernel grow any more). Normally, when intersecting a subspace $W$ with factors of a direct sum decomposition, the sum of the intersections are contained in $W$ but may be smaller than $W$. However, the primary decomposition theorem for $T_W$ says that $W$ is the sum of its primary factors = intersections of $W$ with the primary factors of $T$. Jun 13, 2023 at 14:07

These two concepts are the same. Indeed, the ability to put ANY matrix in its "rational form" is the matrix version of "the cyclic decomposition theorem".

More precisely, let $F$ be (ANY) field and $A\in M_n(F)$. Then there are monic polynomials $p_i\in F[x]$ s.t. $p_{i+1}$ divides $p_i$ and s.t. $A$ is similar over $F$ to $B=diag(B_1,\cdots,B_n)$, where $B_i$ is the companion matrix of $p_i$ (noted $B_i=C_{p_i}$).

For instance $A=\begin{pmatrix}-13&7&-7&28\\-18&10&-8&36\\2&-1&3&-4\\-2&1&-1&5\end{pmatrix}$ admits as rational form $diag(C_{x-1},C_{(x-1)^2(x-2)})$.

• should it be $\mathtt{diag} (C_{(x-1)^2(x-2}, C_{x-1})$, as $p_2 \mid p_1$? Apr 17, 2015 at 1:25
• Yes, of course.
– user91684
Apr 17, 2015 at 2:21
• Can't it be $diag(C_{(x-1)^2 (x-2)}, C_{(x-2)}$ ?
– Our
May 11, 2018 at 11:05
• @onurcanbektas ; no, my result is correct. Indeed, $spectrum(A)=\{1,1,1,2\}$ while the spectrum of your matrix is $\{1,1,2,2\}$.
– user91684
May 11, 2018 at 15:26