How is the determinant related to eigenvalues?

The determinant equals the product of all eigenvalues: det(A) = λ₁·λ₂·...·λₙ. This means A is invertible iff no eigenvalue is zero. A single zero eigenvalue makes the determinant zero and the matrix singular.

What is the difference between algebraic and geometric multiplicity?

Algebraic multiplicity is how many times λ appears as a root of the characteristic polynomial. Geometric multiplicity is the dimension of the eigenspace. Always: 1 ≤ geometric ≤ algebraic. When they differ, the matrix is not diagonalizable.

What are the eigenvalues of the inverse matrix?

If λ is an eigenvalue of invertible A with eigenvector v, then 1/λ is an eigenvalue of A⁻¹ with the same eigenvector. The eigenvalues of A⁻¹ are the reciprocals of those of A.

What are the eigenvalues of A^k?

If Av = λv, then Aᵏv = λᵏv. The eigenvalues of Aᵏ are the k-th powers of the eigenvalues of A, with the same eigenvectors. More generally, for polynomial q(A), the eigenvalues are q(λᵢ).

What happens to eigenvalues when you add cI to a matrix?

Adding cI shifts every eigenvalue by c: eigenvalues of A + cI are λ₁ + c, λ₂ + c, ..., λₙ + c. The eigenvectors remain unchanged. Similarly, cA has eigenvalues cλ₁, cλ₂, ..., cλₙ.

Do A and A^T have the same eigenvalues?

Yes. det(Aᵀ - λI) = det(A - λI) because determinant is transpose-invariant. The characteristic polynomials are identical, so eigenvalues match. However, the eigenvectors are generally different—eigenvectors of Aᵀ are left eigenvectors of A.

What eigenvalues do special matrices have?

Symmetric: all real. Orthogonal: |λ| = 1. Idempotent (A² = A): λ = 0 or 1. Nilpotent (Aᵏ = 0): all λ = 0. Involutory (A² = I): λ = ±1. Positive definite symmetric: all λ > 0. Triangular/diagonal: eigenvalues on diagonal.

Are eigenvectors of distinct eigenvalues independent?

Yes, always. Eigenvectors corresponding to distinct eigenvalues are linearly independent. A matrix with n distinct eigenvalues automatically has n independent eigenvectors and is diagonalizable.

Do similar matrices have the same eigenvalues?

Yes. If B = P⁻¹AP, then A and B share eigenvalues, algebraic multiplicities, geometric multiplicities, and characteristic polynomial. Eigenvalues are properties of the transformation, not the matrix representation. Eigenvectors transform by P⁻¹.

Properties of Eigenvalue and Eigenvector

Q: How is the trace related to eigenvalues?

The trace equals the sum of all eigenvalues (with algebraic multiplicity): tr(A) = λ₁ + λ₂ + ... + λₙ. This follows from the characteristic polynomial coefficients and provides a quick consistency check when computing eigenvalues.

Trace and Eigenvalues

Determinant and Eigenvalues

Algebraic and Geometric Multiplicity

Eigenvalues of the Inverse

Eigenvalues of Powers and Polynomials

Eigenvalue Shifting

Eigenvalues of the Transpose

Eigenvalues of Special Matrix Types

Independence of Eigenvectors

Summary: Spectral Properties at a Glance

How Eigenvalues Behave Under Matrix Operations

Eigenvalues interact in predictable ways with the trace, determinant, transpose, inverse, powers, and special matrix structures. These relationships provide shortcuts for computing eigenvalues, constraints on what eigenvalues are possible for a given matrix type, and structural connections between the eigenvalue spectrum and the algebraic properties of the matrix.

Trace and Eigenvalues

The trace of

A

equals the sum of its eigenvalues, counted with algebraic multiplicity:

\text{tr}(A) = \lambda_1 + \lambda_2 + \cdots + \lambda_n

This follows from the characteristic polynomial. The coefficient of

\lambda^{n-1}

p(\lambda) = \det(A - \lambda I)

(-1)^{n-1}\text{tr}(A)

, and by Vieta's formulas, the sum of the roots of a degree-

n

polynomial equals (up to sign) the coefficient of the

(n-1)

-th power term.

