Linear-Algebra-Done-Right · Unit 11 · Video 4 · Interactive Practice

The Minimal Polynomial: Existence, Uniqueness and Computation

IKey Formulas

Statement	Name	Hypotheses
$a_0 + a_1 z + \cdots + a_{m-1}z^{m-1} + z^m$ (leading coefficient $1$ ); a nonzero $p$ divided by its leading coefficient is monic	Monic polynomial	$p \in \mathcal{P}(\mathbf{F})$
There is a unique monic $p$ of smallest degree with $p(T) = 0$ , and $\deg p \leq \dim V$	Minimal polynomial	$V$ finite-dimensional, $T \in \mathcal{L}(V)$
Smallest $m \geq 1$ with $c_0 I + c_1 T + \cdots + c_{m-1}T^{m-1} = -T^m$ solvable gives $p(z) = c_0 + c_1 z + \cdots + c_{m-1}z^{m-1} + z^m$	Direct method	$V$ finite-dimensional, $V \neq \{0\}$ , $T \in \mathcal{L}(V)$
If $c_0 v + c_1 Tv + \cdots + c_{n-1}T^{n-1}v = -T^n v$ has a unique solution, then $p(z) = c_0 + c_1 z + \cdots + c_{n-1}z^{n-1} + z^n$	One-vector method	$n = \dim V \geq 1$ , $T \in \mathcal{L}(V)$ , $v \neq 0$

Key Insight: The word monic is what makes the minimal polynomial unique: $2p$ also sends $T$ to $0$ with the same degree. The easy argument from $\dim \mathcal{L}(V) = (\dim V)^2$ bounds the degree by $(\dim V)^2$ ; the theorem cuts this to $\dim V$ .

IIVisualization 1 — One Vector Recovers the Polynomial

For the video's operator on $\mathbf{F}^5$ , changing $v$ changes every vector $T^k v$ , yet the solved coefficients never change.

v₁ 1 v₂ 0 v₃ 0 v₄ 0 v₅ 0

IIIVisualization 2 — When One Vector Is Not Enough

The one-vector method finds $p$ exactly when $v, Tv, T^2v$ are independent; the direct method always finds it.

Operator T Vector v

💡 When $\deg p < \dim V$ , as for $I$ and $M$ , no vector can work: an independent list $v, Tv, \dots, T^{n-1}v$ would force $\deg p = n$ .

IVQuiz Questions

Problem 1 · The Cyclic Shift

Given: $C \in \mathcal{L}(\mathbf{F}^3)$ defined by $C(x_1, x_2, x_3) = (x_3, x_1, x_2)$ . Use the one-vector method with $v = e_1 = (1, 0, 0)$ to find the minimal polynomial of $C$ .

✅ Correct! The list

e_1, Ce_1, C^2e_1 = e_1, e_2, e_3

is linearly independent, and

C^3 e_1 = e_1

gives

c_0 = -1

c_1 = c_2 = 0

❌ Check the sign. The equation is

c_0 v + c_1 Cv + c_2 C^2v = -C^3v = -e_1

, so

c_0 = -1

and

p(z) = -1 + z^3

. Indeed

C^3 + I = 2I \neq 0

❌ Not quite.

p(z) = z - 1

would mean

C - I = 0

, i.e.

C = I

, but

Ce_1 = e_2 \neq e_1

. Since

e_1, Ce_1, C^2e_1

are independent, the degree is exactly

3

❌ Not quite.

C^2 + C + I

sends

e_1

e_3 + e_2 + e_1 \neq 0

. A monic polynomial of degree less than

3

cannot kill

e_1

, because

e_1, Ce_1, C^2e_1

are independent.

Show solution

Apply $C$ repeatedly to $v = e_1$ :

Ce_1 = (0, 1, 0) = e_2, \qquad C^2 e_1 = Ce_2 = (0, 0, 1) = e_3, \qquad C^3 e_1 = Ce_3 = (1, 0, 0) = e_1.

