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✓ 7 results · all verified · 2 also independently AI-judged

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Eigenvalues and the Characteristic Polynomial: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

The matrix $(\begin{array}{cc} 2 & 1 \\ 0 & 3 \end{array})$ has characteristic polynomial $(x - 2) (x - 3)$ and two explicitly computed eigenspaces

Example

Over $R$ , let

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

Then $χ_{A} (x) = (x - 2) (x - 3)$ , $E_{2} (A) = span {(1, 0)}$ , and $E_{3} (A) = span {(1, 1)}$ .

Facts & Assumptions

Given: The displayed real matrix $A$ .

[L1]

The characteristic polynomial is $\det (x I - A)$ (For $A \in M_{n} (F)$ , the characteristic polynomial is $χ_{A} (x) = \det (x I_{n} - A)$ when $n \geq 1$ , with $χ_{A} (x) = 1$ for the unique $0 \times 0$ matrix).

[L2]

The eigenspace is $E_{λ} (A) = \ker (A - λ I)$ (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism).

Verification

technique · direct computation

1.1

By [L1], $χ_{A} (x) = \det (\begin{array}{cc} x - 2 & - 1 \\ 0 & x - 3 \end{array}) = (x - 2) (x - 3)$ .

L1algebra

1.2

The equation $(A - 2 I) (u, v) = (v, v) = (0, 0)$ gives $v = 0$ , hence $E_{2} (A) = {(u, 0) : u \in R} = span {(1, 0)}$ .

L2algebra

1.3

The equation $(A - 3 I) (u, v) = (- u + v, 0) = (0, 0)$ gives $v = u$ , hence $E_{3} (A) = span {(1, 1)}$ .

L2algebra

2.1

The two nonzero spanning vectors also directly exhibit both roots as eigenvalues.

step 1.1step 1.2step 1.3L2∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

A quarter-turn of $R^{2}$ has characteristic polynomial $x^{2} + 1$ and no real eigenvalue

Example

The real quarter-turn

$A = (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$

has $χ_{A} (x) = x^{2} + 1$ and empty real spectrum.

Facts & Assumptions

Given: The displayed matrix, acting on $R^{2}$ .

[L1]

The spectrum over the base field is exactly the root set in that field of the characteristic polynomial (For every finite-dimensional space, $σ_{F} (T)$ is exactly the set of roots in $F$ of $χ_{T}$ ).

[L2]

Every real square is nonnegative (Squares of nonzero elements are positive).

Verification

technique · direct computation

1.1

Directly, $χ_{A} (x) = \det (\begin{array}{cc} x & 1 \\ - 1 & x \end{array}) = x^{2} + 1$ .

algebra

2.1

For every $x \in R$ , [L2] gives $x^{2} + 1 \geq 1 > 0$ , so $χ_{A}$ has no real root.

step 1.1L2algebra

3.1

By [L1], $A$ has no real eigenvalue, despite acting on a nonzero real space.

step 2.1L1∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13Open item page →

$(\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array})$ over $F_{2}$ has characteristic polynomial $x^{2} + x + 1$ and no eigenvalue in its base field

Example

Over $F_{2}$ , the matrix

$A = (\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array})$

has characteristic polynomial $x^{2} + x + 1$ and no eigenvalue in $F_{2}$ .

Facts & Assumptions

Given: The displayed matrix over $F_{2} = Z / 2 Z$ .

[L1]

Since $2$ is prime, $Z / 2 Z$ is a field, whose only elements are $0$ and $1$ (For every prime $p$ , the two operations on $Z / p$ make it a field).

[L2]

Verification

technique · exhaustive finite computation

1.1

In $F_{2} [x]$ , $χ_{A} (x) = \det (\begin{array}{cc} x & - 1 \\ - 1 & x - 1 \end{array}) = x (x - 1) - 1 = x^{2} + x + 1$ .

