Eigenvalue Properties: Trace, Determinant & More

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}

in

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

is

(-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

Multiplicity Inequality

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

Multiplicity inequality →

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

A repeated eigenvalue with two directions, then with one

Both matrices have a repeated eigenvalue. The upper one still supplies two independent directions, so geometric multiplicity matches algebraic. The lower one supplies only one: the second direction has been lost and cannot be recovered. That shortfall is the whole meaning of the word defective. Compare the two side by side on the eigenvector explorer.

A defective matrix is precisely one that cannot be diagonalised, which is why this gap is worth naming rather than treating as a curiosity.

Spectral Property Notation

Notation

Spectral Property Notation

Almost every result below is written in one shape, and the part that carries the claim is the part that does not change. Three marks decide how the rest of the page reads.

The letter

\lambda

itself, the enumeration

\lambda_1, \dots, \lambda_n

, the spectrum

\sigma(A)

, the characteristic determinant

\det(A - \lambda I)

, the eigenspace

E_\lambda

and the multiplicity pair

m_a, m_g

are all set out at the characteristic equation. Modulus

|\lambda|

and conjugate pairs belong to complex eigenvalues, and the factorisation

A = PDP^{-1}

to diagonalization.

A\mathbf{v} = \lambda\mathbf{v} \;\Rightarrow\; f(A)\mathbf{v} = f(\lambda)\mathbf{v}

if v is an eigenvector of A with eigenvalue lambda, then v is still an eigenvector after the operation, and only the eigenvalue has moved

The claim sits in the letter that stays put. Both sides carry the same

\mathbf{v}

, and that repetition is the assertion: the operation left the direction alone and rescaled it differently. Readers watch the

\lambda

side, where something visibly happens, and take the fixed

\mathbf{v}

for background.

CasesFive operations on this page fit the shape: the inverse sends

\lambda

to

1/\lambda

, the

k

-th power to

\lambda^k

, a polynomial to

q(\lambda)

, a shift to

\lambda + c

, a scaling to

c\lambda

. The arrow is also read one vector at a time. It says nothing about a vector that was not already an eigenvector of

A

.

Also writtenStated in words as

A

and

f(A)

share eigenvectors, or compressed to the spectrum statement

\sigma(f(A)) = f(\sigma(A))

, which drops the eigenvectors from the record entirely.

Do not confuseRead as a general licence for algebra on matrices. Nothing here covers

A + B

or

AB

: the eigenvalues of a sum are not the sums of the eigenvalues, and the same holds for products, because two matrices need not share a single eigenvector. Every operation in the list acts on one matrix, and that restriction is the reason the pattern holds.

q(\lambda) = c_0 + c_1\lambda + \cdots + c_m\lambda^m \;\longrightarrow\; q(A) = c_0 I + c_1 A + \cdots + c_m A^m

the polynomial q evaluated at the matrix A

One letter

q

is fed two different kinds of object: a number on the left, a square matrix on the right. Substitution is mechanical everywhere except at the constant term. A bare

c_0

is a number and cannot be added to a matrix, so it becomes

c_0 I

— the notation

q(A)

performs that swap without announcing it.

CasesExponents on

A

mean repeated matrix multiplication, so

A

must be square. Any two polynomials in the same

A

commute with each other and with

A

, which is why the expansion in the proof may be reordered freely.

Also writtenWritten

p(A)

when the polynomial in question is the characteristic one, where the result is the zero matrix.

Do not confuseApplying

q

to each entry of

A

separately. The two agree for a diagonal matrix apart from the constant term, which lets the mistake pass unnoticed on exactly the examples used to introduce the idea.

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

w transpose A equals lambda times w transpose — w is a left eigenvector of A

The side of the multiplication is the whole distinction. Unqualified, eigenvector always means the right one, with

A

acting from the left on a column. Here the vector stands to the left of

A

, so it has to be a row, and the

T

is what makes the product legal rather than a statement about

\mathbf{w}

itself.

CasesTransposing the equation gives

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

, so the left eigenvectors of

A

are the ordinary eigenvectors of

A^{T}

, sharing

\lambda

with the right eigenvectors of

A

.

Also writtenRow-vector notation

\mathbf{w}^{T}

is sometimes replaced by

\mathbf{w}^{*}

once complex entries are allowed, where the conjugate travels with the transpose.

Do not confuseTaking equal eigenvalues to mean equal eigenvectors.

A

and

A^{T}

agree on every

\lambda

and generally disagree on the directions — the transpose is the one operation on this page that breaks the fixed-vector shape above.