The list $e_1, Ce_1, C^2e_1 = e_1, e_2, e_3$ is the standard basis, so it is linearly independent and the system

c_0 e_1 + c_1 e_2 + c_2 e_3 = -C^3 e_1 = -e_1

has the unique solution $c_0 = -1$ , $c_1 = c_2 = 0$ . Hence the minimal polynomial is $p(z) = -1 + z^3 = z^3 - 1$ .

Check: applying $C$ three times returns every coordinate to its place, so $C^3 = I$ and $C^3 - I = 0$ .

Problem 2 · A Non-Unique Solution

Given: $P \in \mathcal{L}(\mathbf{F}^3)$ defined by $P(x_1, x_2, x_3) = (x_1 + x_3, x_2, 0)$ , which satisfies $P^2 = P$ . A student takes $v = e_3$ and finds $Pv = P^2v = P^3v = e_1$ . Find the minimal polynomial of $P$ .

✅ Correct!

P^2 - P = 0

, and no monic polynomial of degree

1

works because

P

is not a scalar multiple of

I

❌ Close, but not smallest.

c_0 = 0

c_1 = 0

c_2 = -1

is one of infinitely many solutions of

c_0 e_3 + (c_1 + c_2)e_1 = -e_1

. The polynomial

z^3 - z^2

does send

P

0

, since

P^3 - P^2 = P - P = 0

, but a non-unique solution says nothing about the smallest degree, and

P^2 - P = 0

already has degree

2

❌ Not quite.

P - I = 0

would mean

P = I

, but

Pe_3 = e_1 \neq e_3

❌ Close, but not smallest.

c_0 = 0

c_1 = -1

c_2 = 0

also solves the student's system, and

P^3 - P = P - P = 0

holds. Because the solution is not unique, the one-vector method does not apply;

P^2 - P = 0

has smaller degree.

Show solution

The student's system is $c_0 e_3 + c_1 e_1 + c_2 e_1 = -e_1$ , i.e. $c_0 = 0$ and $c_1 + c_2 = -1$ : infinitely many solutions, because $v, Pv, P^2v = e_3, e_1, e_1$ is linearly dependent. So the one-vector method does not apply to this $v$ ; use the direct method.

$m = 1$ : $c_0 I = -P$ would make $P$ a scalar multiple of $I$ . But $Pe_3 = e_1$ is not a multiple of $e_3$ . No solution.

$m = 2$ : $c_0 I + c_1 P = -P^2 = -P$ is solved by $c_0 = 0$ , $c_1 = -1$ . Since $m = 1$ failed, this solution is the only one.

So the minimal polynomial is $p(z) = 0 - z + z^2 = z^2 - z$ , of degree $2 < 3 = \dim \mathbf{F}^3$ . Both cubic options also send $P$ to $0$ , but they are not of smallest degree.

Problem 3 · The Direct Method Stops Early

Given: $A \in \mathcal{L}(\mathbf{F}^3)$ with matrix $\mathcal{M}(A) = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ with respect to the standard basis. Find the smallest $m$ for which $c_0 I + \cdots + c_{m-1}A^{m-1} = -A^m$ is solvable, and the minimal polynomial of $A$ .

What is the smallest m?

What is the minimal polynomial?

✅ Correct!

A^2 - 2A + I = 0

, and

m = 1

fails because

A

is not a scalar multiple of

I

❌ Not quite.

\dim V = 3

is only an upper bound for the degree. Test

m = 2

first: it already has a solution.

❌ Not quite.

m = 1

needs

c_0 I = -A

, so

A

would be a scalar multiple of

I

; the entry

1

in row 1, column 2 of

\mathcal{M}(A)

rules this out.

❌ Not quite.

9 = (\dim V)^2

is the easy bound from

\dim \mathcal{L}(\mathbf{F}^3) = 9

. The theorem guarantees a solution for some

m \leq 3

, and here

m = 2

works.

