7.4* The Rational Canonical Form
Until now we have used eigenvalues, eigenvectors, and generalized eigenvectors in our analysis of linear operators with characteristic polynomials that split. In general, characteristic polynomials need not split, and indeed, operators need not have eigenvalues! However, the unique factorization theorem for polynomials (see page 562) guarantees that the characteristic polynomial f(t) of any linear operator T on an ndimensional vector space factors uniquely as
where the ${\varphi}_{i}(t)$’s $(1\le i\le k)$ are distinct irreducible monic polynomials and the ${n}_{i}$’s are positive integers. In the case that f(t) splits, each irreducible monic polynomial factor is of the form ${\varphi}_{i}(t)=t{\lambda}_{i}$, where ${\lambda}_{i}$ is an eigenvalue of T, and there is a onetoone correspondence between eigenvalues of T and the irreducible monic factors of the characteristic polynomial. In general, eigenvalues need not exist, but the irreducible monic factors always exist. In this section, we establish structure theorems based on the irreducible monic factors of the characteristic polynomial instead of eigenvalues.
In this context, the following definition is the appropriate replacement for eigenspace and generalized eigenspace.
Definition.
Let T be a linear operator on a finitedimensional vector space V with characteristic polynomial
where the ${\varphi}_{i}(t)$’s $(1\le i\le k)$ are distinct irreducible monic polynomials and the ${n}_{i}$’s are positive integers. For $1\le i\le k$, we define the subset ${\text{K}}_{{\varphi}_{i}}$ of V by
We show that each ${\text{K}}_{{\varphi}_{i}}$ is a nonzero Tinvariant subspace of V. Note that if ${\varphi}_{i}(t)=t\lambda $ is of degree one, then ${\text{K}}_{{\varphi}_{i}}$ is the generalized eigenspace of T corresponding to the eigenvalue $\lambda $.
Having obtained suitable generalizations of the related concepts of eigenvalue and eigenspace, our next task is to describe a canonical form of a linear operator suitable to this context. The one that we study is called the rational canonical form. Since a canonical form is a description of a matrix representation of a linear operator, it can be defined by specifying the form of the ordered bases allowed for these representations.
Here the bases of interest naturally arise from the generators of certain cyclic subspaces. For this reason, the reader should recall the definition of a Tcyclic subspace generated by a vector and Theorem 5.21 (p. 314). We briefly review this concept and introduce some new notation and terminology.
Let T be a linear operator on a finitedimensional vector space V, and let x be a nonzero vector in V. We use the notation ${\text{C}}_{x}$ for the Tcyclic subspace generated by x. Recall (Theorem 5.21) that if $\mathrm{dim}({\text{C}}_{x})=k$, then the set
is an ordered basis for ${\text{C}}_{x}$. To distinguish this basis from all other ordered bases for ${\text{C}}_{x}$, we call it the Tcyclic basis generated by x and denote it by ${\beta}_{x}$. Let A be the matrix representation of the restriction of T to ${\text{C}}_{x}$ in the ordered basis ${\beta}_{x}$. Recall from the proof of Theorem 5.21 that
where
Furthermore, the characteristic polynomial of A is given by
The matrix A is called the companion matrix of the monic polynomial $h(t)={a}_{0}+{a}_{1}t+\cdots +{a}_{k1}{t}^{k1}+{t}^{k}$. Every monic polynomial has a companion matrix, and the characteristic polynomial of the companion matrix of a monic polynomial g(t) of degree k is equal to ${(1)}^{k}g(t)$. (See Exercise 19 of Section 5.4.) By Theorem 7.15 (p. 512), the monic polynomial h(t) is also the minimal polynomial of A. Since A is the matrix representation of the restriction of T to ${\text{C}}_{x},\text{}h(t)$ is also the minimal polynomial of this restriction. By Exercise 15 of Section 7.3, h(t) is also the Tannihilator of x.
It is the object of this section to prove that for every linear operator T on a finitedimensional vector space V, there exists an ordered basis $\beta $ for V such that the matrix representation ${[\text{T}]}_{\beta}$ is of the form
where each ${C}_{i}$ is the companion matrix of a polynomial ${(\varphi (t))}^{m}$ such that $\varphi (t)$ is a monic irreducible divisor of the characteristic polynomial of T and m is a positive integer. A matrix representation of this kind is called a rational canonical form of T. We call the accompanying basis a rational canonical basis for T.
The next theorem is a simple consequence of the following lemma, which relies on the concept of Tannihilator, introduced in the Exercises of Section 7.3.
Lemma.
Let T be a linear operator on a finitedimensional vector space V, let x be a nonzero vector in V, and suppose that the Tannihilator of x is of the form ${(\varphi (t))}^{p}$ for some irreducible monic polynomial $\varphi (t)$. Then $\varphi (t)$ divides the minimal polynomial of T, and $x\in {\text{K}}_{\varphi}$.
Proof.
By Exercise 15(b) of Section 7.3, ${(\varphi (t))}^{p}$ divides the minimal polynomial of T. Therefore $\varphi (t)$ divides the minimal polynomial of T. Furthermore, $x\in {\text{K}}_{\varphi}$ by the definition of ${\text{K}}_{\varphi}$.
Theorem 7.17.
Let T be a linear operator on a finitedimensional vector space V, and let $\beta $ be an ordered basis for V. Then $\beta $ is a rational canonical basis for T if and only if $\beta $ is the disjoint union of Tcyclic bases ${\beta}_{{v}_{i}}$, where each ${v}_{i}$ lies in ${\text{K}}_{\varphi}$ for some irreducible monic divisor $\varphi (t)$ of the characteristic polynomial of T.
Proof.
Exercise.
Example 1
Suppose that T is a linear operator on ${\text{R}}^{8}$ and
is a rational canonical basis for T such that
is a rational canonical form of T. In this case, the submatrices ${C}_{1},\text{}{C}_{2}$, and ${C}_{3}$ are the companion matrices of the polynomials ${\varphi}_{1}(t),\text{}{({\varphi}_{2}(t))}^{2}$, and ${\varphi}_{2}(t)$, respectively, where
In the context of Theorem 7.17, $\beta $ is the disjoint union of the Tcyclic bases; that is,
By Exercise 39 of Section 5.4, the characteristic polynomial f(t) of T is the product of the characteristic polynomials of the companion matrices:
The rational canonical form C of the operator T in Example 1 is constructed from matrices of the form ${C}_{i}$, each of which is the companion matrix of some power of a monic irreducible divisor of the characteristic polynomial of T. Furthermore, each such divisor is used in this way at least once.
In the course of showing that every linear operator T on a finite dimensional vector space has a rational canonical form C, we show that the companion matrices ${C}_{i}$ that constitute C are always constructed from powers of the monic irreducible divisors of the characteristic polynomial of T. A key role in our analysis is played by the subspaces ${\text{K}}_{\varphi}$, where $\varphi (t)$ is an irreducible monic divisor of the minimal polynomial of T. Since the minimal polynomial of an operator divides the characteristic polynomial of the operator, every irreducible divisor of the former is also an irreducible divisor of the latter. We eventually show that the converse is also true; that is, the minimal polynomial and the characteristic polynomial have the same irreducible divisors.
We begin with a result that lists several properties of irreducible divisors of the minimal polynomial. The reader is advised to review the definition of Tannihilator and the accompanying Exercises 15 of Section 7.3.
Theorem 7.18.
Let T be a linear operator on a finitedimensional vector space V, and suppose that
is the minimal polynomial of T, where the ${\varphi}_{i}(t)$‘s $(1\le i\le k)$ are the distinct irreducible monic factors of p(t) and the ${m}_{i}$’s are positive integers. Then the following statements are true.

