Linear-Algebra-Done-Right · Unit 25 · Video 3 · Interactive Practice

Cayley–Hamilton Without Determinants

IKey Formulas

Formula	Name	Hypotheses
$q(z) = (z - \lambda_1)^{d_1} \cdots (z - \lambda_m)^{d_m}$	Characteristic polynomial	$\mathbf{F} = \mathbf{C}$ , $V$ finite-dimensional; $\lambda_k$ the distinct eigenvalues of $T$ , $d_k$ their multiplicities
$\deg q = \dim V$ ; zeros of $q$ $=$ eigenvalues of $T$	Degree and zeros	same
$q(T) = 0$	Cayley–Hamilton theorem	same
$q$ is a polynomial multiple of the minimal polynomial of $T$	Characteristic vs minimal	same; they are equal when the minimal polynomial has degree $\dim V$

Key Insight: $(T - \lambda_k I)^{d_k}$ is already $0$ on $G(\lambda_k, T)$ , because a nilpotent operator to the power of the dimension of its domain is $0$ ; the factors commute, so each piece of $V$ meets its own factor first.

IIWhich Products of the Factors Vanish?

With every exponent equal to its multiplicity the product is $0$ ; sometimes smaller exponents already give $0$ .

Operator exponent a 1 exponent b 1

IIIEach Piece Meets Its Own Factor

Applying the factors of $q(T) = (T - 6I)^2(T - 7I)$ one at a time, each generalized eigenspace component dies at its own factor.

z₁ 1 z₂ 2 z₃ 1 Order Factors applied 0

IVQuiz Questions

Problem 1 · Build the Characteristic Polynomial

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ , $T(z_1, z_2, z_3) = (z_2,\ 0,\ 4z_3)$ — find the characteristic polynomial of $T$ .

✅ Correct!

0

has multiplicity

2

and

4

has multiplicity

1

, so

q(z) = z^2(z - 4) = z^3 - 4z^2

, of degree

3 = \dim \mathbf{C}^3

❌ The exponents are multiplicities, not eigenspace dimensions.

E(0, T)

is a line, but

G(0, T) = \operatorname{null} T^3

is a plane; and

\deg q

must be

\dim \mathbf{C}^3 = 3

❌ Check the sign. The factor for the eigenvalue

4

z - 4

, so

q

has zeros

0

and

4

, not

-4

❌ The exponents are swapped. That is

z(z - 4)^2

; but

4

has multiplicity

1

and

0

has multiplicity

2

Show solution

$T^2(z_1, z_2, z_3) = T(z_2, 0, 4z_3) = (0, 0, 16z_3)$ , and $T^3(z_1, z_2, z_3) = (0, 0, 64z_3)$ . So $G(0, T) = \operatorname{null} T^3 = \{(z_1, z_2, 0)\}$ has dimension $2$ , although $E(0, T) = \operatorname{null} T = \{(z_1, 0, 0)\}$ is a line.

$(T - 4I)(z_1, z_2, z_3) = (z_2 - 4z_1,\ -4z_2,\ 0)$ , whose null space and those of its powers are $\{(0, 0, z_3)\}$ ; so $4$ has multiplicity $1$ . The multiplicities $2 + 1 = 3$ add up to $\dim \mathbf{C}^3$ , and

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

Problem 2 · Characteristic or Minimal?

Given: $T \in \mathcal{L}(\mathbf{C}^4)$ with matrix $\begin{pmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$ in the standard basis, whose characteristic polynomial is $(z - 3)^3(z + 1)$ — find the minimal polynomial of $T$ .

✅ Correct!

(T - 3I)(T + I) \ne 0

but

(T - 3I)^2(T + I) = 0

, and the minimal polynomial divides

q

and has every eigenvalue as a zero.

❌ That is $q$ itself. Cayley–Hamilton says

q(T) = 0

, so the minimal polynomial divides

q

, but here a proper divisor already sends

T

0

❌ $(T - 3I)(T + I) \neq 0$ . It sends

e_2

(T - 3I)(T + I)e_2 = (T - 3I)(e_1 + 4e_2) = 4e_1

❌ Every eigenvalue is a zero of the minimal polynomial.