❌ Not quite.

A^2 - I

has entry

2

in row 1, column 2, so it is not the zero operator.

❌ Close, but not smallest. This polynomial is

(z - 1)(z^2 - 2z + 1)

, so it does send

A

0

, but its degree

3

is not the smallest.

❌ Check the signs. The system gives

c_0 = 1

c_1 = -2

, so

p(z) = 1 - 2z + z^2

. With the signs flipped,

A^2 + 2A + I

has diagonal entries

4

Show solution

First compute $\mathcal{M}(A^2) = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ .

$m = 1$ : $c_0 I = -A$ compares row 1, column 2: $0 = -1$ . No solution.

$m = 2$ : $c_0 I + c_1 A = -A^2$ . Comparing entries:

\text{row 1, column 2: } c_1 = -2, \qquad \text{diagonal: } c_0 + c_1 = -1 \;\Rightarrow\; c_0 = 1,

and every other entry is $0 = 0$ . So the smallest $m$ is $2$ , and the minimal polynomial is

p(z) = 1 - 2z + z^2 = z^2 - 2z + 1.

Check: $A^2 - 2A + I = \begin{pmatrix} 1 - 2 + 1 & 2 - 2 & 0 \\ 0 & 1 - 2 + 1 & 0 \\ 0 & 0 & 1 - 2 + 1 \end{pmatrix} = 0$ . Because $\deg p = 2 < 3$ , the one-vector method fails for every nonzero $v$ here.

Problem 4 · A Shifted Cyclic Operator

Given: $S \in \mathcal{L}(\mathbf{F}^3)$ defined by $S(x_1, x_2, x_3) = (x_1 + x_3, x_1 + x_2, x_2 + x_3)$ , that is, $S = C + I$ for the cyclic shift $C$ of Problem 1. Find the minimal polynomial of $S$ .

✅ Correct! The list

e_1, Se_1, S^2e_1

is linearly independent, and

c_0 = -2

c_1 = 3

c_2 = -3

. As a check,

(S - I)^3 = C^3 = I

❌ Not quite. That is the minimal polynomial of

C

, not of

S

. Applied to

e_1

S^3 e_1 - e_1 = (2, 3, 3) - (1, 0, 0) \neq 0

❌ Close, but… This is

(z - 1)^3

, and

(S - I)^3 = C^3 = I

, not

0

. Subtract one more

I

❌ Not quite. This is

(z + 1)^3 - 1

, which substitutes

C = S + I

instead of

C = S - I

. Applied to

e_1

it gives

(2,3,3) + (3,6,3) + (3,3,0) = (8, 12, 6) \neq 0

Show solution

Take $v = e_1$ and apply $S$ repeatedly:

Se_1 = (1, 1, 0), \qquad S^2 e_1 = S(1, 1, 0) = (1, 2, 1), \qquad S^3 e_1 = S(1, 2, 1) = (2, 3, 3).

The vectors $(1, 0, 0)$ , $(1, 1, 0)$ , $(1, 2, 1)$ are linearly independent (their last nonzero coordinates sit in different positions). Solve $c_0 e_1 + c_1 Se_1 + c_2 S^2 e_1 = -S^3 e_1$ :

\begin{aligned} \text{third coordinate: } & c_2 = -3, \\ \text{second coordinate: } & c_1 + 2c_2 = -3 \;\Rightarrow\; c_1 = 3, \\ \text{first coordinate: } & c_0 + c_1 + c_2 = -2 \;\Rightarrow\; c_0 = -2. \end{aligned}

The solution is unique, so the minimal polynomial is $p(z) = -2 + 3z - 3z^2 + z^3 = z^3 - 3z^2 + 3z - 2$ .

Check: $C = S - I$ and $C^3 = I$ , so $(S - I)^3 - I = S^3 - 3S^2 + 3S - 2I = 0$ .

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