(a) ${\text{K}}_{{\varphi}_{i}}$ is a nonzero Tinvariant subspace of V for each i.

(b) If x is a nonzero vector in some ${\text{K}}_{{\varphi}_{i}}$, then the Tannihilator of x is of the form ${({\varphi}_{i}(t))}^{p}$ for some integer p.

(c) ${\text{K}}_{{\varphi}_{i}}\cap {\text{K}}_{{\varphi}_{j}}=\left\{0\right\}\text{for}i\ne j$.

(d) ${\text{K}}_{{\varphi}_{i}}$ is invariant under ${\varphi}_{j}(\text{T})$ for $i\ne j$, and the restriction of ${\varphi}_{j}(\text{T})$ to ${\text{K}}_{{\varphi}_{i}}$ is onetoone and onto.

(e) ${\text{K}}_{{\varphi}_{i}}=\text{N}({({\varphi}_{i}(\text{T}))}^{{m}_{i}})$ for each i.
Proof.
If $k=1$, then (a), (b), and (e) are obvious, while (c) and (d) are vacuously true. Now suppose that $k>1$.

(a) The proof that ${\text{K}}_{{\varphi}_{i}}$ is a Tinvariant subspace of V is left as an exercise. Let ${f}_{i}(t)$ be the polynomial obtained from p(t) by omitting the factor ${({\varphi}_{i}(t))}^{{m}_{i}}$. To prove that ${\text{K}}_{{\varphi}_{i}}$ is nonzero, first observe that ${f}_{i}(t)$ is a proper divisor of p(t); therefore there exists a vector $z\in \text{V}$ such that $x={f}_{i}(\text{T})(z)\ne 0$. Then $x\in {\text{K}}_{{\varphi}_{i}}$ because
$${({\varphi}_{i}(\text{T}))}^{{m}_{i}}(x)={({\varphi}_{i}(\text{T}))}^{{m}_{i}}{f}_{i}(\text{T})(z)=p(\text{T})(z)=0.$$ 
(b) Assume the hypothesis. Then ${({\varphi}_{i}(\text{T}))}^{q}(x)=0$ for some positive integer q. Hence the Tannihilator of x divides ${({\varphi}_{i}(t))}^{q}$ by Exercise 15(b) of Section 7.3, and the result follows.

(c) Assume $i\ne j$. Let $x\in {\text{K}}_{{\varphi}_{i}}\cap {\text{K}}_{{\varphi}_{j}}$, and suppose that $x\ne 0$. By (b), the Tannihilator of x is a power of both ${\varphi}_{i}(t)$ and ${\varphi}_{j}(t)$. But this is impossible because ${\varphi}_{i}(t)$ and ${\varphi}_{j}(t)$ are relatively prime (see Appendix E). We conclude that $x=0$.

(d) Assume $i\ne j$. Since ${\text{K}}_{{\varphi}_{i}}$ is Tinvariant, it is also ${\varphi}_{j}(\text{T})$invariant. Suppose that ${\varphi}_{j}(\text{T})(x)=0$ for some $x\in {\text{K}}_{{\varphi}_{i}}$. Then $x\in {\text{K}}_{{\varphi}_{i}}\cap {\text{K}}_{{\varphi}_{j}}=\left\{0\right\}$ by (c). Therefore the restriction of ${\varphi}_{j}(\text{T})$ to ${\text{K}}_{{\varphi}_{i}}$ is onetoone. Since V is finitedimensional, this restriction is also onto.

(e) Suppose that $1\le i\le k$. Clearly, $\text{N}({({\varphi}_{i}(\text{T}))}^{{m}_{i}})\subseteq {\text{K}}_{{\varphi}_{i}}$. Let ${f}_{i}(t)$ be the polynomial defined in (a). Since ${f}_{i}(t)$ is a product of polynomials of the form ${\varphi}_{j}(t)$ for $j\ne i$, we have by (d) that the restriction of ${f}_{i}(\text{T})$ to ${\text{K}}_{{\varphi}_{i}}$ is onto. Let $x\in {\text{K}}_{{\varphi}_{i}}$. Then there exists $y\in {\text{K}}_{{\varphi}_{i}}$ such that ${f}_{i}(\text{T})(y)=x$. Therefore
$$({({\varphi}_{i}(\text{T}))}^{{m}_{i}})(x)=({({\varphi}_{i}(\text{T}))}^{{m}_{i}}){f}_{i}(\text{T})(y)=p(\text{T})(y)=0\text{,}$$
and hence $x\in \text{N}({({\varphi}_{i}(\text{T}))}^{{m}_{i}})$. Thus ${\text{K}}_{{\varphi}_{i}}=\text{N}({({\varphi}_{i}(\text{T}))}^{{m}_{i}})$.
Since a rational canonical basis for an operator T is obtained from a union of Tcyclic bases, we need to know when such a union is linearly independent. The next major result, Theorem 7.19, reduces this problem to the study of Tcyclic bases within ${\text{K}}_{\varphi}$, where $\varphi (t)$ is an irreducible monic divisor of the minimal polynomial of T. We begin with the following lemma.
Lemma.
Let T be a linear operator on a finitedimensional vector space V, and suppose that
is the minimal polynomial of T, where the ${\varphi}_{i}$’s $(1\le i\le k)$ are the distinct irreducible monic factors of p(t) and the ${m}_{i}$’s are positive integers. For $1\le i\le k$, let ${v}_{i}\in {\text{K}}_{{\varphi}_{i}}$ be such that
Then ${v}_{i}=0$ for all i.
Proof.
The result is trivial if $k=1$, so suppose that $k>1$. Consider any i. Let ${f}_{i}(t)$ be the polynomial obtained from p(t) by omitting the factor ${({\varphi}_{i}(t))}^{{m}_{i}}$. As a consequence of Theorem 7.18, ${f}_{i}(\text{T})$ is onetoone on ${\text{K}}_{{\varphi}_{i}}$, and ${f}_{i}(\text{T})({v}_{j})=0$ for $i\ne j$. Thus, applying ${f}_{i}(\text{T})$ to (2), we obtain ${f}_{i}(\text{T})({v}_{i})=0$, from which it follows that ${v}_{i}=0$.
Theorem 7.19.
Let T be a linear operator on a finitedimensional vector space V, and suppose that
is the minimal polynomial of T, where the ${\varphi}_{i}$‘s $(1\le i\le k)$ are the distinct irreducible monic factors of p(t) and the ${m}_{i}$’s are positive integers. For $1\le i\le k$, let ${S}_{i}$ be a linearly independent subset of ${\text{K}}_{{\varphi}_{i}}$. Then

