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7.3 The Minimal Polynomial

The Cayley-Hamilton theorem (Theorem 5.22 p. 315) tells us that for any linear operator T on an n-dimensional vector space, there is a polynomial f(t) of degree n such that $f (T) = T_{0}$ $f (T) = T_{0}$ , namely, the characteristic polynomial of T. Hence there is a polynomial of least degree with this property, and this degree is at most n. If g(t) is such a polynomial, we can divide g(t) by its leading coefficient to obtain another polynomial p(t) of the same degree with leading coefficient 1, that is, p(t) is a monic polynomial. (See Appendix E.)

Definition.

Let T be a linear operator on a finite-dimensional vector space. A polynomial p(t) is called a minimal polynomial of T if p(t) is a monic polynomial of least positive degree for which $p (T) = T_{0}$ $p (T) = T_{0}$ .

The preceding discussion shows that every linear operator on a finite-dimensional vector space has a minimal polynomial. The next result shows that it is unique, and hence we can speak of the minimal polynomial of T.

Theorem 7.12.

Let p(t) be a minimal polynomial of a linear operator T on a finite-dimensional vector space V.

(a) For any polynomial g(t), if $g (T) = T_{0}$ $g (T) = T_{0}$ , then p(t) divides g(t). In particular, p(t) divides the characteristic polynomial of T.
(b) The minimal polynomial of T is unique.

Proof.

(a) Let g(t) be a polynomial for which $g (T) = T_{0}$ $g (T) = T_{0}$ . By the division algorithm for polynomials (Theorem E.1 of Appendix E, p. 555), there exist polynomials q(t) and r(t) such that

g (t) = q (t) p (t) + r (t), (1)

$g (t) = q (t) p (t) + r (t), (1)$

where r(t) has degree less than the degree of p(t). Substituting T into (1) and using that $g (T) = p (T) = T_{0}$ $g (T) = p (T) = T_{0}$ , we have $r (T) = T_{0}$ $r (T) = T_{0}$ . Since r(t) has degree less than p(t) and p(t) is the minimal polynomial of T, r(t) must be the zero polynomial. Thus (1) simplifies to $g (t) = q (t) p (t)$ $g (t) = q (t) p (t)$ , proving (a).

(b) Suppose that $p_{1} (t)$ $p_{1} (t)$ and $p_{2} (t)$ $p_{2} (t)$ are each minimal polynomials of T. Then $p_{1} (t)$ $p_{1} (t)$ divides $p_{2} (t)$ $p_{2} (t)$ by (a). Since $p_{1} (t)$ $p_{1} (t)$ and $p_{2} (t)$ $p_{2} (t)$ have the same degree, we have that $p_{2} (t) = c p_{1} (t)$ $p_{2} (t) = c p_{1} (t)$ for some nonzero scalar c. Because $p_{1} (t)$ $p_{1} (t)$ and $p_{2} (t)$ $p_{2} (t)$ are monic, $c = 1$ $c = 1$ ; hence $p_{1} (t) = p_{2} (t)$ $p_{1} (t) = p_{2} (t)$ .

The minimal polynomial of a linear operator has an obvious analog for a matrix.

Definition.

Let $A \in M_{n \times n} (F)$ $A \in M_{n \times n} (F)$ . The minimal polynomial p(t) of A is the monic polynomial of least positive degree for which $p (A) = O$ $p (A) = O$ .

The following results are now immediate.

Theorem 7.13.

Let T be a linear operator on a finite-dimensional vector space V, and let $β$ $β$ be an ordered basis for V. Then the minimal polynomial of T is the same as the minimal polynomial of ${[T]}_{β}$ ${[T]}_{β}$ .

Proof.

Exercise.

Corollary.

For any $A \in M_{n \times n} (F)$ $A \in M_{n \times n} (F)$ , the minimal polynomial of A is the same as the minimal polynomial of $L_{A}$ $L_{A}$ .

Proof.

Exercise.

In view of the preceding theorem and corollary, Theorem 7.12 and all subsequent theorems in this section that are stated for operators are also valid for matrices.

