Eigenvalue and Eigenvector

An eigenvector of a linear transformation $T$ is a nonzero vector $v$ whose direction is preserved by $T$ : the output $Tv$ is a scalar multiple of $v$ . The corresponding scalar $\lambda$ is the eigenvalue. Eigenvalues reveal the intrinsic scaling behavior of a linear map, independent of any choice of basis.

Definition

Definition5.1Eigenvalue and eigenvector

Let $V$ be a vector space over a field $F$ , and let $T: V \to V$ be a linear transformation. A scalar $\lambda \in F$ is an eigenvalue of $T$ if there exists a nonzero vector $v \in V$ such that:

$T(v) = \lambda v.$

Such a vector $v \neq 0$ is called an eigenvector of $T$ corresponding to the eigenvalue $\lambda$ .

For a matrix $A \in M_{n \times n}(F)$ , this becomes $Av = \lambda v$ , or equivalently $(A - \lambda I)v = 0$ with $v \neq 0$ .

Definition5.2Eigenspace

The eigenspace of $T$ corresponding to an eigenvalue $\lambda$ is:

$E_\lambda = \ker(T - \lambda I) = \{ v \in V : T(v) = \lambda v \}.$

This is a subspace of $V$ (it always contains the zero vector). The dimension $\dim E_\lambda$ is the geometric multiplicity of $\lambda$ .

ExampleEigenvalues of a 2x2 matrix

Let $A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$ . To find eigenvalues, solve $\det(A - \lambda I) = 0$ :

$\det \begin{pmatrix} 3 - \lambda & 1 \\ 0 & 2 - \lambda \end{pmatrix} = (3 - \lambda)(2 - \lambda) = 0.$

So $\lambda_1 = 3$ and $\lambda_2 = 2$ .

For $\lambda_1 = 3$ : $(A - 3I)v = 0$ gives $\begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix} v = 0$ , so $E_3 = \operatorname{span}\{(1, 0)\}$ .

For $\lambda_2 = 2$ : $(A - 2I)v = 0$ gives $\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} v = 0$ , so $E_2 = \operatorname{span}\{(-1, 1)\}$ .

ExampleRotation matrix (no real eigenvalues)

The rotation by $\theta$ in $\mathbb{R}^2$ : $R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ .

The characteristic polynomial is $\lambda^2 - 2\cos\theta \cdot \lambda + 1 = 0$ , with discriminant $4\cos^2\theta - 4 = -4\sin^2\theta$ .

For $\theta \neq 0, \pi$ : the discriminant is negative, so there are no real eigenvalues. Over $\mathbb{C}$ , the eigenvalues are $\lambda = e^{\pm i\theta}$ .
For $\theta = 0$ : $\lambda = 1$ (double), every nonzero vector is an eigenvector.
For $\theta = \pi$ : $\lambda = -1$ (double), $R_\pi = -I$ .

Eigenvalues of specific transformations

ExampleDiagonal matrices

If $A = \operatorname{diag}(d_1, d_2, \ldots, d_n)$ , then the eigenvalues are exactly $d_1, d_2, \ldots, d_n$ , with eigenvectors $e_1, e_2, \ldots, e_n$ (the standard basis vectors).

For $A = \operatorname{diag}(2, -1, 5)$ : eigenvalues $2, -1, 5$ with eigenvectors $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ .

ExampleTriangular matrices

If $A$ is upper (or lower) triangular, the eigenvalues are the diagonal entries. This follows from $\det(A - \lambda I)$ being the product of the diagonal entries of $A - \lambda I$ .

For $A = \begin{pmatrix} 1 & 4 & 7 \\ 0 & 2 & 8 \\ 0 & 0 & 3 \end{pmatrix}$ : eigenvalues are $1, 2, 3$ .

ExampleProjection operators

Let $P: V \to V$ be a projection onto a subspace $W$ along a complement $U$ (so $V = W \oplus U$ and $P|_W = \operatorname{id}$ , $P|_U = 0$ ). Then:

$\lambda = 1$ : eigenspace is $W$ (everything in $W$ is fixed by $P$ ).
$\lambda = 0$ : eigenspace is $U$ (everything in $U$ is killed by $P$ ).
These are the only eigenvalues, since $P^2 = P$ implies $\lambda^2 = \lambda$ , so $\lambda \in \{0, 1\}$ .

ExampleNilpotent operators

A linear map $N: V \to V$ is nilpotent if $N^k = 0$ for some $k \geq 1$ . If $Nv = \lambda v$ with $v \neq 0$ , then $N^k v = \lambda^k v = 0$ , so $\lambda^k = 0$ , hence $\lambda = 0$ .

The only eigenvalue of a nilpotent operator is $0$ . For example, $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ has $N^2 = 0$ , eigenvalue $\lambda = 0$ with eigenspace $\operatorname{span}\{(1, 0)\}$ (geometric multiplicity $1$ , algebraic multiplicity $2$ ).