L1algebra

2.1

By [L1], the only candidate roots are $0$ and $1$ , and $χ_{A} (0) = 1$ while $χ_{A} (1) = 1 + 1 + 1 = 1$ in $F_{2}$ .

step 1.1L1algebra

3.1

Thus $χ_{A}$ has no root in its base field, and [L2] gives $σ_{F_{2}} (A) = \emptyset$ .

step 2.1L2∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

$(\begin{array}{cc} 2 & 1 \\ 0 & 2 \end{array})$ has algebraic multiplicity two but geometric multiplicity one at $2$

Example

For

$A = (\begin{array}{cc} 2 & 1 \\ 0 & 2 \end{array}) \in M_{2} (R),$

the eigenvalue $2$ has algebraic multiplicity $2$ and geometric multiplicity $1$ .

Facts & Assumptions

Given: The displayed real matrix $A$ .

[L1]

Algebraic multiplicity is the exponent of $x - λ$ in the characteristic polynomial, and geometric multiplicity is $\dim \ker (A - λ I)$ (Algebraic multiplicity as the exponent of $x - λ$ in $χ_{T}$ , and geometric multiplicity as $\dim E_{λ} (T)$ ).

Verification

technique · direct computation

1.1

One has $χ_{A} (x) = \det (\begin{array}{cc} x - 2 & - 1 \\ 0 & x - 2 \end{array}) = (x - 2)^{2}$ , so [L1] gives algebraic multiplicity $2$ .

L1algebra

1.2

Since $(A - 2 I) (u, v) = (v, 0)$ , its kernel is ${(u, 0) : u \in R}$ , a one-dimensional space.

L1algebra

2.1

Thus the geometric multiplicity is $1 < 2$ , so the general multiplicity inequality can be strict.

step 1.1step 1.2L1∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

The scalar matrix $2 I_{2}$ has algebraic and geometric multiplicity two at $2$

Example

Over any field $F$ , the scalar matrix $A = 2 I_{2}$ has algebraic and geometric multiplicity $2$ at the scalar $2 \in F$ .

Facts & Assumptions

Given: The matrix $A = 2 I_{2} \in M_{2} (F)$ .

[L1]

[L2]

The two standard coordinate vectors form a basis of $F^{2}$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

Verification

technique · direct computation

1.1

Since $x I_{2} - A = (x - 2) I_{2}$ , one has $χ_{A} (x) = (x - 2)^{2}$ , so the algebraic multiplicity is $2$ .

L1algebra

1.2

Since $A - 2 I_{2} = 0$ , its kernel is all of $F^{2}$ , which has dimension $2$ by [L2]. Thus the geometric multiplicity is $2$ .

L1L2algebra

2.1

Both multiplicities therefore equal $2$ , including in characteristic $2$ , where the scalar denoted $2$ is $0$ .

step 1.1step 1.2∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

Cayley-Hamilton reduces every power of $(\begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array})$ to a linear combination of $A$ and $I$

Example

Let $A = (\begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array})$ , and define $F_{0} = 0$ , $F_{1} = 1$ , and $F_{m + 1} = F_{m} + F_{m - 1}$ for $m \geq 1$ . Then, over any field and for every $m \geq 1$ ,

$A^{m} = F_{m} A + F_{m - 1} I_{2},$

with the integer coefficients interpreted in the field.

Facts & Assumptions

Given: The displayed matrix and recurrence.

[L1]

Cayley-Hamilton states that a matrix satisfies its characteristic polynomial (Cayley-Hamilton: every finite-dimensional endomorphism satisfies its characteristic polynomial, $χ_{T} (T) = 0$ ).

Verification

technique · induction on $m$

1.1

Direct calculation gives $χ_{A} (x) = x^{2} - x - 1$ and $A^{2} = (\begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array}) = A + I_{2}$ , agreeing with [L1]. For $m = 1$ , $A = F_{1} A + F_{0} I_{2}$ ; for $m = 2$ , $A^{2} = F_{2} A + F_{1} I_{2}$ .

baseL1algebra

2.1

Assume $m \geq 2$ and $A^{m} = F_{m} A + F_{m - 1} I_{2}$ . Multiplication by $A$ and $A^{2} = A + I_{2}$ give $A^{m + 1} = F_{m} A^{2} + F_{m - 1} A = (F_{m} + F_{m - 1}) A + F_{m} I_{2} = F_{m + 1} A + F_{m} I_{2}$ .

ihstep 1.1algebra

3.1

The base cases and recurrence prove the formula for every $m \geq 1$ , so every positive power lies in $span {A, I_{2}}$ .

step 1.1step 2.1discharge-induction∎

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

FALSE: substituting a fixed matrix $A$ for $x$ defines a ring homomorphism $M_{n} (F [x]) \to M_{n} (F)$

Statement refuted

Refuted claim: For a fixed $A \in M_{n} (F)$ , the coefficientwise rule $\sum_{k} C_{k} x^{k} \mapsto \sum_{k} C_{k} A^{k}$ is a ring homomorphism $M_{n} (F [x]) \to M_{n} (F)$ .