For the remainder of this section, we study primarily minimal polynomials of operators (and hence matrices) whose characteristic polynomials split. A more general treatment of minimal polynomials is given in Section 7.4.

Theorem 7.14.

Let T be a linear operator on a finite-dimensional vector space V, and let p(t) be the minimal polynomial of T. A scalar $λ$ $λ$ is an eigenvalue of T if and only if $p (λ) = 0$ $p (λ) = 0$ . Hence the characteristic polynomial and the minimal polynomial of T have the same zeros.

Proof.

Let f(t) be the characteristic polynomial of T. Since p(t) divides f(t), there exists a polynomial q(t) such that $f (t) = q (t) p (t)$ $f (t) = q (t) p (t)$ . If $λ$ $λ$ is a zero of p(t), then

f (λ) = q (λ) p (λ) = q (λ) \cdot 0 = 0.

$f (λ) = q (λ) p (λ) = q (λ) \cdot 0 = 0.$

So $λ$ $λ$ is a zero of f(t); that is, $λ$ $λ$ is an eigenvalue of T.

Conversely, suppose that $λ$ $λ$ is an eigenvalue of T, and let $x \in V$ $x \in V$ be an eigenvector corresponding to $λ$ $λ$ . By Exercise 22 of Section 5.1, we have

0 = T_{0} (x) = p (T) (x) = p (λ) x .

$0 = T_{0} (x) = p (T) (x) = p (λ) x .$

Since $x \neq 0$ $x \neq 0$ , it follows that $p (λ) = 0$ $p (λ) = 0$ , and so $λ$ $λ$ is a zero of p(t).

The following corollary is immediate.

Corollary.

Let T be a linear operator on a finite-dimensional vector space V with minimal polynomial p(t) and characteristic polynomial f(t). Suppose that f(t) factors as

f (t) = {(λ_{1} - t)}^{n_{1}} {(λ_{2} - t)}^{n_{2}} \dots {(λ_{k} - t)}^{n_{k}},

$f (t) = {(λ_{1} - t)}^{n_{1}} {(λ_{2} - t)}^{n_{2}} \dots {(λ_{k} - t)}^{n_{k}},$

where $λ_{1}, λ_{2}, \dots, λ_{k}$ $λ_{1}, λ_{2}, \dots, λ_{k}$ are the distinct eigenvalues of T. Then there exist integers $m_{1}, m_{2}, \dots, m_{k}$ $m_{1}, m_{2}, \dots, m_{k}$ such that $1 \leq m_{i} \leq n_{i}$ $1 \leq m_{i} \leq n_{i}$ for all i and

p (t) = {(t - λ_{1})}^{m_{1}} {(t - λ_{2})}^{m_{2}} \dots {(t - λ_{k})}^{m_{k}} .

$p (t) = {(t - λ_{1})}^{m_{1}} {(t - λ_{2})}^{m_{2}} \dots {(t - λ_{k})}^{m_{k}} .$

Example 1

We compute the minimal polynomial of the matrix

A = (\begin{array}{r} 3 & - 1 & 0 \\ 0 & 2 & 0 \\ 1 & - 1 & 2 \end{array}) .

$A = (\begin{array}{r} 3 & - 1 & 0 \\ 0 & 2 & 0 \\ 1 & - 1 & 2 \end{array}) .$

Since A has the characteristic polynomial

f (t) = \det (\begin{array}{c} 3 - t & - 1 & 0 \\ 0 & 2 - t & 0 \\ 1 & - 1 & 2 - t \end{array}) = - {(t - 2)}^{2} (t - 3),

$f (t) = \det (\begin{array}{c} 3 - t & - 1 & 0 \\ 0 & 2 - t & 0 \\ 1 & - 1 & 2 - t \end{array}) = - {(t - 2)}^{2} (t - 3),$

the minimal polynomial of A must be either $(t - 2) (t - 3)$ $(t - 2) (t - 3)$ or ${(t - 2)}^{2} (t - 3)$ ${(t - 2)}^{2} (t - 3)$ by the corollary to Theorem 7.14. Substituting A into $p (t) = (t - 2) (t - 3)$ $p (t) = (t - 2) (t - 3)$ , we find that $p (A) = O$ $p (A) = O$ ; hence p(t) is the minimal polynomial of A.