Same glyph elsewhereTwo transposes sit in the same sentence and do different work: the

T

on

\mathbf{w}

turns a column into a row, while the

T

on

A

names the matrix being studied.

All linear algebra symbols →↑ Eigenvalues and eigenvectors

Eigenvalues of the Inverse

If

\lambda

is an eigenvalue of an invertible matrix

A

with eigenvector

\mathbf{v}

, then the eigenvector is preserved and the eigenvalue inverts:

Eigenvalue of Inverse

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

Eigenvalue of inverse →

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}

is

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

Powers of

A

preserve the eigenvector and raise the eigenvalue:

Eigenvalue of Power

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

Eigenvalue of power →

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:

Eigenvalue of Polynomial

q(A)\mathbf{v} = q(\lambda)\mathbf{v}

Eigenvalue of polynomial →

This follows by expansion:

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}

If

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:

Eigenvalue Shift

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

Eigenvalue shift →

The proof:

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

Special Matrix Eigenvalue Restrictions

\begin{aligned} \text{symmetric (real)}: \;& \lambda \in \mathbb{R} \\ \text{skew-symmetric (real)}: \;& \lambda = 0 \text{ or } \lambda \in i\mathbb{R} \\ \text{orthogonal}: \;&|\lambda|= 1 \\ \text{idempotent}: \;& \lambda \in \{0, 1\} \\ \text{nilpotent}: \;& \lambda = 0 \\ \text{involutory}: \;& \lambda \in \{-1, +1\} \\ \text{positive definite}: \;& \lambda > 0 \end{aligned}

Special matrix eigenvalue restrictions →

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

or

\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:

Independence of Distinct Eigenvectors

\lambda_1, \ldots, \lambda_k \text{ distinct} \Rightarrow \{\mathbf{v}_1, \ldots, \mathbf{v}_k\} \text{ linearly independent}

Independence of distinct eigenvectors →

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.

Distinct eigenvalues, independent directions

The two eigenvectors point along genuinely different lines, and that is guaranteed whenever the eigenvalues differ — no amount of tuning can make eigenvectors from distinct eigenvalues line up. It is this guarantee that makes a matrix with all-distinct eigenvalues automatically diagonalisable. Test it by moving the eigenvalues together on the eigenvector explorer.

Distinct eigenvalues are therefore a sufficient condition for diagonalisability, 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.

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Eigen · summary

Spectral properties at a glance

Nine operations, each linked to the section that derives it. The column worth watching is whether the eigenvectors survive — most operations move the eigenvalues while leaving the directions fixed, and the two exceptions are the informative ones.

09operations

Unconditional4

iv

Powers§ 5

A^k

has eigenvalues

\lambda_i^k

Eigenvectors unchanged.

Unconditional

Requires: $k$ a positive integer

Eigenvectors unchanged. Apply $A$ repeatedly to $\mathbf{v}$ and each pass multiplies by $\lambda$ again. This is why $|\lambda| < 1$ means $A^k \to O$ and $|\lambda| > 1$ means it blows up — the whole of iterative stability sits in this line.

v

Polynomials§ 5

q(A)

has eigenvalues

q(\lambda_i)

Eigenvectors unchanged.

Unconditional

Requires: any polynomial $q$

Eigenvectors unchanged. Generalises the two entries above — powers and shifts are both special cases. It is also what makes the Cayley–Hamilton theorem believable: the characteristic polynomial sends every eigenvalue to zero.

vi

Shift§ 6

A + cI

has eigenvalues

\lambda_i + c

Eigenvectors unchanged.

Unconditional

Requires: any scalar $c$

Eigenvectors unchanged. Adding a multiple of the identity slides the whole spectrum along the number line, leaving the directions alone — which is what makes shifted power iteration and the shifted QR algorithm work.

vii

Scalar multiple§ 6

cA

has eigenvalues

c\lambda_i

Eigenvectors unchanged.

Unconditional

Requires: any scalar $c$

Eigenvectors unchanged. Contrast the shift directly above: scaling stretches the spectrum about zero, shifting translates it. Neither disturbs the eigenvectors.

Conditional5

i

Trace§ 1

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

Spectral information read off the diagonal without solving anything.

Conditional

Holds when: eigenvalues counted with algebraic multiplicity, over $\mathbb{C}$

Spectral information read off the diagonal without solving anything. Over $\mathbb{R}$ the sum can appear to fail because some eigenvalues live in $\mathbb{C}$ — a rotation has trace $2\cos\theta$ and no real eigenvalues at all. See trace properties.

ii

Determinant§ 2

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

The product rather than the sum.