This is the invalid substitution step in a familiar pseudo-proof of Cayley-Hamilton; the actual theorem Cayley-Hamilton: every finite-dimensional endomorphism satisfies its characteristic polynomial, $χ_{T} (T) = 0$ requires a coefficient-comparison argument.

Facts & Assumptions

Given: Work over any field $F$ and take $A = diag (1, 0)$ and $B = E_{12}$ in $M_{2} (F)$ .

[L1]

Matrix multiplication over a commutative ring is defined by finite row-column sums, and the identity matrix has $1$ on its diagonal and $0$ elsewhere (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

Refutation

technique · explicit counterexample

1.1

In $M_{2} (F [x])$ , the row-column formula [L1] and commutativity of $F [x]$ give $(x I_{2}) B = B (x I_{2}) = x B$ .

L1algebra

2.1

The coefficientwise rule has $E (x I_{2}) = A$ , $E (B) = B$ , and $E ((x I_{2}) B) = E (x B) = B A = 0$ , whereas $E (x I_{2}) E (B) = A B = B$ .

step 1.1givenalgebra

3.1

Since $B \neq 0$ , step 2.1 shows $E ((x I_{2}) B) \neq E (x I_{2}) E (B)$ . The coefficientwise substitution rule is therefore not multiplicative and hence is not a ring homomorphism.

step 2.1∎

Sources

Standard references

Recommended treatments; not extraction sources.

Eigenvalues and the Characteristic Polynomial: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The matrix (2103)(20​13​) has characteristic polynomial (x−2)(x−3)(x−2)(x−3) and two explicitly computed eigenspaces

Example

Facts & Assumptions

Verification

A quarter-turn of R2R2 has characteristic polynomial x2+1x2+1 and no real eigenvalue

Example

Facts & Assumptions

Verification

(0111)(01​11​) over F2F2​ has characteristic polynomial x2+x+1x2+x+1 and no eigenvalue in its base field

Example

Facts & Assumptions

Verification

(2102)(20​12​) has algebraic multiplicity two but geometric multiplicity one at 22

Example

Facts & Assumptions

Verification

The scalar matrix 2I22I2​ has algebraic and geometric multiplicity two at 22

Example

Facts & Assumptions

Verification

Cayley-Hamilton reduces every power of (1110)(11​10​) to a linear combination of AA and II

Example

Facts & Assumptions

Verification

FALSE: substituting a fixed matrix AA for xx defines a ring homomorphism Mn(F[x])→Mn(F)Mn​(F[x])→Mn​(F)

Statement refuted

Facts & Assumptions

Refutation

Sources

Standard references

The matrix $(\begin{array}{cc} 2 & 1 \\ 0 & 3 \end{array})$ has characteristic polynomial $(x - 2) (x - 3)$ and two explicitly computed eigenspaces

A quarter-turn of $R^{2}$ has characteristic polynomial $x^{2} + 1$ and no real eigenvalue

$(\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array})$ over $F_{2}$ has characteristic polynomial $x^{2} + x + 1$ and no eigenvalue in its base field

$(\begin{array}{cc} 2 & 1 \\ 0 & 2 \end{array})$ has algebraic multiplicity two but geometric multiplicity one at $2$

The scalar matrix $2 I_{2}$ has algebraic and geometric multiplicity two at $2$

Cayley-Hamilton reduces every power of $(\begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array})$ to a linear combination of $A$ and $I$

FALSE: substituting a fixed matrix $A$ for $x$ defines a ring homomorphism $M_{n} (F [x]) \to M_{n} (F)$