The trace provides a quick consistency check. A

3 \times 3

matrix with diagonal entries

7, -2, 4

has trace

9

. If the eigenvalues are computed as

5, 3, 1

, the sum is

9

— consistent. If the sum does not match the trace, a computation error has occurred.

Determinant and Eigenvalues

The determinant of

A

equals the product of its eigenvalues:

\det(A) = \lambda_1 \cdot \lambda_2 \cdots \lambda_n

This follows from evaluating the characteristic polynomial at

\lambda = 0

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

, and since the roots of

p

are

\lambda_1, \dots, \lambda_n

, the constant term is (up to sign) their product.

The most immediate consequence is that

A

is invertible if and only if no eigenvalue is zero. A single vanishing eigenvalue makes the product zero, collapsing the determinant and rendering

A

singular. Conversely, all eigenvalues nonzero means

\det(A) \neq 0

and

A

is invertible.

Algebraic and Geometric Multiplicity

Every eigenvalue

\lambda

has two multiplicity measures. The algebraic multiplicity

m_a(\lambda)

is the number of times

\lambda

appears as a root of the characteristic polynomial. The geometric multiplicity

m_g(\lambda)

is the dimension of the eigenspace

E_\lambda = \text{Null}(A - \lambda I)

.

These two numbers always satisfy

1 \leq m_g(\lambda) \leq m_a(\lambda)

The geometric multiplicity is at least

1

because the eigenspace must contain at least one nonzero eigenvector. It cannot exceed the algebraic multiplicity — a fact whose proof requires the Jordan normal form or the theory of invariant subspaces.

When

m_g = m_a

for every eigenvalue, the matrix is diagonalizable — there are enough independent eigenvectors to form a basis. When

m_g < m_a

for any eigenvalue, the matrix is defective and cannot be diagonalized.

For

A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}

, the eigenvalue

\lambda = 2

has

m_a = 2

but

m_g = 1

(the eigenspace is one-dimensional, spanned by

(1, 0)

). This matrix is not diagonalizable.

Attribute	Algebraic multiplicity m_a(λ)	Geometric multiplicity m_g(λ)
Definition	number of times λ appears as a root of the characteristic polynomial	dimension of the eigenspace E_λ = Null(A − λI)
Range	1 ≤ m_a(λ) ≤ n	1 ≤ m_g(λ) ≤ m_a(λ)
Computed from	factoring p(λ) = det(A − λI)	row reducing A − λI; counting free variables
Diagonalizability test	total Σ m_a(λ) is always n	A is diagonalizable ⟺ m_g(λ) = m_a(λ) for every λ

Eigenvalues of the Inverse

\lambda

is an eigenvalue of an invertible matrix

A

with eigenvector

\mathbf{v}

, then

1/\lambda

is an eigenvalue of

A^{-1}

with the same eigenvector.

The proof is one line:

A\mathbf{v} = \lambda\mathbf{v}

implies

\mathbf{v} = \lambda A^{-1}\mathbf{v}

, so

A^{-1}\mathbf{v} = (1/\lambda)\mathbf{v}

.

The eigenvalues of

A^{-1}

are the reciprocals of the eigenvalues of

A

, and the eigenvectors are unchanged. This requires

\lambda \neq 0

, which is guaranteed by the invertibility of

A

.

If

A

has eigenvalues

2, -3, 5

, then

A^{-1}

has eigenvalues

1/2, -1/3, 1/5

. The trace of

A^{-1}

1/2 - 1/3 + 1/5 = 11/30

, and

\det(A^{-1}) = 1/(2 \cdot (-3) \cdot 5) = -1/30

Eigenvalues of Powers and Polynomials

A\mathbf{v} = \lambda\mathbf{v}

, then

A^k\mathbf{v} = \lambda^k\mathbf{v}

for every positive integer

k

. The proof is induction:

A^{k+1}\mathbf{v} = A(A^k\mathbf{v}) = A(\lambda^k\mathbf{v}) = \lambda^k A\mathbf{v} = \lambda^{k+1}\mathbf{v}

.