Conditional

Holds when: with algebraic multiplicity, over $\mathbb{C}$

The product rather than the sum. It gives the invertibility criterion directly: $A$ is invertible exactly when no eigenvalue is zero, since a single zero factor kills the whole product.

iii

Inverse§ 4

A^{-1}

has eigenvalues

1/\lambda_i

Eigenvectors unchanged.

Conditional

Holds when: $A$ invertible — no $\lambda_i = 0$

Eigenvectors unchanged. From $A\mathbf{v} = \lambda\mathbf{v}$ , multiply both sides by $A^{-1}/\lambda$ and the same $\mathbf{v}$ comes back with the reciprocal. The condition is not a technicality: a zero eigenvalue is precisely what makes the inverse fail to exist.

viii

Transpose§ 7

A^{\mathsf{T}}

has the same eigenvalues as

A

Same characteristic polynomial, since (A^T - λ I) = (A - λ I).

Eigenvectors differ

Holds when: always for the eigenvalues
Fails when: for the eigenvectors — generally different
Common error: assuming the eigenvectors carry over because the eigenvalues do

Same characteristic polynomial, since $\det(A^{\mathsf{T}} - \lambda I) = \det(A - \lambda I)$ . But the eigenvectors of $A^{\mathsf{T}}$ are the left eigenvectors of $A$ , a different set. One of only two entries here where the spectrum survives and the directions do not.

WitnessA = [[1, 1], [0, 2]] — eigenvalues 1, 2
eigenvector for λ=1: A → (1, 0), Aᵀ → (1, −1)

ix

Similarity§ 10

P^{-1}AP

has the same eigenvalues, multiplicities included

Both algebraic and geometric multiplicities are preserved, so the entire spectral picture is…

Eigenvectors map

Holds when: $P$ invertible

Both algebraic and geometric multiplicities are preserved, so the entire spectral picture is basis-independent. The eigenvectors are not lost but relabelled: $\mathbf{v}$ becomes $P^{-1}\mathbf{v}$ . That is the whole content of similarity — the same transformation, described from a different basis.

Unconditional

Conditional

Trace§ 1Conditional

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

Holds when: eigenvalues counted with algebraic multiplicity, over $\mathbb{C}$

Spectral information read off the diagonal without solving anything. Over

\mathbb{R}

the sum can appear to fail because some eigenvalues live in

\mathbb{C}

— a rotation has trace

2\cos\theta

and no real eigenvalues at all. See trace properties.

Read the full section

Determinant§ 2Conditional

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

Holds when: with algebraic multiplicity, over $\mathbb{C}$

The product rather than the sum. It gives the invertibility criterion directly:

A

is invertible exactly when no eigenvalue is zero, since a single zero factor kills the whole product.

Read the full section

Inverse§ 4Conditional

A^{-1}

has eigenvalues

1/\lambda_i

Holds when: $A$ invertible — no $\lambda_i = 0$

Eigenvectors unchanged. From

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

, multiply both sides by

A^{-1}/\lambda

and the same

\mathbf{v}

comes back with the reciprocal. The condition is not a technicality: a zero eigenvalue is precisely what makes the inverse fail to exist.

Read the full section

Powers§ 5Unconditional

A^k

has eigenvalues

\lambda_i^k

Requires: $k$ a positive integer

Eigenvectors unchanged. Apply

A

repeatedly to

\mathbf{v}

and each pass multiplies by

\lambda

again. This is why

|\lambda| < 1

means

A^k \to O

and

|\lambda| > 1

means it blows up — the whole of iterative stability sits in this line.

Read the full section

Polynomials§ 5Unconditional

q(A)

has eigenvalues

q(\lambda_i)

Requires: any polynomial $q$

Eigenvectors unchanged. Generalises the two entries above — powers and shifts are both special cases. It is also what makes the Cayley–Hamilton theorem believable: the characteristic polynomial sends every eigenvalue to zero.

Read the full section

Shift§ 6Unconditional

A + cI

has eigenvalues

\lambda_i + c

Requires: any scalar $c$

Eigenvectors unchanged. Adding a multiple of the identity slides the whole spectrum along the number line, leaving the directions alone — which is what makes shifted power iteration and the shifted QR algorithm work.

Read the full section

Scalar multiple§ 6Unconditional

cA

has eigenvalues

c\lambda_i

Requires: any scalar $c$

Eigenvectors unchanged. Contrast the shift directly above: scaling stretches the spectrum about zero, shifting translates it. Neither disturbs the eigenvectors.