Example 2

Let T be the linear operator on $R^{2}$ $R^{2}$ defined by

T (a, b) = (2 a + 5 b, 6 a + b)

$T (a, b) = (2 a + 5 b, 6 a + b)$

and $β$ $β$ be the standard ordered basis for $R^{2}$ $R^{2}$ . Then

{[T]}_{β} = (\begin{array}{c} 2 & 5 \\ 6 & 1 \end{array}),

${[T]}_{β} = (\begin{array}{c} 2 & 5 \\ 6 & 1 \end{array}),$

and hence the characteristic polynomial of T is

f (t) = \det (\begin{matrix} 2 - t & 5 \\ 6 & 1 - t \end{matrix}) = (t - 7) (t + 4) .

$f (t) = \det (\begin{matrix} 2 - t & 5 \\ 6 & 1 - t \end{matrix}) = (t - 7) (t + 4) .$

Thus the minimal polynomial of T is also $(t - 7) (t + 4)$ $(t - 7) (t + 4)$ .

Example 3

Let D be the linear operator on $P_{2} (R)$ $P_{2} (R)$ defined by $D (g (x)) = g^{'} (x)$ $D (g (x)) = g^{'} (x)$ , the derivative of g(x). We compute the minimal polynomial of T. Let $β$ $β$ be the standard ordered basis for $P_{2} (R)$ $P_{2} (R)$ . Then

{[D]}_{β} = (\begin{array}{r} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{array}),

${[D]}_{β} = (\begin{array}{r} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{array}),$

and it follows that the characteristic polynomial of D is $- t^{3}$ $- t^{3}$ . So by the corollary to Theorem 7.14, the minimal polynomial of D is $t, t^{2}$ $t, t^{2}$ , or $t^{3}$ $t^{3}$ . Since $D^{2} (x^{2}) = 2 \neq 0$ $D^{2} (x^{2}) = 2 \neq 0$ , it follows that $D^{2} \neq T_{0}$ $D^{2} \neq T_{0}$ ; hence the minimal polynomial of D must be $t^{3}$ $t^{3}$ .

In Example 3, it is easily verified that $P_{2} (R)$ $P_{2} (R)$ is a D-cyclic subspace (of itself). Here the minimal and characteristic polynomials are of the same degree. This is no coincidence.

Theorem 7.15.

Let T be a linear operator on an n-dimensional vector space V such that V is a T-cyclic subspace of itself. Then the characteristic polynomial f(t) and the minimal polynomial p(t) have the same degree, and hence $f (t) = {(- 1)}^{n} p (t)$ $f (t) = {(- 1)}^{n} p (t)$ .

Proof.

Since V is a T-cyclic space, there exists an $x \in V$ $x \in V$ such that

β = {x, T (x), \dots, T^{n - 1} (x)}

$β = {x, T (x), \dots, T^{n - 1} (x)}$

is a basis for V (Theorem 5.21 p. 314). Let

g (t) = a_{0} + a_{1} t + \dots + a_{k} t^{k},

$g (t) = a_{0} + a_{1} t + \dots + a_{k} t^{k},$

be a polynomial of degree $k < n$ $k < n$ . Then $a_{k} \neq 0$ $a_{k} \neq 0$ and

g (T) (x) = a_{0} x + a_{1} T (x) + \dots + a_{k} T^{k} (x),

$g (T) (x) = a_{0} x + a_{1} T (x) + \dots + a_{k} T^{k} (x),$

and so g(T)(x) is a linear combination of the vectors of $β$ $β$ having at least one nonzero coefficient, namely, $a_{k}$ $a_{k}$ . Since $β$ $β$ is linearly independent, it follows that $g (T) (x) \neq 0$ $g (T) (x) \neq 0$ ; hence $g (T) \neq T_{0}$ $g (T) \neq T_{0}$ . Therefore the minimal polynomial of T has degree n, which is also the degree of the characteristic polynomial of T.

