Linear-Algebra-Done-Right · Unit 11 · Video 5 · Interactive Practice

Eigenvalues Are the Zeros of the Minimal Polynomial

IKey Formulas

Formula	Name	What it says
$\{\text{zeros of } p \text{ in } \mathbf{F}\} = \{\text{eigenvalues of } T\}$	Eigenvalues are the zeros of the minimal polynomial	$V$ finite-dimensional, $T \in \mathcal{L}(V)$ , $p$ = minimal polynomial of $T$ . If $\mathbf{F} = \mathbf{C}$ and $V \neq \{0\}$ : $p(z) = (z - \lambda_1)\cdots(z - \lambda_m)$ , where $\lambda_1, \ldots, \lambda_m$ lists all eigenvalues of $T$ , repeats allowed
$q(T) = 0 \iff q = ps$ for some $s \in \mathcal{P}(\mathbf{F})$	Polynomials that send $T$ to $0$	For $q \in \mathcal{P}(\mathbf{F})$ : divide, $q = ps + r$ with $\deg r < \deg p$ ; then $q(T) = r(T)$ , and $r(T) = 0$ forces $r = 0$ by minimality of $p$
$p = (\text{minimal polynomial of } T\|_U)\cdot s$	Restriction to an invariant subspace	$U$ a subspace with $Tu \in U$ for all $u \in U$ , so $T\|_U \in \mathcal{L}(U)$ ; every eigenvalue of $T\|_U$ is an eigenvalue of $T$
$T$ not invertible $\iff p(0) = 0$	Invertibility from the constant term	$T$ not invertible $\iff 0$ is an eigenvalue $\iff 0$ is a zero of $p$ , and $p(0)$ is the constant term of $p$

Key Insight: If $Tv = \lambda v$ then $T^k v = \lambda^k v$ for every $k$ , so $p(T)v = p(\lambda)v$ : the operator equation $p(T) = 0$ becomes the scalar equation $p(\lambda) = 0$ on each eigenvector.

IIWhich Zeros Are Eigenvalues?

The eigenvalues of $T$ are exactly the zeros of its minimal polynomial that lie in $\mathbf{F}$ .

Operator Field Zero λ

IIIWhich Polynomials Send $T$ to $0$ ?

Dividing $q$ by $p$ gives $q = ps + r$ with $q(T) = r(T)$ , so $q(T) = 0$ exactly when $r = 0$ .

z⁶ 0 z⁵ 1 z⁴ 0 z³ 0 z² 0 z −6 1 0

IVThe Constant Term Decides Invertibility

$T$ is invertible exactly when $p(0) \neq 0$ , and then $p(T) = 0$ can be solved for $T^{-1}$ .

z² coefficient 0 z coefficient −3 constant term 1

💡 Every monic polynomial is the minimal polynomial of some operator (the video's construction with ones below the diagonal and the negated lower coefficients in the last column), so each cubic here describes an actual operator on $\mathbf{F}^3$ .

VQuiz Questions

Problem 1 · Read Off the Eigenvalues

Given: $T \in \mathcal{L}(\mathbf{C}^4)$ has minimal polynomial $p(z) = (z - 1)(z + 2)^2$ . Find the eigenvalues of $T$ .

✅ Correct! The zeros of

p

are

1

and

-2

, and the zeros of the minimal polynomial are exactly the eigenvalues.

❌ Not quite. The factor

z + 2

vanishes at

z = -2

, not at

z = 2

❌ Not quite. Both signs are flipped:

z - 1

vanishes at

1

and

z + 2

vanishes at

-2

❌ Not quite. The repeated factor does not crowd out the other one:

p(1) = 0 \cdot 9 = 0

, so

1

is an eigenvalue too.

Show solution

Here $V = \mathbf{C}^4$ is finite-dimensional, so the zeros of the minimal polynomial are exactly the eigenvalues of $T$ .

p(z) = (z - 1)(z + 2)^2 = 0 \iff z = 1 \ \text{or}\ z = -2.