Read the full section

Transpose§ 7Eigenvectors differ

A^{\mathsf{T}}

has the same eigenvalues as

A

Holds when: always for the eigenvalues
Fails when: for the eigenvectors — generally different
Common error: assuming the eigenvectors carry over because the eigenvalues do

Same characteristic polynomial, since

\det(A^{\mathsf{T}} - \lambda I) = \det(A - \lambda I)

. But the eigenvectors of

A^{\mathsf{T}}

are the left eigenvectors of

A

, a different set. One of only two entries here where the spectrum survives and the directions do not.

WitnessA = [[1, 1], [0, 2]] — eigenvalues 1, 2
eigenvector for λ=1: A → (1, 0), Aᵀ → (1, −1)

Read the full section

Similarity§ 10Eigenvectors map

P^{-1}AP

has the same eigenvalues, multiplicities included

Holds when: $P$ invertible

Both algebraic and geometric multiplicities are preserved, so the entire spectral picture is basis-independent. The eigenvectors are not lost but relabelled:

\mathbf{v}

becomes

P^{-1}\mathbf{v}

. That is the whole content of similarity — the same transformation, described from a different basis.

Read the full section

The pattern: anything built from

A

by polynomials, inversion or shifting keeps the eigenvectors and transforms the eigenvalues by the same rule. Transposition and similarity are the exceptions — they preserve the spectrum but change or relabel the directions, which is exactly why diagonalization is a statement about a basis rather than about numbers.

Spectral properties at a glance·/linear-algebra/eigen/propertiesLearn Math Class

Eigenvalue Properties FAQ

Do the eigenvalue rules extend to $A + B$ and $AB$ ?

+

No, and this is the boundary worth memorising. Every rule of this kind acts on a single matrix: inverses, powers, polynomials, shifts and scalings all preserve the eigenvector while transforming the eigenvalue. Sums and products of two matrices do not, because

A

and

B

need not share any eigenvector at all.Read more →

Is $q(A)$ the same as applying $q$ to each entry of $A$ ?

+

No. In

q(A)

the powers mean repeated matrix multiplication, which mixes entries throughout, whereas applying

q

entrywise leaves each entry isolated. The two happen to agree on diagonal matrices apart from the constant term, which is unfortunate: the mistake survives exactly the simple examples used to introduce the idea.Read more →

Do $A$ and $A^{T}$ have the same eigenvectors?

+

They share every eigenvalue but generally not the eigenvectors. Transposing preserves the characteristic polynomial, so the spectrum is identical, yet the directions differ. The eigenvectors of

A^{T}

are known as the left eigenvectors of

A

, and the transpose is the one operation here that changes the vector rather than the value.Read more →

Properties of Eigenvalue and Eigenvector

How Eigenvalues Behave Under Matrix Operations

Trace and Eigenvalues

Determinant and Eigenvalues

Algebraic and Geometric Multiplicity

Spectral Property Notation

Spectral Property Notation

Eigenvalues of the Inverse

Eigenvalues of Powers and Polynomials

Eigenvalue Shifting

Eigenvalues of the Transpose

Eigenvalues of Special Matrix Types

Independence of Eigenvectors

Similar Matrices and Spectral Invariants

Summary: Spectral Properties at a Glance

Spectral properties at a glance

Eigenvalue Properties FAQ

Do the eigenvalue rules extend to $A + B$ and $AB$ ?

Is $q(A)$ the same as applying $q$ to each entry of $A$ ?

Do $A$ and $A^{T}$ have the same eigenvectors?

Properties of Eigenvalue and Eigenvector

How Eigenvalues Behave Under Matrix Operations

Trace and Eigenvalues

Determinant and Eigenvalues

Algebraic and Geometric Multiplicity

Spectral Property Notation

Spectral Property Notation

Eigenvalues of the Inverse

Eigenvalues of Powers and Polynomials

Eigenvalue Shifting

Eigenvalues of the Transpose

Eigenvalues of Special Matrix Types

Independence of Eigenvectors

Similar Matrices and Spectral Invariants

Summary: Spectral Properties at a Glance

Spectral properties at a glance

Eigenvalue Properties FAQ

Do the eigenvalue rules extend to A+BA + BA+B and ABABAB?

Is q(A)q(A)q(A) the same as applying qqq to each entry of AAA?

Do AAA and ATA^{T}AT have the same eigenvectors?

Do the eigenvalue rules extend to $A + B$ and $AB$ ?

Is $q(A)$ the same as applying $q$ to each entry of $A$ ?

Do $A$ and $A^{T}$ have the same eigenvectors?