${S}_{i}\cap {S}_{j}=\varnothing $ for $i\ne j$

${S}_{1}\cup {S}_{2}\cup \cdots \cup {S}_{k}$ is linearly independent.
Proof.
If $k=1$, then (a) is vacuously true and (b) is obvious. Now suppose that $k>1$. Then (a) follows immediately from Theorem 7.18(c). Furthermore, the proof of (b) is identical to the proof of Theorem 5.5 (p. 261) with the eigenvectors replaced by the generalized eigenvectors.
In view of Theorem 7.19, we can focus on bases of individual spaces of the form ${\text{K}}_{\varphi}$, where $\varphi (t)$ is an irreducible monic divisor of the minimal polynomial of T. The next several results give us ways to construct bases for these spaces that are unions of Tcyclic bases. These results serve the dual purposes of leading to the existence theorem for the rational canonical form and of providing methods for constructing rational canonical bases.
For Theorems 7.20 and 7.21 and the latter’s corollary, we fix a linear operator T on a finitedimensional vector space V and an irreducible monic divisor $\varphi (t)$ of the minimal polynomial of T.
Theorem 7.20.
Let ${v}_{1},\text{}{v}_{2},\text{}\dots ,\text{}{v}_{k}$ be distinct vectors in ${\text{K}}_{\varphi}$ such that
is linearly independent. For each i, suppose there exists ${w}_{i}\in \text{V}$ such that $\varphi (\text{T})({w}_{i})={v}_{i}$. Then
is also linearly independent.
Proof.
Consider any linear combination of vectors in ${S}_{2}$ that sums to zero, say,
For each i, let ${f}_{i}(t)$ be the polynomial defined by
Then (3) can be rewritten as
Apply $\varphi (\text{T})$ to both sides of (4) to obtain
This last sum can be rewritten as a linear combination of the vectors in ${S}_{1}$ so that each ${f}_{i}(\text{T})({v}_{i})$ is a linear combination of the vectors in ${\beta}_{{v}_{i}}$. Since ${S}_{1}$ is linearly independent, it follows that
Therefore the Tannihilator of ${v}_{i}$ divides ${f}_{i}(t)$ for all i. (See Exercise 15 of Section 7.3.) By Theorem 7.18(b), $\varphi (t)$ divides the Tannihilator of ${v}_{i}$, and hence $\varphi (t)$ divides ${f}_{i}(t)$ for all i. Thus, for each i, there exists a polynomial ${g}_{i}(t)$ such that ${f}_{i}(t)={g}_{i}(t)\varphi (t)$. So (4) becomes
Again, linear independence of ${S}_{1}$ requires that
But ${f}_{i}(\text{T})({w}_{i})$ is the result of grouping the terms of the linear combination in (3) that arise from the linearly independent set ${\beta}_{{w}_{i}}$. We conclude that for each i, ${a}_{ij}=0$ for all j. Therefore ${S}_{2}$ is linearly independent.
We now show that ${\text{K}}_{\varphi}$ has a basis consisting of a union of Tcycles.
Lemma.
Let W be a Tinvariant subspace of ${\text{K}}_{\varphi}$, and let $\beta $ be a basis for W. Then the following statements are true.

(a) Suppose that $x\in \text{N}(\varphi (\text{T}))$, but $x\notin \text{W}$. Then $\beta \cup {\beta}_{x}$ is linearly independent.

(b) For some ${w}_{1},\text{}{w}_{2},\text{}\dots ,\text{}{w}_{s}$ in $\text{N}(\varphi (\text{T}))$, $\beta $ can be extended to the linearly independent set
$${\beta}^{\prime}=\beta \cup {\beta}_{{w}_{1}}\cup {\beta}_{{w}_{2}}\cup \cdots \cup {\beta}_{{w}_{s}},\text{}$$whose span contains $\text{N}(\varphi (\text{T}))$.
Proof.
(a) Let $\beta =\{{v}_{1},\text{}{v}_{2},\text{}\dots ,\text{}{v}_{k}\}$, and suppose that
where d is the degree of $\varphi (t)$. Then $z\in {\text{C}}_{x}\cap \text{W}$, and hence ${\text{C}}_{z}\subseteq {\text{C}}_{x}\cap \text{W}$. Suppose that $z\ne 0$. Then z has $\varphi (t)$ as its Tannihilator, and therefore
It follows that ${\text{C}}_{x}\cap \text{W}={\text{C}}_{x}$, and consequently $x\in \text{W}$, contrary to hypothesis. Therefore $z=0$, from which it follows that ${b}_{j}=0$ for all j. Since $\beta $ is linearly independent, it follows that ${a}_{i}=0$ for all i. Thus $\beta \cup {\beta}_{x}$ is linearly independent.
(b) Suppose that W does not contain $\text{N}(\varphi (\text{T}))$. Choose a vector ${w}_{1}\in \text{N}(\varphi (\text{T}))$ that is not in W. By (a), ${\beta}_{1}=\beta \cup {\beta}_{{w}_{1}}$ is linearly independent. Let ${\text{W}}_{1}=\text{span}({\beta}_{1})$. If ${\text{W}}_{1}$ does not contain $\text{N}(\varphi (\text{T}))$, choose a vector ${w}_{2}$ in $\text{N}(\varphi (\text{T}))$, but not in ${\text{W}}_{1}$, so that ${\beta}_{2}={\beta}_{1}\cup {\beta}_{{w}_{2}}=\beta \cup {\beta}_{{w}_{1}}\cup {\beta}_{{w}_{2}}$ is linearly independent. Continuing this process, we eventually obtain vectors ${w}_{1},\text{}{w}_{2},\text{}\dots ,\text{}{w}_{s}$ in $\text{N}(\varphi (\text{T}))$ such that the union
is a linearly independent set whose span contains $\text{N}(\varphi (\text{T}))$.
Theorem 7.21.
If the minimal polynomial of T is of the form $p(t)={(\varphi (t))}^{m}$, then there exists a rational canonical basis for T.
Proof.