So the eigenvalues are $1$ and $-2$ . The square on $z + 2$ only means that $-2$ appears twice in the list $\lambda_1, \lambda_2, \lambda_3$ of the factorization $p(z) = (z - 1)(z + 2)(z + 2)$ ; it adds no new eigenvalue. Two distinct eigenvalues is consistent with the bound $\#\{\text{eigenvalues}\} \leq \deg p = 3 \leq \dim V = 4$ .

Problem 2 · Over the Real Numbers

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ has minimal polynomial $p(z) = z^3 - 2z + 4$ . Find the true statement.

✅ Correct!

p(0) = 4 \neq 0

gives invertibility, and

-2

is the only zero of

p

\mathbf{R}

❌ Not quite. Invertibility fails only when

0

is a zero of

p

. Here

p(0) = 4 \neq 0

; the zero

-2

means

T + 2I

is not invertible, not

T

❌ Not quite. An eigenvalue of an operator on a real vector space must be a real number. The zeros

1 \pm i

are not in

\mathbf{F} = \mathbf{R}

❌ Not quite. Test small integers:

p(-2) = -8 + 4 + 4 = 0

, so

-2 \in \mathbf{R}

is a zero and hence an eigenvalue.

Show solution

Zeros. $p(-2) = -8 + 4 + 4 = 0$ , so $z + 2$ is a factor:

z^3 - 2z + 4 = (z + 2)(z^2 - 2z + 2).

The quadratic has $b^2 - 4c = 4 - 8 = -4 < 0$ , so its zeros $1 \pm i$ are not real. The only zero of $p$ in $\mathbf{R}$ is $-2$ .

Eigenvalues. With $\mathbf{F} = \mathbf{R}$ the eigenvalues of $T$ are the zeros of $p$ in $\mathbf{R}$ : only $-2$ .

Invertibility. The constant term is $p(0) = 4 \neq 0$ , so $0$ is not a zero of $p$ , $0$ is not an eigenvalue, and $T$ is invertible.

Problem 3 · The Inverse as a Polynomial in $T$

Given: $V$ is finite-dimensional and $T \in \mathcal{L}(V)$ has minimal polynomial $p(z) = z^3 + 2z^2 - 5z + 2$ . Decide whether $T$ is invertible, then find $T^{-1}$ as a polynomial in $T$ .

Is $T$ invertible?

Which formula gives $T^{-1}$ ?

✅ Correct!

T \cdot \left(-\tfrac{1}{2}\right)(T^2 + 2T - 5I) = -\tfrac{1}{2}\left(p(T) - 2I\right) = I

❌ Not quite.

p(1) = 0

is true, so

1

is an eigenvalue — that makes

T - I

non-invertible. Invertibility of

T

itself depends on whether

0

is a zero of

p

❌ Not quite. The minimal polynomial alone decides it:

T

is not invertible exactly when the constant term of

p

0

❌ Not quite. Look at the constant term of

p

❌ Check the sign. From

T^3 + 2T^2 - 5T = -2I

you divide by

-2

, not

2

; this choice gives

T \cdot \tfrac{1}{2}(T^2 + 2T - 5I) = -I

❌ Not quite. Since

T^3 + 2T^2 - 5T = -2I

, this expression equals

I

, not

T^{-1}

. Factor one

T

out first:

T(T^2 + 2T - 5I) = -2I

❌ Not quite. The coefficient to divide by is the constant term

2

, and the factor left after pulling out

T

T^2 + 2T - 5I

❌ Not quite. Move the constant term to the other side of

p(T) = 0

and factor out

T

Show solution

Invertibility. The constant term of $p$ is $p(0) = 2 \neq 0$ , so $0$ is not a zero of $p$ , hence not an eigenvalue of $T$ , and $T$ is invertible. (That $p(1) = 1 + 2 - 5 + 2 = 0$ only says $1$ is an eigenvalue.)

Inverse. Since $p(T) = 0$ ,

T^3 + 2T^2 - 5T + 2I = 0 \implies T\left(T^2 + 2T - 5I\right) = -2I.

Divide by $-2$ :

T \cdot \left(-\tfrac{1}{2}\right)\left(T^2 + 2T - 5I\right) = I.

Polynomials in $T$ commute with $T$ , so the same product in the other order is also $I$ , and

T^{-1} = -\tfrac{1}{2}\left(T^2 + 2T - 5I\right) = -\tfrac{1}{2}T^2 - T + \tfrac{5}{2}I.