The eigenvectors are preserved; only the eigenvalues change by raising to the

k

-th power.

More generally, if

q(\lambda) = c_0 + c_1\lambda + \cdots + c_m\lambda^m

is any polynomial, then

q(A)

has eigenvalues

q(\lambda_i)

with the same eigenvectors:

q(A)\mathbf{v} = (c_0 I + c_1 A + \cdots + c_m A^m)\mathbf{v} = (c_0 + c_1\lambda + \cdots + c_m\lambda^m)\mathbf{v} = q(\lambda)\mathbf{v}

A

has eigenvalue

3

, then

2A^2 - A + 4I

has eigenvalue

2(9) - 3 + 4 = 19

for the same eigenvector.

Eigenvalue Shifting

Adding a scalar multiple of the identity to

A

shifts every eigenvalue by that scalar while leaving the eigenvectors unchanged.

If

A\mathbf{v} = \lambda\mathbf{v}

, then

(A + cI)\mathbf{v} = A\mathbf{v} + c\mathbf{v} = (\lambda + c)\mathbf{v}

.

The eigenvalues of

A + cI

are

\lambda_1 + c, \lambda_2 + c, \dots, \lambda_n + c

. Scaling works similarly:

cA

has eigenvalues

c\lambda_1, c\lambda_2, \dots, c\lambda_n

with the same eigenvectors.

These operations are useful in practice. Adding

cI

can shift all eigenvalues to be positive (making a matrix positive definite for numerical purposes), or shift a known eigenvalue to zero (making

A - \lambda_0 I

singular, which is exactly how the eigenvalue equation is set up).

Eigenvalues of the Transpose

A matrix

A

and its transpose

A^T

have the same eigenvalues. The characteristic polynomials are identical:

\det(A^T - \lambda I) = \det((A - \lambda I)^T) = \det(A - \lambda I)

The second equality uses transpose invariance of the determinant.

The eigenvectors are generally different. If

\mathbf{v}

is a right eigenvector of

A

(

A\mathbf{v} = \lambda\mathbf{v}

), the corresponding left eigenvector

\mathbf{w}

satisfies

\mathbf{w}^T A = \lambda \mathbf{w}^T

, which is the same as

A^T\mathbf{w} = \lambda\mathbf{w}

. So the left eigenvectors of

A

are the (right) eigenvectors of

A^T

. The eigenvalues match, but the directions are different.

Eigenvalues of Special Matrix Types

The structure of a matrix constrains which eigenvalues are possible.

Diagonal and triangular matrices have their eigenvalues on the diagonal — immediately visible.

Real symmetric matrices have all real eigenvalues, and eigenvectors for distinct eigenvalues are orthogonal. This is the spectral theorem's prerequisite.

Real skew-symmetric matrices have eigenvalues that are zero or purely imaginary — they come in conjugate pairs

\pm bi

.

Orthogonal matrices have eigenvalues on the unit circle:

|\lambda| = 1

. For real orthogonal matrices, real eigenvalues are restricted to

\pm 1

, and complex eigenvalues come in conjugate pairs of modulus

1

.

Idempotent matrices (

A^2 = A

) have eigenvalues satisfying

\lambda^2 = \lambda

, so

\lambda = 0

\lambda = 1

.

Nilpotent matrices (

A^k = 0

) have all eigenvalues equal to zero.

Involutory matrices (

A^2 = I

) have eigenvalues satisfying

\lambda^2 = 1

, so

\lambda = \pm 1

.

Positive definite symmetric matrices have all eigenvalues strictly positive.

Matrix type	Defining property	Eigenvalue constraint
Diagonal / triangular	zeros above or below (or both) the main diagonal	eigenvalues are the diagonal entries (visible by inspection)
Real symmetric	A = A^T	all eigenvalues are real; eigenvectors for distinct λ are orthogonal
Real skew-symmetric	A = −A^T	eigenvalues are zero or purely imaginary, in conjugate pairs ± bi
Orthogonal	A^TA = I	\|λ\| = 1; real eigenvalues are ±1; complex eigenvalues in conjugate pairs on the unit circle
Idempotent	A² = A	λ ∈ {0, 1}
Nilpotent	A^k = 0 for some positive integer k	all eigenvalues equal zero
Involutory	A² = I	λ ∈ {+1, −1}
Positive definite (symmetric)	x^TA x > 0 for every nonzero x, with A symmetric	all eigenvalues strictly positive

Independence of Eigenvectors

Eigenvectors corresponding to distinct eigenvalues are always linearly independent.

