[Math] Cyclic Decompositions and the Rational form

linear algebra

Given $T$ be a linear operator on the the finite dimensional space $V$, let $R$ be the range of $T$. Then, I have to show that

(a) $R$ has a complementary $T$-invariant subspace if and only if $R$ is independent of the null space $N$ of $T$.

(b) If $R$ and $N$ are independent, I need to prove that $N$ is the unique $T$-invariant subspace complementary to $R$.

Please suggest how to proceed.

I supposed $R$ has a complementary $T$-invariant subspace, say, $W$. Then $R$ should be $T$-admissible. I assumed to the contrary, that $R$ intersection $T$ is not equal to $\{\mathbf{0}\}$. I took a point in the intersection but could not proceed further. Please suggest what to do.

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(a) If $R\cap N =\{\mathbf{0}\}$, then you should be able to use the Rank-Nullity Theorem to show that $R+N=V$, and you are done.

Conversely, if there exists $W$ such that $R\cap W=\{\mathbf{0}\}$, $R+W=V$, and $W$ is $T$-invariant, what can you say about $T(\mathbf{w})$ for any $\mathbf{w}\in W$? It is certainly in $W$... is it in $R$ as well?

(b) As above, suppose that $R+W=V$, with $R\cap W=\{\mathbf{0}\}$ and $T(W)\subseteq W$. Show that $W\subseteq N$ and conclude, for example, using dimension, that $W=N$.

P.S. This has nothing to do with the rational canonical form, nor with cyclic decompositions.

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[Math] Showing no non-trivial t-invariant subspace has a t-invariant complement.

An essential hypothesis is (b), implying that no nonzero polynomial of degree less than the dimension of the vector space annihilates$~T$ (as such a degree is incompatible with being a multiple of the characteristic polynomial). Let me call this condition, which has many equivalent statements, that $T$ is cyclic (actually it is the $K[X]$-module defined by$~T$ that is cyclic, but I don't want to mention $K[X]$-modules here). One basic fact is that the restriction of a cyclic operator$~T$ to any $T$-invariant subspace is still cyclic (as by the way is the operator that $T$ induces in the quotient modulo this subspace); let me prove that first.

If the restriction of$~T$ to a $T$-invariant subspace of dimension$~d$ were annihilated by a polynomial$~P[T]$ with $\deg(P)<d$, then the image$~W$ of$~P[T]$ would be a $T$-invariant subspace of dimension at most$~\dim V-d$ (by rank-nullity), and annihilated by some$~Q$ with $\deg(Q)\leq\dim W$ (for instance the characteristic polynomial of $T|_W$). But then $QP$ annihilates$~T$ (as $P[T]$ maps $V$ into $W$ which is contained in the kernel of $Q[T]$), and $\deg(QP)<\dim V$, contradicting the hypothesis that $T$ is cyclic.

Now for the actual question. Let $P$ be the irreducible polynomial of point (a), which I may suppose monic, and $P^k$ the minimal polynomial of$~T$. Suppose for a contradiction that $V$ decomposes as a direct sum $W_1\oplus W_2$ of two proper $T$-invariant subspaces. The minimal polynomials of the restrictions of$~T$ to the summands both divide the minimal polynomial $P^k$, and since by the above these restrictions are cyclic, their degrees are $\dim W_1$ respectively $\dim W_2$; in particular they are proper monic divisors of$~P^k$. But from the irreducibility of$~P$ this implies they are of the form$~P^l$ with $l<k$. But then their least common multiple, which gives the minimal polynomial of$~T$, cannot be $P^k$, a contradiction.

Linear Algebra – Matrix Cyclic Decomposition and Rational Form

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).

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[Math] Showing no non-trivial t-invariant subspace has a t-invariant complement.

Linear Algebra – Matrix Cyclic Decomposition and Rational Form

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