Problem 4 · Multiples and Divisors

Given: $V$ is finite-dimensional and $T \in \mathcal{L}(V)$ has minimal polynomial $p(z) = z^2(z - 3)$ . Find which polynomial cannot be the minimal polynomial of $T|_U$ for an invariant subspace $U \neq \{0\}$ , and which polynomial $q$ satisfies $q(T) = 0$ .

Which cannot be the minimal polynomial of $T|_U$ ?

Which $q$ satisfies $q(T) = 0$ ?

✅ Correct!

(z - 3)^2

does not divide

p

, while

z^4 - 6z^3 + 9z^2 = z^2(z - 3)^2 = p(z)\,(z - 3)

is a multiple of

p

❌ Not quite.

3

is a zero of

p

, so it is an eigenvalue; for an eigenvector

v

U = \operatorname{span}(v)

is invariant and

T|_U = 3I

has minimal polynomial

z - 3

❌ Not quite.

0

is a zero of

p

, so it is an eigenvalue; on the span of an eigenvector for

0

T|_U = 0

has minimal polynomial

z

❌ Not quite. Take eigenvectors

u

for

0

and

w

for

3

and

U = \operatorname{span}(u, w)

T|_U

is not a scalar multiple of the identity, and

T|_U(T|_U - 3I) = 0

, so its minimal polynomial is

z(z - 3)

❌ Not quite. The minimal polynomial of

T|_U

must divide

p

❌ Not quite.

z^2 - 3z = z(z - 3)

divides

p

but is not a multiple of it: it is nonzero of degree

2 < 3 = \deg p

, so

q(T) \neq 0

by minimality.

❌ Not quite. Having the zeros

0

and

3

is not enough:

z^4 - 3z^3 + z^2 - 3z = z(z - 3)(z^2 + 1)

contains only one factor

z

, so

z^2(z - 3)

does not divide it.

❌ Not quite.

z^3 - 3z^2 + 1 = p(z) + 1

, so

q(T) = p(T) + I = I \neq 0

❌ Not quite.

q(T) = 0

exactly when

q

is a polynomial multiple of

p

Show solution

Restriction. For an invariant $U$ , $p$ is a polynomial multiple of the minimal polynomial of $T|_U$ , so that minimal polynomial divides $z^2(z - 3)$ . The polynomial $(z - 3)^2$ does not divide $z^2(z - 3)$ : the quotient would have to be a polynomial $s$ with $z^2(z - 3) = (z - 3)^2 s(z)$ , and dividing by $z - 3$ gives $z^2 = (z - 3)s(z)$ , which fails at $z = 3$ ( $9 \neq 0$ ). So $(z - 3)^2$ is impossible.

The other three all occur. The zeros $0$ and $3$ of $p$ are eigenvalues of $T$ ; pick eigenvectors $u$ (for $0$ ) and $w$ (for $3$ ). Then $\operatorname{span}(u)$ , $\operatorname{span}(w)$ and $\operatorname{span}(u, w)$ are invariant, with restrictions having minimal polynomials $z$ , $z - 3$ and $z(z - 3)$ .

Polynomials sending $T$ to $0$ . $q(T) = 0 \iff q = ps$ for some $s$ .

z^4 - 6z^3 + 9z^2 = z^2(z - 3)^2 = \underbrace{z^2(z - 3)}_{p(z)}\,(z - 3) \quad\checkmark

z^2 - 3z = z(z - 3), \qquad z^4 - 3z^3 + z^2 - 3z = z(z - 3)(z^2 + 1), \qquad z^3 - 3z^2 + 1 = p(z) + 1.

None of the last three is a multiple of $p$ : the first two contain only one factor $z$ , and the third leaves remainder $1$ , giving $q(T) = I$ .

Solved: 0 / 4

Eigenvalues Are the Zeros of the Minimal Polynomial

IKey Formulas

IIWhich Zeros Are Eigenvalues?

IIIWhich Polynomials Send TTT to 000?

IVThe Constant Term Decides Invertibility

VQuiz Questions

IIIWhich Polynomials Send $T$ to $0$ ?