Theorem 7.15 gives a condition under which the degree of the minimal polynomial of an operator is as large as possible. We now investigate the other extreme. By Theorem 7.14, the degree of the minimal polynomial of an operator must be greater than or equal to the number of distinct eigenvalues of the operator. The next result shows that the operators for which the degree of the minimal polynomial is as small as possible are precisely the diagonalizable operators.

Theorem 7.16.

Let T be a linear operator on a finite-dimensional vector space V. Then T is diagonalizable if and only if the minimal polynomial of T is of the form

p (t) = (t - λ_{1}) (t - λ_{2}) \dots (t - λ_{k}),

$p (t) = (t - λ_{1}) (t - λ_{2}) \dots (t - λ_{k}),$

where $λ_{1}, λ_{2}, \dots, λ_{k}$ $λ_{1}, λ_{2}, \dots, λ_{k}$ are the distinct eigenvalues of T.

Proof.

Suppose that T is diagonalizable. Let $λ_{1}, λ_{2}, \dots, λ_{k}$ $λ_{1}, λ_{2}, \dots, λ_{k}$ be the distinct eigenvalues of T, and define

p (t) = (t - λ_{1}) (t - λ_{2}) \dots (t - λ_{k}) .

$p (t) = (t - λ_{1}) (t - λ_{2}) \dots (t - λ_{k}) .$

By Theorem 7.14, p(t) divides the minimal polynomial of T. Let $β = {v_{1}, v_{2}, \dots, v_{n}}$ $β = {v_{1}, v_{2}, \dots, v_{n}}$ be a basis for V consisting of eigenvectors of T, and consider any $v_{i} \in β$ $v_{i} \in β$ . Then $(T - λ_{j} I) (v_{i}) = 0$ $(T - λ_{j} I) (v_{i}) = 0$ for some eigenvalue $λ_{j}$ $λ_{j}$ . Since $t - λ_{j}$ $t - λ_{j}$ divides p(t), there is a polynomial $q_{j} (t)$ $q_{j} (t)$ such that $p (t) = q_{j} (t) (t - λ_{j})$ $p (t) = q_{j} (t) (t - λ_{j})$ . Hence

p (T) (v_{i}) = q_{j} (T) (T - λ_{j} I) (v_{i}) = 0.

$p (T) (v_{i}) = q_{j} (T) (T - λ_{j} I) (v_{i}) = 0.$

It follows that $p (T) = T_{0}$ $p (T) = T_{0}$ , since p(T) takes each vector in a basis for V into 0. Therefore p(t) is the minimal polynomial of T.

Conversely, suppose that there are distinct scalars $λ_{1}, λ_{2}, \dots, λ_{k}$ $λ_{1}, λ_{2}, \dots, λ_{k}$ such that the minimal polynomial p(t) of T factors as

p (t) = (t - λ_{1}) (t - λ_{2}) \dots (t - λ_{k}) .

$p (t) = (t - λ_{1}) (t - λ_{2}) \dots (t - λ_{k}) .$

By Theorem 7.14, the $λ_{i}$ $λ_{i}$ ’s are eigenvalues of T. We apply mathematical induction on $n = \dim (V)$ $n = \dim (V)$ . Clearly T is diagonalizable for $n = 1$ $n = 1$ . Now assume that T is diagonalizable whenever $\dim (V) < n$ $\dim (V) < n$ for some $n > 1$ $n > 1$ , and let $\dim (V) = n$ $\dim (V) = n$ and $W = R (T - λ_{k} I)$ $W = R (T - λ_{k} I)$ . Obviously $W \neq V$ $W \neq V$ , because $λ_{k}$ $λ_{k}$ is an eigenvalue of T. If $W = {0}$ $W = {0}$ , then $T = λ_{k} I$ $T = λ_{k} I$ , which is clearly diagonalizable. So suppose that $0 < \dim (W) < n$ $0 < \dim (W) < n$ . Then W is T-invariant, and for any $x \in W$ $x \in W$ ,

(T - λ_{1} I) (T - λ_{2} I) \dots (T - λ_{k - 1} I) (x) = 0.