The proof is by mathematical induction on m. Suppose that $m=1$. Apply (b) of the lemma to $\text{W}=\left\{0\right\}$ to obtain a linearly independent subset of V of the form ${\beta}_{{v}_{1}}\cup {\beta}_{{v}_{2}}\cup \cdots \cup {\beta}_{{v}_{k}}$, whose span contains $\text{N}(\varphi (\text{T}))$. Since $\text{V}=\text{N}(\varphi (\text{T}))$, this set is a rational canonical basis for V.
Now suppose that, for some integer $m>1$, the result is valid whenever the minimal polynomial of T is of the form ${(\varphi (t))}^{k}$, where $k<m$, and assume that the minimal polynomial of T is $p(t)={(\varphi (t))}^{m}$. Let $r=\text{rank}(\varphi (\text{T}))$. Then $\text{R}(\varphi (\text{T}))$ is a Tinvariant subspace of V, and the restriction of T to this subspace has ${(\varphi (t))}^{m1}$ as its minimal polynomial. Therefore we may apply the induction hypothesis to obtain a rational canonical basis for the restriction of T to R(T). Suppose that ${v}_{1},\text{}{v}_{2},\text{}\dots ,\text{}{v}_{k}$ are the generating vectors of the Tcyclic bases that constitute this rational canonical basis. For each i, choose ${w}_{i}$ in V such that ${v}_{i}=\varphi (\text{T})({w}_{i})$. By Theorem 7.20, the union $\beta $ of the sets ${\beta}_{{w}_{i}}$ is linearly independent. Let $\text{W}=\text{span}(\beta )$. Then W contains $\text{R}(\varphi (\text{T}))$. Apply (b) of the lemma and adjoin additional Tcyclic bases ${\beta}_{{w}_{k+1}},\text{}{\beta}_{{w}_{k+2}},\text{}\dots ,\text{}{\beta}_{{w}_{s}}$ to $\beta $, if necessary, where ${w}_{i}$ is in $\text{N}(\varphi (\text{T}))$ for $i\ge k$, to obtain a linearly independent set
whose span ${\text{W}}^{\prime}$ contains both W and $\text{N}(\varphi (\text{T}))$.
We show that ${\text{W}}^{\prime}=\text{V}$. Let U denote the restriction of $\varphi (\text{T})$ to ${\text{W}}^{\prime}$, which is $\varphi \text{(T)}$invariant. By the way in which ${\text{W}}^{\prime}$ was obtained from $\text{R}(\varphi (\text{T}))$, it follows that $\text{R}(\text{U})=\text{R}(\varphi (\text{T}))$ and $\text{N}(\text{U})=\text{N}(\varphi (\text{T}))$. Therefore
Thus ${\text{W}}^{\prime}=\text{V}$, and ${\beta}^{\prime}$ is a rational canonical basis for T.
Corollary.
${\text{K}}_{\varphi}$ has a basis consisting of the union of Tcyclic bases.
Proof.
Apply Theorem 7.21 to the restriction of T to ${\text{K}}_{\varphi}$.
We are now ready to study the general case.
Theorem 7.22.
Every linear operator on a finitedimensional vector space has a rational canonical basis and, hence, a rational canonical form.
Proof.
Let T be a linear operator on a finitedimensional vector space V, and let $p(t)={({\varphi}_{1}(t))}^{{m}_{1}}{({\varphi}_{2}(t))}^{{m}_{2}}\cdots {({\varphi}_{k}(t))}^{{m}_{k}}$ be the minimal polynomial of T, where the ${\varphi}_{i}(t)$’s are the distinct irreducible monic factors of p(t) and ${m}_{i}>0$ for all i. The proof is by mathematical induction on k. The case $k=1$ is proved in Theorem 7.21.
Suppose that the result is valid whenever the minimal polynomial contains fewer than k distinct irreducible factors for some $k>1$, and suppose that p(t) contains k distinct factors. Let U be the restriction of T to the Tinvariant subspace $\text{W}=\text{R}({({\varphi}_{k}(\text{T}))}^{{m}_{k}})$, and let q(t) be the minimal polynomial of U. Then q(t) divides p(t) by Exercise 10 of Section 7.3. Furthermore, ${\varphi}_{k}(t)$ does not divide q(t). For otherwise, there would exist a nonzero vector $x\in \text{W}$ such that ${\varphi}_{k}(\text{U})(x)=0$ and a vector $y\in \text{V}$ such that $x={({\varphi}_{k}(\text{T}))}^{{m}_{k}}(y)$. It follows that ${({\varphi}_{k}(\text{T}))}^{{m}_{k+1}}(y)=0$, and hence $y\in {\text{K}}_{{\varphi}_{k}}$ and $x={({\varphi}_{k}(\text{T}))}^{{m}_{k}}(y)=0$ by Theorem 7.18(e), a contradiction. Thus q(t) contains fewer than k distinct irreducible divisors. So by the induction hypothesis, U has a rational canonical basis ${\beta}_{1}$ consisting of a union of Ucyclic bases (and hence Tcyclic bases) of vectors from some of the subspaces ${\text{K}}_{{\varphi}_{i}},\text{}1\le i\le k1$. By the corollary to Theorem 7.21, ${\text{K}}_{{\varphi}_{k}}$ has a basis ${\beta}_{2}$ consisting of a union of Tcyclic bases. By Theorem 7.19, ${\beta}_{1}$ and ${\beta}_{2}$ are disjoint, and $\beta ={\beta}_{1}\cup {\beta}_{2}$ is linearly independent. Let s denote the number of vectors in $\beta $.Then
We conclude that $\beta $ is a basis for V. Therefore $\beta $ is a rational canonical basis, and T has a rational canonical form.
In our study of the rational canonical form, we relied on the minimal polynomial. We are now able to relate the rational canonical form to the characteristic polynomial.
Theorem 7.23.
Let T be a linear operator on an ndimensional vector space V with characteristic polynomial
where the ${\varphi}_{i}(t)$’s $(1\le i\le k)$ are distinct irreducible monic polynomials and the ${n}_{i}$’s are positive integers. Then the following statements are true.