-1

is an eigenvalue (

Te_4 = -e_4

), so

z + 1

must divide it.

Show solution

The minimal polynomial $p$ divides $q = (z - 3)^3(z + 1)$ and has zeros $3$ and $-1$ , so $p = (z - 3)^j (z + 1)$ with $1 \le j \le 3$ .

$(T - 3I)$ sends $e_1 \mapsto 0$ , $e_2 \mapsto e_1$ , $e_3 \mapsto 0$ , $e_4 \mapsto -4e_4$ ; and $(T + I)$ sends $e_1 \mapsto 4e_1$ , $e_2 \mapsto e_1 + 4e_2$ , $e_3 \mapsto 4e_3$ , $e_4 \mapsto 0$ .

For $j = 1$ : $(T - 3I)(T + I)e_2 = (T - 3I)(e_1 + 4e_2) = 4e_1 \neq 0$ . For $j = 2$ : $(T - 3I)^2$ sends $e_1, e_2, e_3$ to $0$ , and $(T + I)$ sends $e_4$ to $0$ ; since the factors commute, $(T - 3I)^2(T + I)$ kills every basis vector. So

p(z) = (z - 3)^2(z + 1) \neq q(z).

Problem 3 · Reduce a Power with $q(T) = 0$

Given: $T \in \mathcal{L}(\mathbf{C}^2)$ has the eigenvalues $1$ and $2$ — find $T^3$ as a combination of $T$ and $I$ .

✅ Correct!

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

, so

T^2 = 3T - 2I

and

T^3 = 3T^2 - 2T = 7T - 6I

❌ That is $T^2$ . Multiply

T^2 = 3T - 2I

T

once more, then replace the new

T^2

❌ The $-2T$ term was dropped.

T^3 = T \cdot T^2 = 3T^2 - 2T

, and

3T^2 = 9T - 6I

❌ Check the sign.

3T^2 = 3(3T - 2I) = 9T - 6I

, with

-6I

Show solution

Two distinct eigenvalues on a $2$ -dimensional space each have multiplicity $1$ , so $q(z) = (z - 1)(z - 2) = z^2 - 3z + 2$ . By Cayley–Hamilton,

T^2 - 3T + 2I = 0 \quad\Rightarrow\quad T^2 = 3T - 2I.

T^3 = T \cdot T^2 = 3T^2 - 2T = 3(3T - 2I) - 2T = 7T - 6I.

Check on the eigenvalues: $1^3 = 7 - 6$ and $2^3 = 14 - 6$ .

Problem 4 · A Polynomial That Must Vanish

Given: $T \in \mathcal{L}(\mathbf{C}^5)$ has exactly the eigenvalues $0$ and $2$ , with $\dim G(0, T) = 2$ — find the polynomial $p$ for which $p(T) = 0$ is guaranteed.

✅ Correct! The multiplicity of

2

5 - 2 = 3

, so

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

, and Cayley–Hamilton gives

q(T) = 0

❌ Only if $T$ has a basis of eigenvectors. If

T|_{G(0, T)}

is a nonzero nilpotent operator, then

T(T - 2I) \neq 0

❌ The exponents are swapped.

\dim G(2, T) = 3

, and

(T - 2I)^2

need not vanish on a

3

-dimensional generalized eigenspace.

❌ Not guaranteed.

(T - 2I)|_{G(2, T)}

is nilpotent on a

3

-dimensional space, so its cube is

0

, but its square can be nonzero.

Show solution

Over $\mathbf{C}$ the multiplicities add up to $5$ , so $\dim G(2, T) = 3$ and

q(z) = z^2 (z - 2)^3, \qquad q(T) = 0.

The other choices can fail. Take $T$ with matrix $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ on $G(0, T) = \operatorname{span}(e_1, e_2)$ and $\begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}$ on $\operatorname{span}(e_3, e_4, e_5)$ . Then $(T - 2I)^2 e_5 = e_3 \neq 0$ , and $T^3 e_3 = 8e_3$ , so $z^3(z - 2)^2$ and $z^2(z - 2)^2$ do not send $T$ to $0$ ; nor does $z(z - 2)$ .

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