Eigenvalues and eigenvectors — characteristic polynomial: Question bank

Before the traps, one picture to keep in mind for every "same line" question below.

Figure — Eigenvalues and eigenvectors — characteristic polynomial

The eigenvector is a direction that survives the matrix unturned — $A x$ lands back on the same dashed line as $x$ , only rescaled by $λ$ . A generic vector (grey) gets knocked off its line. Hold this image against each claim.

True or false — justify

True or false: If

det (A) = 0

then

0

is an eigenvalue of

A

True.

det (A) = det (A - 0 \cdot I) = p (0)

, so

det (A) = 0

means

λ = 0

is a root of the characteristic polynomial — equivalently the null space is nontrivial, giving a nonzero

x

with

A x = 0

True or false: Every

2 \times 2

real matrix has two real eigenvalues.

False. The polynomial

λ^{2} - tr (A) λ + det (A)

can have a negative discriminant; the rotation

(01 - 1 0)

gives

λ = \pm i

, real eigenvalues do not exist there (see Complex Numbers).

True or false: An eigenvalue can have several eigenvectors, but an eigenvector belongs to only one eigenvalue.

True. If

A x = λ x

and

A x = μ x

then

(λ - μ) x = 0

; since

x \neq = 0

we get

λ = μ

. But one

λ

can own a whole subspace of eigenvectors.

True or false: If

x

is an eigenvector then so is

5 x

, with the same eigenvalue.

True.

A (5 x) = 5 A x = 5 λ x = λ (5 x)

. Any nonzero scalar multiple stays on the same magic line — eigenvectors are directions, not fixed-length arrows.

True or false: The characteristic polynomial of an

n \times n

matrix always has degree exactly

n

True. The

λ

terms come only from the diagonal factors

(a_{ii} - λ)

; their product contributes a leading

(- λ)^{n}

, so degree is exactly

n

and never drops.

True or false: A matrix and its transpose have the same eigenvalues.

True.

det (A^{⊤} - λ I) = det ((A - λ I)^{⊤}) = det (A - λ I)

, since transposing leaves the determinant unchanged. Same polynomial ⇒ same eigenvalues (though eigenvectors may differ).

True or false: If

A

is invertible with eigenvalue

λ

, then

A^{- 1}

has eigenvalue

1/ λ

True. From

A x = λ x

, multiply by

A^{- 1}

x = λ A^{- 1} x

, so

A^{- 1} x = (1/ λ) x

. Invertibility guarantees

λ \neq = 0

, so dividing is legal.

True or false: Two matrices with the same characteristic polynomial must be the same matrix.

False.

(3013)

and

3 I

both give

(3 - λ)^{2}

, yet the first is defective and the second is not. The polynomial fixes eigenvalues, not the whole matrix.

True or false: The eigenvalues of

A^{2}

are the squares of the eigenvalues of

A

True.

A x = λ x \Rightarrow A^{2} x = A (λ x) = λ^{2} x

. The same eigenvector persists with eigenvalue

λ^{2}

(the converse — every eigenvalue of

A^{2}

being some

λ^{2}

— also holds over

C

Spot the error

Find the flaw: "

A - λ

is singular, so set

det (A - λ) = 0

You cannot subtract a scalar

λ

from a matrix

A

— they live in different worlds. The correct object is

A - λ I

; the identity

I

turns the scalar into a matrix so the subtraction is defined.

Find the flaw: "I found

λ = 3

, then plugged

3

back into

det (A - λ I)

to read off the eigenvector."

A determinant is a single number, it stores no direction. Eigenvectors come from solving the system

(A - 3 I) x = 0

— the null space of

A - 3 I

Find the flaw: "

A 0 = λ 0

is true, so

0

is an eigenvector for every

λ

The equation is true but vacuous. Eigenvectors are defined to be nonzero; allowing

0

would let every scalar "work", collapsing the concept into nothing.

Find the flaw: "

λ = 2

is a double root, so there must be two independent eigenvectors."

That confuses algebraic with geometric multiplicity. A double root guarantees algebraic multiplicity

2

, but geometric multiplicity can be just

1

(a defective matrix like

(2012)

), blocking Diagonalization.