The proof proceeds by induction. For a single eigenvector, independence is trivial (one nonzero vector is independent). Suppose

\{\mathbf{v}_1, \dots, \mathbf{v}_{k-1}\}

are independent eigenvectors with distinct eigenvalues. If

c_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k = \mathbf{0}

, multiply both sides by

A

to get

c_1\lambda_1\mathbf{v}_1 + \cdots + c_k\lambda_k\mathbf{v}_k = \mathbf{0}

. Subtract

\lambda_k

times the original equation:

c_1(\lambda_1 - \lambda_k)\mathbf{v}_1 + \cdots + c_{k-1}(\lambda_{k-1} - \lambda_k)\mathbf{v}_{k-1} = \mathbf{0}

. By the induction hypothesis, all coefficients

c_i(\lambda_i - \lambda_k) = 0

. Since the eigenvalues are distinct,

\lambda_i - \lambda_k \neq 0

, forcing

c_i = 0

for all

i < k

. Then the original equation gives

c_k\mathbf{v}_k = \mathbf{0}

, so

c_k = 0

.

The immediate consequence: a matrix with

n

distinct eigenvalues has

n

independent eigenvectors and is automatically diagonalizable. Distinctness of eigenvalues is a sufficient condition for diagonalizability, though not a necessary one.

Similar Matrices and Spectral Invariants

Similar matrices share every spectral property: eigenvalues, algebraic multiplicities, geometric multiplicities, and the characteristic polynomial are all identical.

If

B = P^{-1}AP

and

\mathbf{v}

is an eigenvector of

A

with eigenvalue

\lambda

, then

P^{-1}\mathbf{v}

is an eigenvector of

B

with the same eigenvalue:

B(P^{-1}\mathbf{v}) = P^{-1}AP(P^{-1}\mathbf{v}) = P^{-1}A\mathbf{v} = P^{-1}\lambda\mathbf{v} = \lambda(P^{-1}\mathbf{v})

.

The eigenvalues stay the same; the eigenvectors transform by

P^{-1}

. This is consistent with the interpretation that eigenvalues are properties of the transformation, not of the matrix. Changing the basis changes the matrix and the eigenvector coordinates, but the eigenvalues — the intrinsic scaling factors — are invariant.

The trace, determinant, and rank are all derivable from the eigenvalues, so their invariance under similarity is a corollary of eigenvalue invariance.

Summary: Spectral Properties at a Glance

The trace, determinant, and the seven transformations of A introduced across this page all interact with the eigenvalue spectrum in predictable ways. The table below collects every relationship into a single reference card, pairing each quantity or operation with what happens to the eigenvalues and how the eigenvectors behave.

Quantity / Operation on A	Eigenvalue relationship	Eigenvector behavior
Trace tr(A)	tr(A) = λ₁ + λ₂ + ⋯ + λₙ (with algebraic multiplicity)	—
Determinant det(A)	det(A) = λ₁ · λ₂ ⋯ λₙ; A invertible ⟺ no λᵢ = 0	—
Inverse A⁻¹	eigenvalues become 1 / λᵢ (requires A invertible)	unchanged
Transpose A^T	same eigenvalues λᵢ	generally different — these are the left eigenvectors of A
Power A^k	eigenvalues become λᵢ^k	unchanged
Polynomial q(A)	eigenvalues become q(λᵢ)	unchanged
Shift A + cI	eigenvalues become λᵢ + c	unchanged
Scalar multiple cA	eigenvalues become c · λᵢ	unchanged
Similar P⁻¹AP	same eigenvalues λᵢ (with same algebraic and geometric multiplicities)	transformed by P⁻¹