$(T - λ_{1} I) (T - λ_{2} I) \dots (T - λ_{k - 1} I) (x) = 0.$

It follows that the minimal polynomial of $T_{W}$ $T_{W}$ divides the polynomial $(t - λ_{1}) (t - λ_{2}) \dots (t - λ_{k - 1})$ $(t - λ_{1}) (t - λ_{2}) \dots (t - λ_{k - 1})$ . Hence by the induction hypothesis, $T_{W}$ $T_{W}$ is diagonalizable. Furthermore, $λ_{k}$ $λ_{k}$ is not an eigenvalue of $T_{W}$ $T_{W}$ by Theorem 7.14. Therefore $W \cap N (T - λ_{k} I) = {0}$ $W \cap N (T - λ_{k} I) = {0}$ . Now let $β_{1} = {v_{1}, v_{2}, \dots, v_{m}}$ $β_{1} = {v_{1}, v_{2}, \dots, v_{m}}$ be a basis for W consisting of eigenvectors of Tw (and hence of T), and let $β_{2} = {w_{1}, w_{2}, \dots, w_{p}}$ $β_{2} = {w_{1}, w_{2}, \dots, w_{p}}$ be a basis for $N (T - λ_{k} I)$ $N (T - λ_{k} I)$ , the eigenspace of T corresponding to $λ_{k}$ $λ_{k}$ . Then $β_{1}$ $β_{1}$ and $β_{2}$ $β_{2}$ are disjoint by the previous comment. Moreover, $m + p = n$ $m + p = n$ by the dimension theorem applied to $T - λ_{k} I$ $T - λ_{k} I$ . We show that $β = β_{1} \cup β_{2}$ $β = β_{1} \cup β_{2}$ is linearly independent. Consider scalars $a_{1}, a_{2}, \dots, a_{m}$ $a_{1}, a_{2}, \dots, a_{m}$ and $b_{1}, b_{2}, \dots, b_{p}$ $b_{1}, b_{2}, \dots, b_{p}$ such that

a_{1} v_{1} + a_{2} v_{2} + \dots + a_{m} v_{m} + b_{1} w_{1} + b_{2} w_{2} + \dots + b_{p} w_{p} = 0.

$a_{1} v_{1} + a_{2} v_{2} + \dots + a_{m} v_{m} + b_{1} w_{1} + b_{2} w_{2} + \dots + b_{p} w_{p} = 0.$

Let

x = \sum_{i = 1}^{m} a_{i} v_{i} and y = \sum_{i = 1}^{p} b_{i} w_{i} .

$x = \sum_{i = 1}^{m} a_{i} v_{i} and y = \sum_{i = 1}^{p} b_{i} w_{i} .$

Then $x \in W, y \in N (T - λ_{k} I)$ $x \in W, y \in N (T - λ_{k} I)$ , and $x + y = 0$ $x + y = 0$ . It follows that $x = - y \in W \cap N (T - λ_{k} I)$ $x = - y \in W \cap N (T - λ_{k} I)$ , and therefore $x = 0$ $x = 0$ . Since $β_{1}$ $β_{1}$ is linearly independent, we have that $a_{1} = a_{2} = \dots = a_{m} = 0$ $a_{1} = a_{2} = \dots = a_{m} = 0$ . Similarly, $b_{1} = b_{2} = \dots = b_{p} = 0$ $b_{1} = b_{2} = \dots = b_{p} = 0$ , and we conclude that $β$ $β$ is a linearly independent subset of V consisting of n eigenvectors. It follows that $β$ $β$ is a basis for V consisting of eigenvectors of T, and consequently T is diagonalizable.

In addition to diagonalizable operators, there are methods for determining the minimal polynomial of any linear operator on a finite-dimensional vector space. In the case that the characteristic polynomial of the operator splits, the minimal polynomial can be described using the Jordan canonical form of the operator. (See Exercise 13.) In the case that the characteristic polynomial does not split, the minimal polynomial can be described using the rational canonical form, which we study in the next section. (See Exercise 7 of Section 7.4.)