(a) ${\varphi}_{1}(t),\text{}{\varphi}_{2}(t),\text{}\dots ,\text{}{\varphi}_{k}(t)$ are the irreducible monic factors of the minimal polynomial.

(b) For each i, $\mathrm{dim}({\text{K}}_{{\varphi}_{i}})={d}_{i}{n}_{i}$, where ${d}_{i}$ is the degree of ${\varphi}_{i}(t)$.

(c) If $\beta $ is a rational canonical basis for T, then ${\beta}_{i}=\beta \cap {\text{K}}_{{\varphi}_{i}}$ is a basis for ${\text{K}}_{{\varphi}_{i}}$ for each i.

(d) If ${\gamma}_{i}$ is a basis for ${\text{K}}_{{\varphi}_{i}}$ for each i, then $\gamma ={\gamma}_{1}\cup {\gamma}_{2}\cup \cdots \cup {\gamma}_{k}$ is a basis for V. In particular, if each ${\gamma}_{i}$ is a disjoint union of Tcyclic bases, then $\gamma $ is a rational canonical basis for T.
Proof.
(a) By Theorem 7.22, T has a rational canonical form C. By Exercise 39 of Section 5.4, the characteristic polynomial of C, and hence of T, is the product of the characteristic polynomials of the companion matrices that compose C. Therefore each irreducible monic divisor ${\varphi}_{i}(t)$ of f(t) divides the characteristic polynomial of at least one of the companion matrices, and hence for some integer p, ${({\varphi}_{i}(t))}^{p}$ is the Tannihilator of a nonzero vector of V. We conclude that ${({\varphi}_{i}(t))}^{p}$, and so ${\varphi}_{i}(t)$, divides the minimal polynomial of T. Conversely, if $\varphi (t)$ is an irreducible monic polynomial that divides the minimal polynomial of T, then $\varphi (t)$ divides the characteristic polynomial of T because the minimal polynomial divides the characteristic polynomial.
(b), (c), and (d) Let $C={[\text{T}]}_{\beta}$, which is a rational canonical form of T. Consider any i $i(1\le i\le k)$. Since f(t) is the product of the characteristic polynomials of the companion matrices that compose C, we may multiply those characteristic polynomials that arise from the Tcyclic bases in ${\beta}_{i}$ to obtain the factor ${({\varphi}_{i}(t))}^{{n}_{i}}$ of f(t). Since this polynomial has degree ${n}_{i}{d}_{i}$, and the union of these bases is a linearly independent subset ${\beta}_{i}$ of ${\text{K}}_{{\varphi}_{i}}$, we have
Furthermore, $n={\displaystyle \sum _{i=1}^{k}{d}_{i}{n}_{i},\text{}}$ because this sum is equal to the degree of f(t).
Now let s denote the number of vectors in $\gamma $. By Theorem 7.19, $\gamma $ is linearly independent, and therefore
Hence $n=s$, and ${d}_{i}{n}_{i}=\mathrm{dim}({\text{K}}_{{\varphi}_{i}})$ for all i. It follows that $\gamma $ is a basis for V and ${\beta}_{i}$ is a basis for ${\text{K}}_{{\varphi}_{i}}$ for each i.
Uniqueness of the Rational Canonical Form
Having shown that a rational canonical form exists, we are now in a position to ask about the extent to which it is unique. Certainly, the rational canonical form of a linear operator T can be modified by permuting the Tcyclic bases that constitute the corresponding rational canonical basis. This has the effect of permuting the companion matrices that make up the rational canonical form. As in the case of the Jordan canonical form, we show that except for these permutations, the rational canonical form is unique, although the rational canonical bases are not.
To simplify this task, we adopt the convention of ordering every rational canonical basis so that all the Tcyclic bases associated with the same irreducible monic divisor of the characteristic polynomial are grouped together. Furthermore, within each such grouping, we arrange the Tcyclic bases in decreasing order of size. Our task is to show that, subject to this order, the rational canonical form of a linear operator is unique up to the arrangement of the irreducible monic divisors.
As in the case of the Jordan canonical form, we introduce arrays of dots from which we can reconstruct the rational canonical form. For the Jordan canonical form, we devised a dot diagram for each eigenvalue of the given operator. In the case of the rational canonical form, we define a dot diagram for each irreducible monic divisor of the characteristic polynomial of the given operator. A proof that the resulting dot diagrams are completely determined by the operator is also a proof that the rational canonical form is unique.
In what follows, T is a linear operator on a finitedimensional vector space with rational canonical basis $\beta ,\text{}\varphi (t)$ is an irreducible monic divisor of the characteristic polynomial of ${\beta}_{{v}_{1}},\text{}{\beta}_{{v}_{2}},\text{}\dots ,\text{}{\beta}_{{v}_{k}}$ are the Tcyclic bases of $\beta $ that are contained in ${\text{K}}_{\varphi}$; and d is the degree of $\varphi (t)$. For each j, let ${(\varphi (t))}^{{p}_{j}}$ be the annihilator of ${v}_{j}$. This polynomial has degree $d{p}_{j}$; therefore, by Exercise 15 of Section 7.3, ${\beta}_{{v}_{j}}$ contains ${d}_{{p}_{j}}$ vectors. Furthermore, ${p}_{1}\ge {p}_{2}\ge \cdots \ge {p}_{k}$ since the Tcyclic bases are arranged in decreasing order of size. We define the dot diagram of $\varphi (t)$ to be the array consisting of k columns of dots with ${p}_{j}$ dots in the jth column, arranged so that the jth column begins at the top and terminates after ${p}_{j}$ dots. For example, if $k=3,\text{}{p}_{1}=4,\text{}{p}_{2}=2$, and ${p}_{3}=2$, then the dot diagram is
Although each column of a dot diagram corresponds to a Tcyclic basis ${\beta}_{{v}_{i}}$ in ${\text{K}}_{\varphi}$, there are fewer dots in the column than there are vectors in the basis.
Example 2
Recall the linear operator T of Example 1 with the rational canonical basis $\beta $ and the rational canonical form $C={[\text{T}]}_{\beta}$. Since there are two irreducible monic divisors of the characteristic polynomial of T, ${\varphi}_{1}(t)={t}^{2}t+3$ and ${\varphi}_{2}(t)={t}^{2}+1$, there are two dot diagrams to consider. Because ${\varphi}_{1}(t)$ is the Tannihilator of ${v}_{1}$ and ${\beta}_{{v}_{1}}$ is a basis for ${\text{K}}_{{\varphi}_{1}}$, the dot diagram for ${\varphi}_{1}(t)$ consists of a single dot. The other two Tcyclic bases, ${\beta}_{{v}_{3}}$ and ${\beta}_{{v}_{7}}$, lie in ${\text{K}}_{{\varphi}_{2}}$. Since ${v}_{3}$ has Tannihilator ${({\varphi}_{2}(t))}^{2}$ and ${v}_{7}$ has Tannihilator ${\varphi}_{2}(t)$, in the dot diagram of ${\varphi}_{2}(t)$ we have ${p}_{1}=2$ and ${p}_{2}=1$. These diagrams are as follows:
In practice, we obtain the rational canonical form of a linear operator from the information provided by dot diagrams. This is illustrated in the next example.
Example 3
Let T be a linear operator on a finitedimensional vector space over R, and suppose that the irreducible monic divisors of the characteristic polynomial of T are
Suppose, furthermore, that the dot diagrams associated with these divisors are as follows:
Since the dot diagram for ${\varphi}_{1}(t)$ has two columns, it contributes two companion matrices to the rational canonical form. The first column has two dots, and therefore corresponds to the $2\times 2$ companion matrix of ${({\varphi}_{1}(t))}^{2}={(t1)}^{2}$. The second column, with only one dot, corresponds to the $1\times 1$ companion matrix of ${\varphi}_{1}(t)=t1$. These two companion matrices are given by
The dot diagram for ${\varphi}_{2}(t)={t}^{2}+2$ consists of two columns. each containing a single dot; hence this diagram contributes two copies of the $2\times 2$ companion matrix for ${\varphi}_{2}(t)$, namely,
The dot diagram for ${\varphi}_{3}(t)={t}^{2}+t+1$ consists of a single column with a single dot contributing the single $2\times 2$ companion matrix
Therefore the rational canonical form of T is the $9\times 9$ matrix
We return to the general problem of finding dot diagrams. As we did before, we fix a linear operator T on a finitedimensional vector space and an irreducible monic divisor $\varphi (t)$ of the characteristic polynomial of T. Let U denote the restriction of the linear operator $\varphi (\text{T})$ to ${\text{K}}_{\varphi}$. By Theorem 7.18(d), ${\text{U}}^{q}={\text{T}}_{0}$ for some positive integer q. Consequently, by Exercise 12 of Section 7.2, the characteristic polynomial of U is ${(1)}^{m}{t}^{m}$, where $m=\text{dim}({\text{K}}_{\varphi})$. Therefore ${\text{K}}_{\varphi}$ is the generalized eigenspace of U corresponding to $\lambda =0$, and U has a Jordan canonical form. The dot diagram associated with the Jordan canonical form of U gives us a key to understanding the dot diagram of T that is associated with $\varphi (t)$. We now relate the two diagrams.
Let $\beta $ be a rational canonical basis for T, and ${\beta}_{{v}_{1}},\text{}{\beta}_{{v}_{2}},\text{}\dots ,\text{}{\beta}_{{v}_{k}}$ be the Tcyclic bases of $\beta $ that are contained in ${\text{K}}_{\varphi}$. Consider one of these Tcyclic bases ${\beta}_{{v}_{j}}$, and suppose again that the Tannihilator of ${v}_{j}$ is ${(\varphi (t))}^{{p}_{j}}$. Then ${\beta}_{{v}_{j}}$ consists of $d{p}_{j}$ vectors in $\beta $. For $0\le i<d$, let ${\gamma}_{i}$ be the cycle of generalized eigenvectors of U corresponding to $\lambda =0$ with end vector ${\text{T}}^{i}({v}_{j})$, where ${\text{T}}^{0}({v}_{j})={b}_{j}$. Then
By Theorem 7.1 (p. 478), ${\gamma}_{i}$ is a linearly independent subset of ${\text{C}}_{{v}_{i}}$. Now let
Notice that ${\alpha}_{j}$ contains $d{p}_{j}$ vectors.
Lemma 1.
${\alpha}_{j}$ is an ordered basis for ${\text{C}}_{{v}_{j}}$.
Proof. The key to this proof is Theorem 7.4 (p. 480). Since ${\alpha}_{j}$ is the union of cycles of generalized eigenvectors of U corresponding to $\lambda =0$, it suffices to show that the set of initial vectors of these cycles
is linearly independent. Consider any linear combination of these vectors
where not all of the coefficients are zero. Let g(t) be the polynomial defined by $g(t)={a}_{0}+{a}_{1}t+\cdots +{a}_{d1}{t}^{d1}$. Then g(t) is a nonzero polynomial of degree less than d, and hence ${(\varphi (t))}^{{p}_{j}1}g(t)$ is a nonzero polynomial with degree less than $d{p}_{j}$. Since ${(\varphi (t))}^{{p}_{j}}$ is the Tannihilator of ${v}_{j}$, it follows that ${(\varphi (\text{T}))}^{{p}_{j}1}g(\text{T})({v}_{j})\ne 0$. Therefore the set of initial vectors is linearly independent. So by Theorem 7.4, ${\alpha}_{j}$ is linearly independent, and the ${\gamma}_{i}$’s are disjoint. Consequently, ${\alpha}_{j}$ consists of $d{p}_{j}$ linearly independent vectors in ${\text{C}}_{{v}_{j}}$, which has dimension $d{p}_{j}$. We conclude that ${\alpha}_{j}$ is a basis for ${\text{C}}_{{v}_{j}}$.
Thus we may replace ${\beta}_{{v}_{j}}$ by ${\alpha}_{j}$ as a basis for ${\text{C}}_{{v}_{j}}$. We do this for each j to obtain a subset $\alpha ={\alpha}_{1}\cup {\alpha}_{2}\cup \cdots \cup {\alpha}_{k}$ of ${\text{K}}_{\varphi}$.
Lemma 2.
$\alpha $ is a Jordan canonical basis for ${\text{K}}_{\varphi}$.
Proof. Since ${\beta}_{{v}_{1}}\cup {\beta}_{{v}_{2}}\cup \cdots \cup {\beta}_{{v}_{k}}$ is a basis for ${\text{K}}_{\varphi}$, and since $\text{span}({\alpha}_{i})=\text{span}({\beta}_{{v}_{i}})={\text{C}}_{{v}_{i}}$, Exercise 9 implies that $\alpha $ is a basis for ${\text{K}}_{\varphi}$. Because $\alpha $ is a union of cycles of generalized eigenvectors of U, we conclude that $\alpha $ is a Jordan canonical basis.
We are now in a position to relate the dot diagram of T corresponding to $\varphi (t)$ to the dot diagram of U, bearing in mind that in the first case we are considering a rational canonical form and in the second case we are considering a Jordan canonical form. For convenience, we designate the first diagram ${D}_{1}$, and the second diagram ${D}_{2}$. For each j, the presence of the Tcyclic basis ${\beta}_{{x}_{j}}$ results in a column of ${p}_{j}$ dots in ${D}_{1}$. By Lemma 1, this basis is replaced by the union ${\alpha}_{j}$ of d cycles of generalized eigenvectors of U, each of length ${p}_{j}$, which becomes part of the Jordan canonical basis for U. In effect, ${\alpha}_{j}$ determines d columns each containing ${p}_{j}$ dots in ${D}_{2}$. So each column in ${D}_{1}$ determines d columns in ${D}_{2}$ of the same length, and all columns in ${D}_{2}$ are obtained in this way. Alternatively, each row in ${D}_{2}$ has d times as many dots as the corresponding row in ${D}_{1}$. Since Theorem 7.10 (p. 493) gives us the number of dots in any row of ${D}_{2}$, we may divide the appropriate expression in this theorem by d to obtain the number of dots in the corresponding row of ${D}_{1}$. Thus we have the following result.
Theorem 7.24.
Let T be a linear operator on a finitedimensional vector space V, let $\varphi (t)$ be an irreducible monic divisor of the characteristic polynomial of T of degree d, and let ${r}_{i}$ denote the number of dots in the ith row of the dot diagram for $\varphi (t)$ with respect to a rational canonical basis for T. Then

