Linear-Algebra-Done-Right · Unit 29 · Video 5 · Interactive Practice

The Characteristic Polynomial det(zI − T) over R and C

IKey Formulas

Formula	Name	Hypotheses
$q(z) = \det(zI - T)$	Characteristic polynomial	$T \in \mathcal{L}(V)$ , $\mathbf{F} = \mathbf{R}$ or $\mathbf{C}$ ; monic of degree $n = \dim V$
$\det(zI - T) = (z - \lambda_1)^{d_1} \cdots (z - \lambda_m)^{d_m}$	Agrees with the earlier definition	$\mathbf{F} = \mathbf{C}$ ; $\lambda_k$ distinct eigenvalues, $d_k$ multiplicities
$q(T) = 0$	Cayley–Hamilton	Now over $\mathbf{R}$ as well as $\mathbf{C}$
$q(z) = z^n - (\operatorname{tr} T)z^{n-1} + \cdots + (-1)^n \det T$	Trace and determinant in $q$	Any $T \in \mathcal{L}(V)$

Key Insight: On a real space, $T$ may have no eigenvalues at all, but $\det(zI - T)$ is still a polynomial; Cayley–Hamilton over $\mathbf{R}$ follows by passing to the complex operator with the same matrix.

IIWhere Trace and Determinant Sit

For a $2$ -by- $2$ matrix, $\det(zI - A) = z^2 - (\operatorname{tr} A)z + \det A$ : its real zeros are the real eigenvalues, and its value at $0$ is $\det A$ .

a 0 b −1 c 1 d 0

IIICayley–Hamilton over R

Even when a real matrix has no real eigenvalues, $q(A) = A^2 - (\operatorname{tr} A)A + (\det A)I$ is the zero matrix.

A Step 1

IVQuiz Questions

Problem 1 · The Constant Term

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ with $\operatorname{tr} T = 4$ and $\det T = -6$ . Find the constant term of the characteristic polynomial $\det(zI - T)$ .

✅ Correct! The constant term is

\det(0I - T) = \det(-T) = (-1)^3 \det T = 6

❌ Check the sign. Setting

z = 0

gives

\det(-T)

, and

-T = (-I)T

with

\det(-I) = (-1)^3

❌ Not quite. The trace sits in the

z^{2}

coefficient (as

-\operatorname{tr} T

); the constant term comes from the determinant.

Show solution

The constant term of $q(z) = \det(zI - T)$ is $q(0)$ :

q(0) = \det(-T) = \det(-I)\det T = (-1)^3(-6) = 6.

So $q(z) = z^3 - 4z^2 + cz + 6$ for some $c$ .

Problem 2 · A 2-by-2 Characteristic Polynomial

Given: $A = \begin{pmatrix} 0 & -4 \\ 1 & 5 \end{pmatrix}$ . Find $\det(zI - A)$ .

✅ Correct!

\operatorname{tr} A = 5

\det A = 0 \cdot 5 - (-4)(1) = 4

; the zeros

1, 4

are the eigenvalues.

❌ Check the $z$ coefficient. It is

-\operatorname{tr} A = -5

❌ Check the constant term.

\det A = 0 - (-4)(1) = +4

❌ Two signs are off. The

z

coefficient is

-\operatorname{tr} A

and the constant term is

+\det A

for a

2

-by-

2

matrix.

Show solution

\det(zI - A) = \det\begin{pmatrix} z & 4 \\ -1 & z - 5 \end{pmatrix} = z(z - 5) - (4)(-1) = z^2 - 5z + 4 = (z - 1)(z - 4).

Problem 3 · Reducing a Power

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ has characteristic polynomial $z^3 - z^2 + z - 1$ . Which identity holds?

✅ Correct! Cayley–Hamilton holds over

\mathbf{R}

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

❌ Check the signs. Moving

-T^2 + T - I

to the other side changes every sign.

❌ Not quite. Only

1

is a real zero, but Cayley–Hamilton uses the whole polynomial, including the factor

z^2 + 1

❌ Check the $T$ term.

+T

q(T) = 0

becomes

-T

on the other side.

Show solution

The characteristic polynomial $q(z) = \det(zI - T)$ works over $\mathbf{R}$ , and Cayley–Hamilton gives $q(T) = 0$ :

T^3 - T^2 + T - I = 0 \implies T^3 = T^2 - T + I.

Note $q(z) = (z - 1)(z^2 + 1)$ , so $T$ has the single real eigenvalue $1$ ; the identity holds anyway.

Problem 4 · No Real Eigenvalues, Positive Determinant

Given: $T \in \mathcal{L}(\mathbf{R}^4)$ has characteristic polynomial $(z^2 + 1)(z^2 - 4z + 5)$ . Find $\det T$ .

✅ Correct! The constant term

1 \cdot 5 = 5

equals

(-1)^4\det T = \det T

❌ Check the sign. In dimension

4

, the constant term is

(-1)^4 \det T = +\det T

❌ Not quite.

4

is the trace (minus the

z^3

coefficient), not the determinant.

❌ Not quite. Both factors contribute to the constant term:

1 \cdot 5

Show solution

The constant term of $q(z) = \det(zI - T)$ is $q(0) = \det(-T) = (-1)^4 \det T = \det T$ . From the factored form,

q(0) = (0 + 1)(0 - 0 + 5) = 5,

so $\det T = 5$ . Neither quadratic factor has a real zero ( $z^2 - 4z + 5 = (z - 2)^2 + 1$ ), so $T$ has no eigenvalues, yet its characteristic polynomial and determinant are perfectly defined.

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