Example 4

We determine all matrices $A \in M_{2 \times 2} (R)$ $A \in M_{2 \times 2} (R)$ for which $A^{2} - 3 A + 2 I = O$ $A^{2} - 3 A + 2 I = O$ . Let $g (t) = t^{2} - 3 t + 2 = (t - 1) (t - 2)$ $g (t) = t^{2} - 3 t + 2 = (t - 1) (t - 2)$ . Since $g (A) = O$ $g (A) = O$ , the minimal polynomial p(t) of A divides g(t). Hence the only possible candidates for p(t) are $t - 1, t - 2$ $t - 1, t - 2$ , and $(t - 1) (t - 2)$ $(t - 1) (t - 2)$ . If $p (t) = t - 1$ $p (t) = t - 1$ or $p (t) = t - 2$ $p (t) = t - 2$ , then $A = I$ $A = I$ or $A = 2 I$ $A = 2 I$ , respectively. If $p (t) = (t - 1) (t - 2)$ $p (t) = (t - 1) (t - 2)$ , then A is diagonalizable with eigenvalues 1 and 2, and hence A is similar to

D = (\begin{array}{c} 1 & 0 \\ 0 & 2 \end{array}),

$D = (\begin{array}{c} 1 & 0 \\ 0 & 2 \end{array}),$

that is, $A = Q D Q^{- 1}$ $A = Q D Q^{- 1}$ for some invertible matrix Q.

Example 5

Let $A \in M_{n \times n} (R)$ $A \in M_{n \times n} (R)$ satisfy $A^{3} = A$ $A^{3} = A$ . We show that A is diagonalizable. Let $g (t) = t^{3} - t = t (t + 1) (t - 1)$ $g (t) = t^{3} - t = t (t + 1) (t - 1)$ . Then $g (A) = O$ $g (A) = O$ , and hence the minimal polynomial p(t) of A divides g(t). Since g(t) has no repeated factors, neither does p(t). Thus A is diagonalizable by Theorem 7.16.

Example 6

In Example 3, we saw that the minimal polynomial of the differential operator D on $P_{2} (R)$ $P_{2} (R)$ is $t^{3}$ $t^{3}$ . Hence, by Theorem 7.16, D is not diagonalizable.