Find the flaw: "The trace is

7

, so one eigenvalue must be

7

The trace is the sum of all eigenvalues, not any single one. For

(5002)

the trace

7

splits as

5 + 2

; no eigenvalue equals

7

(see Trace of a Matrix).

Find the flaw: "I computed with

det (λ I - A)

and my friend used

det (A - λ I)

; we got different polynomials so one of us is wrong."

Both are fine.

det (λ I - A) = (- 1)^{n} det (A - λ I)

, so for odd

n

the polynomials differ by a global sign — but a sign flip never changes the roots, so the eigenvalues match.

Why questions

Why do we demand a nonzero solution rather than any solution of

(A - λ I) x = 0

x = 0

always solves a homogeneous system, so it carries no information. Only a nonzero solution reveals a genuine invariant direction — and requiring it forces

A - λ I

to be singular.

Why does "singular" translate exactly into "

det = 0

A square matrix is invertible iff its determinant is nonzero. So "not invertible" (singular) and "

det = 0

" are literally the same statement, which is what turns the eigenvalue hunt into a solvable polynomial equation.

Why are the eigenvalues of a triangular matrix just its diagonal entries?

For a triangular

A

A - λ I

stays triangular, and a triangular determinant is the product of its diagonal:

\prod_{i} (a_{ii} - λ)

. Setting the product to zero forces

λ = a_{ii}

for each

i

Why must a real matrix ever be allowed complex eigenvalues?

Because

det (A - λ I)

is a real polynomial that may have no real roots (e.g.

λ^{2} + 1

). The Fundamental Theorem of Algebra guarantees

n

roots only over

C

; refusing them would leave rotations "eigenvalue-less".

Why does the sum of eigenvalues equal the trace?

Writing

p (λ) = λ^{2} - tr (A) λ + det (A)

for

2 \times 2

and applying Vieta's formulas, the coefficient of

λ

(up to sign) is the root sum. In general the

λ^{n - 1}

coefficient collects the diagonal, giving the trace.

Why does the Cayley–Hamilton theorem let us compute

A^{- 1}

from the characteristic polynomial?

Cayley–Hamilton says

A

satisfies its own

p (λ)

. Rearranging

p (A) = 0

isolates

I

as a polynomial in

A

; dividing by the constant term

det (A)

(nonzero when invertible) expresses

A^{- 1}

using only powers of

A

Edge cases

What are the eigenvalues of the zero matrix

0_{n \times n}

Only

λ = 0

, with algebraic multiplicity

n

;

p (λ) = (- λ)^{n}

. Every nonzero vector is an eigenvector, so geometric multiplicity is also

n

— as un-defective as possible.

What are the eigenvalues of the identity matrix

I

Just

λ = 1

repeated

n

times, since

I x = 1 \cdot x

for all

x

. Every direction is fixed, so the whole space is the eigenspace — geometric multiplicity

n

A

is a projection (

A^{2} = A

), what eigenvalues are possible?

Only

0

and

1

. From

A^{2} x = A x

we get

λ^{2} = λ

, so

λ (λ - 1) = 0

— vectors are either killed (

λ = 0

) or kept (

λ = 1

Can a

3 \times 3

real matrix have exactly two real eigenvalues and no third real one?

Not "exactly two isolated" leaving one missing — complex roots come in conjugate pairs, so a real cubic has either 3 real roots or 1 real + a conjugate pair. A lone unpaired complex root is impossible.

What happens to eigenvectors when an eigenvalue is repeated but the matrix is diagonalizable?

The repeated

λ

owns a full-dimensional eigenspace (geometric = algebraic multiplicity), e.g.

2 I

has

λ = 2

twice with the entire plane as eigenspace. This equality is exactly the Diagonalization condition.

What does an eigenvalue of

0

tell you about the matrix's rank?

It means the null space is nontrivial, so the matrix is singular and rank-deficient. The dimension of the eigenspace for

λ = 0

equals the nullity, i.e.

n - rank (A)

Recall One-line self-check

Q: Same characteristic polynomial ⇒ same matrix? → No; defective vs diagonalizable can share it.
Q: Zero determinant ⇒ which eigenvalue? → $λ = 0$ .
Q: Projection eigenvalues? → Only $0$ and $1$ .