(a) ${r}_{1}={\displaystyle \frac{1}{d}}[\text{dim}(\text{V})\text{rank}(\varphi (\text{T}))]$;

(b) ${r}_{i}={\displaystyle \frac{1}{d}}[\text{rank}({(\varphi (\text{T}))}^{i1})\text{rank}({(\varphi (\text{T}))}^{i})]$ for $i>1$.
Thus the dot diagrams associated with a rational canonical form of an operator are completely determined by the operator. Since the rational canonical form is completely determined by its dot diagrams, we have the following uniqueness condition.
Corollary.
Under the conventions described earlier, the rational canonical form of a linear operator is unique up to the arrangement of the irreducible monic divisors of the characteristic polynomial.
Since the rational canonical form of a linear operator is unique, the polynomials corresponding to the companion matrices that determine this form are also unique. These polynomials, which are powers of the irreducible monic divisors, are called the elementary divisors of the linear operator. Since a companion matrix may occur more than once in a rational canonical form, the same is true for the elementary divisors. We call the number of such occurrences the multiplicity of the elementary divisor.
Conversely, the elementary divisors and their multiplicities determine the companion matrices and, therefore, the rational canonical form of a linear operator.
Example 4
Let
be viewed as a subset of $\mathcal{F}(R,\text{}R)$, the space of all realvalued functions defined on R, and let $\text{V}=\text{span}(\beta )$. Then V is a fourdimensional subspace of $\mathcal{F}(R,\text{}R)$, and $\beta $ is an ordered basis for V. Let D be the linear operator on V defined by $\text{D}(y)={y}^{\prime}$, the derivative of y, and let $A={[\text{D}]}_{\beta}$. Then
and the characteristic polynomial of D, and hence of A, is
Thus $\varphi (t)={t}^{2}2t+5$ is the only irreducible monic divisor of f(t). Since $\varphi (t)$ has degree 2 and V is fourdimensional, the dot diagram for $\varphi (t)$ contains only two dots. Therefore the dot diagram is determined by ${r}_{1}$, the number of dots in the first row. Because ranks are preserved under matrix representations, we can use A in place of D in the formula given in Theorem 7.24. Now
and so
It follows that the second dot lies in the second row, and the dot diagram is as follows:
Hence V is a Dcyclic space generated by a single function with Dannihilator ${(\varphi (t))}^{2}$. Furthermore, its rational canonical form is given by the companion matrix of ${(\varphi (t))}^{2}={t}^{4}4{t}^{3}+14{t}^{2}20t+25$, which is
Thus ${(\varphi (t))}^{2}$ is the only elementary divisor of D, and it has multiplicity 1. For the cyclic generator, it suffices to find a function g in V for which $\varphi (\text{D})(g)\ne 0$. Since $\varphi (A)({e}_{3})\ne 0$, it follows that $\varphi (\text{D})(x{e}^{x}\mathrm{cos}\text{}2x)\ne 0$; therefore $g(x)=x{e}^{x}\mathrm{cos}\text{}2x$ can be chosen as the cyclic generator. Hence
is a rational canonical basis for D. Notice that the function h defined by $h(x)=x{e}^{x}\mathrm{sin}2x$ can be chosen in place of g. This shows that the rational canonical basis is not unique.
It is convenient to refer to the rational canonical form and elementary divisors of a matrix, which are defined in the obvious way.
Definitions.
Let $A\in {\text{M}}_{n\times n}(F)$. The rational canonical form of A is defined to be the rational canonical form of ${\text{L}}_{A}$. Likewise, for A, the elementary divisors and their multiplicities are the same as those of ${\text{L}}_{A}$.
Let A be an $n\times n$ matrix, let C be a rational canonical form of A, and let $\beta $ be the appropriate rational canonical basis for ${\text{L}}_{A}$. Then $C={[{\text{L}}_{A}]}_{\beta}$, and therefore A is similar to C. In fact, if Q is the matrix whose columns are the vectors of $\beta $ in the same order, then ${Q}^{1}AQ=C$.
Example 5
For the following real matrix A, we find the rational canonical form C of A and a matrix Q such that ${Q}^{1}AQ=C$.
The characteristic polynomial of A is $f(t)={({t}^{2}+2)}^{2}(t2)$; therefore ${\varphi}_{1}(t)={t}^{2}+2$ and ${\varphi}_{2}(t)=t2$ are the distinct irreducible monic divisors of f(t). By Theorem 7.23, $\text{dim}({\text{K}}_{{\varphi}_{1}})=4$ and $\text{dim}({\text{K}}_{{\varphi}_{2}})=1$. Since the degree of ${\varphi}_{1}(t)$ is 2, the total number of dots in the dot diagram of ${\varphi}_{1}(t)$ is $4/2=2$, and the number of dots ${r}_{1}$ in the first row is given by
Thus the dot diagram of ${\varphi}_{1}(t)$ is
and each column contributes the companion matrix
for ${\varphi}_{1}(t)={t}^{2}+2$ to the rational canonical form C. Consequently ${\varphi}_{1}(t)$ is an elementary divisor with multiplicity 2. Since $\text{dim}({\text{K}}_{{\varphi}_{2}})=1$, the dot diagram of ${\varphi}_{2}(t)=t2$ consists of a single dot, which contributes the $1\times 1$ matrix (2). Hence ${\varphi}_{2}(t)$ is an elementary divisor with multiplicity 1. Therefore the rational canonical form C is
We can infer from the dot diagram of ${\varphi}_{1}(t)$ that if $\beta $ is a rational canonical basis for ${\text{L}}_{A}$, then $\beta \cap {\text{K}}_{{\varphi}_{1}}$ is the union of two cyclic bases ${\beta}_{{v}_{1}}$ and ${\beta}_{{v}_{2}}$, where ${v}_{1}$ and ${v}_{2}$ each have annihilator ${\varphi}_{1}(t)$. It follows that both ${v}_{1}$ and ${v}_{2}$ lie in $\text{N}({\varphi}_{1}({\text{L}}_{A}))$. It can be shown that
is a basis for $\text{N}({\varphi}_{1}({\text{L}}_{A}))$. Setting ${v}_{1}={e}_{1}$, we see that
Next choose ${v}_{2}$ in ${\text{K}}_{{\varphi}_{1}}=\text{N}(\varphi ({\text{L}}_{A}))$, but not in the span of ${\beta}_{{v}_{1}}=\{{v}_{1},\text{}A{v}_{1}\}$. For example, ${v}_{2}={e}_{2}$. Then it can be seen that
and ${\beta}_{{v}_{1}}\cup {\beta}_{{v}_{2}}$ is a basis for ${\text{K}}_{{\varphi}_{1}}$.
Since the dot diagram of ${\varphi}_{2}(t)=t2$ consists of a single dot, any nonzero vector in ${\text{K}}_{{\varphi}_{2}}$ is an eigenvector of A corresponding to the eigenvalue $\lambda =2$. For example, choose
By Theorem 7.23, $\beta =\{{v}_{1},\text{}A{v}_{1},\text{}{v}_{2},\text{}A{v}_{2},\text{}{v}_{3}\}$ is a rational canonical basis for ${\text{L}}_{A}$. So setting
we have ${Q}^{1}AQ=C$.
Example 6
For the following matrix A, we find the rational canonical form C and a matrix Q such that ${Q}^{1}AQ=C$.
Since the characteristic polynomial of A is $f(t)={(t2)}^{4}$, the only irreducible monic divisor of f(t) is $\varphi (t)=t2$, and so ${\text{K}}_{\varphi}={\text{R}}^{4}$. In this case, $\varphi (t)$ has degree 1; hence in applying Theorem 7.24 to compute the dot diagram for $\varphi (t)$, we obtain
and
where ${r}_{i}$ is the number of dots in the ith row of the dot diagram. Since there are $\text{dim}({\text{R}}^{4})=4$ dots in the diagram, we may terminate these computations with ${r}_{3}$. Thus the dot diagram for A is
Since ${(t2)}^{3}$ has the companion matrix
and $(t2)$ has the companion matrix (2), the rational canonical form of A is given by
Next we find a rational canonical basis for ${\text{L}}_{A}$. The preceding dot diagram indicates that there are two vectors ${v}_{1}$ and ${v}_{2}$ in ${\text{R}}^{4}$ with annihilators ${(\varphi (t))}^{3}$ and $\varphi (t)$, respectively, and such that
is a rational canonical basis for ${\text{L}}_{A}$. Furthermore, ${v}_{1}\notin \text{N}({({\text{L}}_{A}2\text{I})}^{2})$, and ${v}_{2}\in \text{N}({\text{L}}_{A}2\text{I})$. It can easily be shown that
and
The standard vector ${e}_{3}$ meets the criteria for ${v}_{1}$; so we set ${v}_{1}={e}_{3}$. It follows that
Next we choose a vector ${v}_{2}\in \text{N}({\text{L}}_{A}2\text{I})$ that is not in the span of ${\beta}_{{v}_{1}}$. Clearly, ${v}_{2}={e}_{4}$ satisfies this condition. Thus
is a rational canonical basis for ${\text{L}}_{A}$.
Finally, let Q be the matrix whose columns are the vectors of $\beta $ in the same order:
Then $C={Q}^{1}AQ$.
Direct Sums*
The next theorem is a simple consequence of Theorem 7.23.
Theorem 7.25. (Primary Decomposition Theorem)
Let T be a linear operator on an ndimensional vector space V with characteristic polynomial
where the ${\varphi}_{i}(t)\text{'}s(1\le i\le k)$ are distinct irreducible monic polynomials and the ${n}_{i}$’s are positive integers. Then the following statements are true.

$\text{V}={\text{K}}_{{\varphi}_{1}}\oplus {\text{K}}_{{\varphi}_{2}}\oplus \cdots \oplus {\text{K}}_{{\varphi}_{k}}$.

If ${\text{T}}_{i}(1\le i\le k)$ is the restriction of T to ${\text{K}}_{{\varphi}_{i}}$ and ${C}_{i}$ is the rational canonical form of ${\text{T}}_{i}$, then ${C}_{1}\oplus {C}_{2}\oplus \cdots \oplus {C}_{k}$ is the rational canonical form of T.
Proof.
Exercise.
The next theorem is a simple consequence of Theorem 7.17.
Theorem 7.26.
Let T be a linear operator on a finitedimensional vector space V. Then V is a direct sum of Tcyclic subspaces ${\text{C}}_{{v}_{i}}$, where each ${v}_{i}$ lies in ${\text{K}}_{\varphi}$ for some irreducible monic divisor $\varphi (t)$ of the characteristic polynomial of T.
Proof.
Exercise.
Exercises

Label the following statements as true or false.

(a) Every rational canonical basis for a linear operator T is the union of Tcyclic bases.

(b) If a basis is the union of Tcyclic bases for a linear operator T, then it is a rational canonical basis for T.