Exercises

Label the following statements as true or false. Assume that all vector spaces are finite-dimensional.
1. (a) Every linear operator T has a polynomial p(t) of largest degree for which $p (T) = T_{0}$ $p (T) = T_{0}$ .
2. (b) Every linear operator has a unique minimal polynomial.
3. (c) The characteristic polynomial of a linear operator divides the minimal polynomial of that operator.
4. (d) The minimal and the characteristic polynomials of any diagonalizable operator are equal.
5. (e) Let T be a linear operator on an n-dimensional vector space V, p(t) be the minimal polynomial of T, and f(t) be the characteristic polynomial of T. Suppose that f(t) splits. Then f(t) divides ${[p (t)]}^{n}$ ${[p (t)]}^{n}$ .
6. (f) The minimal polynomial of a linear operator always has the same degree as the characteristic polynomial of the operator.
7. (g) A linear operator is diagonalizable if its minimal polynomial splits.
8. (h) Let T be a linear operator on a vector space V such that V is a T-cyclic subspace of itself. Then the degree of the minimal polynomial of T equals dim(V).
9. (i) Let T be a linear operator on a vector space V such that T has n distinct eigenvalues, where $n = \dim (V)$ $n = \dim (V)$ . Then the degree of the minimal polynomial of T equals n.
Find the minimal polynomial of each of the following matrices.
1. (a) $(\begin{array}{r} 2 & 1 \\ 1 & 2 \end{array})$ $(\begin{array}{r} 2 & 1 \\ 1 & 2 \end{array})$
2. (b) $(\begin{array}{r} 1 & 1 \\ 0 & 1 \end{array})$ $(\begin{array}{r} 1 & 1 \\ 0 & 1 \end{array})$
3. (c) $(\begin{array}{r} 4 & - 14 & 5 \\ 1 & - 4 & 2 \\ 1 & - 6 & 4 \end{array})$ $(\begin{array}{r} 4 & - 14 & 5 \\ 1 & - 4 & 2 \\ 1 & - 6 & 4 \end{array})$
4. (d) $(\begin{array}{r} 3 & 0 & 1 \\ 2 & 2 & 2 \\ - 1 & 0 & 1 \end{array})$ $(\begin{array}{r} 3 & 0 & 1 \\ 2 & 2 & 2 \\ - 1 & 0 & 1 \end{array})$
For each linear operator T on V, find the minimal polynomial of T.
1. (a) $V = R^{2}$ $V = R^{2}$ and $T (a, b) = (a + b, a - b)$ $T (a, b) = (a + b, a - b)$
2. (b) $V = P_{2} (R)$ $V = P_{2} (R)$ and $T (g (x)) = g^{'} (x) + 2 g (x)$ $T (g (x)) = g^{'} (x) + 2 g (x)$
3. (c) $V = P_{2} (R)$ $V = P_{2} (R)$ and $T (f (x)) = - x f^{″} (x) + f^{'} (x) + 2 f (x)$ $T (f (x)) = - x f^{″} (x) + f^{'} (x) + 2 f (x)$
4. (d) $V = M_{n \times n} (R)$ $V = M_{n \times n} (R)$ and $T (A) = A^{t}$ $T (A) = A^{t}$ Hint: Note that $T^{2} = I$ $T^{2} = I$ .
Determine which of the matrices and operators in Exercises 2 and 3 are diagonalizable.
Describe all linear operators T on $R^{2}$ $R^{2}$ such that T is diagonalizable and $T^{3} - 2 T^{2} + T = T_{0}$ $T^{3} - 2 T^{2} + T = T_{0}$ .
Prove Theorem 7.13 and its corollary.
Prove the corollary to Theorem 7.14.
Let T be a linear operator on a finite-dimensional vector space, and let p(t) be the minimal polynomial of T. Prove the following results.
1. (a) T is invertible if and only if $p (0) \neq 0$ $p (0) \neq 0$ .
2. (b) If T is invertible and $p (t) = t^{m} + a_{m - 1} t^{m - 1} + \dots + a_{1} t + a_{0}$ $p (t) = t^{m} + a_{m - 1} t^{m - 1} + \dots + a_{1} t + a_{0}$ , then
  
  $T^{- 1} = \frac{1}{a_{0}} (T^{m - 1} + a_{m - 1} T^{m - 2} + \dots + a_{2} T + a_{1} I) .$ $T^{- 1} = \frac{1}{a_{0}} (T^{m - 1} + a_{m - 1} T^{m - 2} + \dots + a_{2} T + a_{1} I) .$
Let T be a diagonalizable linear operator on a finite-dimensional vector space V. Prove that V is a T-cyclic subspace if and only if each of the eigenspaces of T is one-dimensional.
Let T be a linear operator on a finite-dimensional vector space V, and suppose that W is a T-invariant subspace of V. Prove that the minimal polynomial of $T_{W}$ $T_{W}$ divides the minimal polynomial of T.
Let g(t) be the auxiliary polynomial associated with a homogeneous linear differential equation with constant coefficients (as defined in Section 2.7), and let V denote the solution space of this differential equation. Prove the following results.
1. (a) V is a D-invariant subspace, where D is the differentiation operator on $C^{\infty}$ $C^{\infty}$ .
2. (b) The minimal polynomial of $D_{V}$ $D_{V}$ (the restriction of D to V) is g(t).
3. (c) If the degree of g(t) is n, then the characteristic polynomial of $D_{V}$ $D_{V}$ is ${(- 1)}^{n} g (t)$ ${(- 1)}^{n} g (t)$ .
Hint: Use Theorem 2.32 (p. 135) for (b) and (c).
Let D be the differentiation operator on P(R), the space of polynomials over R. Prove that there exists no polynomial g(t) for which $g (D) = T_{0}$ $g (D) = T_{0}$ . Hence D has no minimal polynomial.
Let T be a linear operator on a finite-dimensional vector space, and suppose that the characteristic polynomial of T splits. Let $λ_{1}, λ_{2}, \dots, λ_{k}$ $λ_{1}, λ_{2}, \dots, λ_{k}$ be the distinct eigenvalues of T, and for each i let $p_{i}$ $p_{i}$ be the number of rows in the largest Jordan block corresponding to $λ_{i}$ $λ_{i}$ in a Jordan canonical form of T. Prove that the minimal polynomial of T is

${(t - λ_{1})}^{p_{i}} {(t - λ_{2})}^{p_{2}} \dots {(t - λ_{k})}^{p_{k}} .$ ${(t - λ_{1})}^{p_{i}} {(t - λ_{2})}^{p_{2}} \dots {(t - λ_{k})}^{p_{k}} .$

The following exercise requires knowledge of direct sums (see Section 5.2).
Let T be a linear operator on a finite-dimensional vector space V, and let $W_{1}$ $W_{1}$ and $W_{2}$ $W_{2}$ be T-invariant subspaces of V such that $V = W_{1} \oplus W_{2}$ $V = W_{1} \oplus W_{2}$ . Suppose that $p_{1} (t)$ $p_{1} (t)$ and $p_{2} (t)$ $p_{2} (t)$ are the minimal polynomials of $T_{W_{1}}$ $T_{W_{1}}$ and $T_{W_{2}}$ $T_{W_{2}}$ , respectively. Either prove that the minimal polynomial f(t) of T always equals $p_{1} (t) p_{2} (t)$ $p_{1} (t) p_{2} (t)$ or give an example in which $f (t) \neq p_{1} (t) p_{2} (t)$ $f (t) \neq p_{1} (t) p_{2} (t)$ .

Exercise 15 uses the following definition.

Definition.

Let T be a linear operator on a finite-dimensional vector space V, and let x be a nonzero vector in V. The polynomial p(t) is called a T-annihilator of x if p(t) is a monic polynomial of least degree for which $p (T) (x) = 0$ $p (T) (x) = 0$ .

†Let T be a linear operator on a finite-dimensional vector space V, and let x be a nonzero vector in V. Prove the following results.
1. (a) The vector x has a unique T-annihilator.
2. (b) The T-annihilator of x divides any polynomial g(t) for which $g (T) = T_{0}$ $g (T) = T_{0}$ .
3. (c) If p(t) is the T-annihilator of x and W is the T-cyclic subspace generated by x, then p(t) is the minimal polynomial of $T_{W}$ $T_{W}$ , and dim(W) equals the degree of p(t).
4. (d) The degree of the T-annihilator of x is 1 if and only if x is an eigenvector of T.
Visit goo.gl/8KD6Gw for a solution.
Let T be a linear operator on a finite-dimensional vector space V, and let $W_{1}$ $W_{1}$ be a T-invariant subspace of V. Let $x \in V$ $x \in V$ such that $x \notin W_{1}$ $x \notin W_{1}$ . Prove the following results.
1. (a) There exists a unique monic polynomial $g_{1} (t)$ $g_{1} (t)$ of least positive degree such that $g_{1} (T) (x) \in W_{1}$ $g_{1} (T) (x) \in W_{1}$ .
2. (b) If h(t) is a polynomial for which $h (T) (x) \in W_{1}$ $h (T) (x) \in W_{1}$ , then $g_{1} (t)$ $g_{1} (t)$ divides h(t).
3. (c) $g_{1} (t)$ $g_{1} (t)$ divides the minimal and the characteristic polynomials of T.
4. (d) Let $W_{2}$ $W_{2}$ be a T-invariant subspace of V such that $W_{2} \subseteq W_{1}$ $W_{2} \subseteq W_{1}$ , and let $g_{2} (t)$ $g_{2} (t)$ be the unique monic polynomial of least degree such that $g_{2} (T) (x) \in W_{2}$ $g_{2} (T) (x) \in W_{2}$ . Then $g_{1} (t)$ $g_{1} (t)$ divides $g_{2} (t)$ $g_{2} (t)$ .

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Table of Contents for 7.3 The Minimal Polynomial

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Table of Contents for
7.3 The Minimal Polynomial