(c) There exist square matrices having no rational canonical form.

(d) A square matrix is similar to its rational canonical form.

(e) For any linear operator T on a finitedimensional vector space, any irreducible factor of the characteristic polynomial of T divides the minimal polynomial of T.

(f) Let $\varphi (t)$ be an irreducible monic divisor of the characteristic polynomial of a linear operator T. The dots in the diagram used to compute the rational canonical form of the restriction of T to ${\text{K}}_{\varphi}$ are in onetoone correspondence with the vectors in a basis for ${\text{K}}_{\varphi}$.

(g) If a matrix has a Jordan canonical form, then its Jordan canonical form and rational canonical form are similar.


For each of the following matrices $A\in {\text{M}}_{n\times n}(F)$, find the rational canonical form C of A and a matrix $Q\in {\text{M}}_{n\times n}(F)$ such that ${Q}^{1}AQ=C$.

(a) $A=\left(\begin{array}{rrr}3& 1& 0\\ 0& 3& 1\\ 0& 0& 3\end{array}\right)\phantom{\rule{1em}{0ex}}F=R$

(b) $A=\left(\begin{array}{rr}0& 1\\ 1& 1\end{array}\right)\phantom{\rule{1em}{0ex}}F=R$

(c) $A=\left(\begin{array}{rr}0& 1\\ 1& 1\end{array}\right)\phantom{\rule{1em}{0ex}}F=C$

(d) $A=\left(\begin{array}{rrrr}0& 7& 14& 6\\ 1& 4& 6& 3\\ 0& 4& 9& 4\\ 0& 4& 11& 5\end{array}\right)\phantom{\rule{1em}{0ex}}F=R$

(e) $A=\left(\begin{array}{rrrr}0& 4& 12& 7\\ 1& 1& 3& 3\\ 0& 1& 6& 4\\ 0& 1& 8& 5\end{array}\right)\phantom{\rule{1em}{0ex}}F=R$


For each of the following linear operators T, find the elementary divisors, the rational canonical form C, and a rational canonical basis $\beta $.

(a) T is the linear operator on ${\text{P}}_{3}(R)$ defined by
$$\text{T}(f(x))=f(0)x{f}^{\prime}(1).$$ 
(b) Let $S=\{\mathrm{sin}x,\text{}\mathrm{cos}x,\text{}x\mathrm{sin}x,\text{}x\mathrm{cos}x\}$, a subset of $\mathcal{F}(R,\text{}R)$, and let $\text{V}=\text{span}(S)$. Define T to be the linear operator on V such that
$$\text{T}(f)={f}^{\prime}.$$ 
(c) T is the linear operator on ${\text{M}}_{2\times 2}\text{(}R\text{)}$ defined by
$$\text{T}(A)=\left(\begin{array}{rr}0& 1\\ 1& 1\end{array}\right)\xb7A.$$ 
(d) Let $S=\{\mathrm{sin}x\mathrm{sin}y,\text{}\mathrm{sin}x\mathrm{cos}y,\text{}\mathrm{cos}x\mathrm{sin}y,\text{}\mathrm{cos}x\mathrm{cos}y\}$, a subset of $\mathcal{F}(R\times R,\text{}R)$, and let $\text{V}=\text{span}(S)$. Define T to be the linear operator on V such that
$$\text{T}(f)(x,\text{}y)={\displaystyle \frac{\partial f(x,\text{}y)}{\partial x}}+{\displaystyle \frac{\partial f(x,\text{}y)}{\partial y}}.$$


Let T be a linear operator on a finitedimensional vector space V with minimal polynomial ${(\varphi (t))}^{m}$ for some positive integer m.

(a) Prove that $\text{R}(\varphi (\text{T}))\subseteq \text{N}({(\varphi (\text{T}))}^{m1})$.

(b) Give an example to show that the subspaces in (a) need not be equal.

(c) Prove that the minimal polynomial of the restriction of T to $\text{R}(\varphi (\text{T}))$ equals ${(\varphi (t))}^{m1}$.


Let T be a linear operator on a finitedimensional vector space. Prove that the rational canonical form of T is a diagonal matrix if and only if T is diagonalizable. Visit goo.gl/
tK8pru for a solution. 
Let T be a linear operator on a finitedimensional vector space V with characteristic polynomial $f(t)={(1)}^{n}{\varphi}_{1}(t){\varphi}_{2}(t)$, where ${\varphi}_{1}(t)$ and ${\varphi}_{2}(t)$ are distinct irreducible monic polynomials and $n=\text{dim}(\text{V})$.

(a) Prove that there exist ${v}_{1},\text{}{v}_{2}\in \text{V}$ such that ${v}_{1}$ has Tannihilator ${\varphi}_{1}(t),\text{}{v}_{2}$ has Tannihilator ${\varphi}_{2}(t)$, and ${\beta}_{{v}_{1}}\cup {\beta}_{{v}_{2}}$ is a basis for V.

(b) Prove that there is a vector ${v}_{3}\in \text{V}$ with Tannihilator ${\varphi}_{1}(t){\varphi}_{2}(t)$ such that ${\beta}_{{v}_{3}}$ is a basis for V.

(c) Describe the difference between the matrix representation of T with respect to ${\beta}_{{v}_{1}}\cup {\beta}_{{v}_{2}}$ and the matrix representation of T with respect to ${\beta}_{{v}_{3}}$.
Thus, to assure the uniqueness of the rational canonical form, we require that the generators of the Tcyclic bases that constitute a rational canonical basis have Tannihilators equal to powers of irreducible monic factors of the characteristic polynomial of T.


Let T be a linear operator on a finitedimensional vector space with minimal polynomial
$$f(t)={({\varphi}_{1}(t))}^{{m}_{1}}{({\varphi}_{2}(t))}^{{m}_{2}}\cdots {({\varphi}_{k}(t))}^{{m}_{k}},\text{}$$where the ${\varphi}_{i}(t)$’s are distinct irreducible monic factors of f(t). Prove that for each i, ${m}_{i}$ is the number of entries in the first column of the dot diagram for ${\varphi}_{i}(t)$.

Let T be a linear operator on a finitedimensional vector space V. Prove that for any irreducible polynomial $\varphi (t)$, if $\varphi (\text{T})$ is not onetoone, then $\varphi (t)$ divides the characteristic polynomial of T. Hint: Apply Exercise 15 of Section 7.3.

Let V be a vector space and ${\beta}_{1},\text{}{\beta}_{2},\text{}\dots ,\text{}{\beta}_{k}$ be disjoint subsets of V whose union is a basis for V. Now suppose that ${\gamma}_{1},\text{}{\gamma}_{2},\text{}\dots ,\text{}{\gamma}_{k}$ are linearly independent subsets of V such that $\text{span}({\gamma}_{i})=\text{span}({\beta}_{i})$ for all i. Prove that ${\gamma}_{1}\cup {\gamma}_{2}\cup \cdots \cup {\gamma}_{k}$ is also a basis for V.

Let T be a linear operator on a finitedimensional vector space, and suppose that $\varphi (t)$ is an irreducible monic factor of the characteristic polynomial of T. Prove that if $\varphi (t)$ is the Tannihilator of vectors x and y, then $x\in {\text{C}}_{y}$ if and only if ${\text{C}}_{x}={\text{C}}_{y}$.
Exercises 11 and 12 are concerned with direct sums.

Prove Theorem 7.25.

Prove Theorem 7.26.