Key properties

TheoremLinear independence of eigenvectors

Eigenvectors corresponding to distinct eigenvalues are linearly independent. That is, if $\lambda_1, \ldots, \lambda_k$ are distinct eigenvalues of $T$ with eigenvectors $v_1, \ldots, v_k$ , then $v_1, \ldots, v_k$ are linearly independent.

ProofProof by induction

Base case $k = 1$ : a single eigenvector is nonzero, hence linearly independent.

Inductive step: suppose $c_1 v_1 + \cdots + c_k v_k = 0$ . Apply $T$ : $c_1 \lambda_1 v_1 + \cdots + c_k \lambda_k v_k = 0$ . Subtract $\lambda_k$ times the original: $c_1(\lambda_1 - \lambda_k)v_1 + \cdots + c_{k-1}(\lambda_{k-1} - \lambda_k)v_{k-1} = 0$ . By the inductive hypothesis, $c_i(\lambda_i - \lambda_k) = 0$ for $i = 1, \ldots, k-1$ . Since $\lambda_i \neq \lambda_k$ , we get $c_i = 0$ . Then $c_k v_k = 0$ gives $c_k = 0$ .

■

ExampleThree distinct eigenvalues in R^3

Let $A = \begin{pmatrix} 2 & 0 & 0 \\ 1 & 3 & 0 \\ 0 & 0 & -1 \end{pmatrix}$ . The eigenvalues (diagonal of a triangular matrix) are $2, 3, -1$ .

Computing eigenvectors: $v_1 = (-1, 1, 0)$ for $\lambda = 2$ ; $v_2 = (0, 1, 0)$ for $\lambda = 3$ ; $v_3 = (0, 0, 1)$ for $\lambda = -1$ .

These three eigenvectors are linearly independent (they form a basis of $\mathbb{R}^3$ ), confirming the theorem.

Spectrum and trace/determinant relations

Definition5.3Spectrum

The spectrum of $T$ (or of the matrix $A$ ) is the set of all eigenvalues:

$\sigma(T) = \{ \lambda \in F : T - \lambda I \text{ is not invertible} \}.$

TheoremTrace and determinant from eigenvalues

If $A \in M_{n \times n}(F)$ has eigenvalues $\lambda_1, \ldots, \lambda_n$ (counted with algebraic multiplicity, over the algebraic closure of $F$ ), then:

$\operatorname{tr}(A) = \lambda_1 + \lambda_2 + \cdots + \lambda_n, \quad \det(A) = \lambda_1 \lambda_2 \cdots \lambda_n.$

ExampleTrace and determinant verification

For $A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}$ : $\operatorname{tr}(A) = 7$ , $\det(A) = 10$ .

Characteristic polynomial: $\lambda^2 - 7\lambda + 10 = (\lambda - 5)(\lambda - 2) = 0$ , so $\lambda_1 = 5, \lambda_2 = 2$ .

Check: $5 + 2 = 7 = \operatorname{tr}(A)$ and $5 \cdot 2 = 10 = \det(A)$ .

Example3x3 verification

For $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ : eigenvalues $1, 2, 3$ .

$\operatorname{tr}(A) = 1 + 2 + 3 = 6$ , $\det(A) = 1 \cdot 2 \cdot 3 = 6$ .

Eigenvalues of special matrices

ExampleIdempotent matrices

$A^2 = A$ implies $\lambda^2 = \lambda$ , so $\lambda \in \{0, 1\}$ .

For $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ : eigenvalues $1, 0$ . For $A = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ : eigenvalues $1, 0$ (check: $\operatorname{tr} = 1$ , $\det = 0$ ).

ExampleInvolutions

$A^2 = I$ implies $\lambda^2 = 1$ , so $\lambda \in \{1, -1\}$ .

For $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ (permutation matrix): $\det(A - \lambda I) = \lambda^2 - 1$ , eigenvalues $1, -1$ , eigenvectors $(1,1)$ and $(1,-1)$ .

ExampleStochastic matrices

A (right) stochastic matrix has nonnegative entries with each row summing to $1$ . Then $A \mathbf{1} = \mathbf{1}$ where $\mathbf{1} = (1, 1, \ldots, 1)^T$ ... actually, for a right stochastic matrix, columns sum to $1$ , so $\mathbf{1}^T A = \mathbf{1}^T$ . For a left stochastic matrix (rows sum to $1$ ): $A \mathbf{1} = \mathbf{1}$ , so $\lambda = 1$ is always an eigenvalue with eigenvector $\mathbf{1}$ .

For $A = \begin{pmatrix} 0.7 & 0.3 \\ 0.4 & 0.6 \end{pmatrix}$ : eigenvalues $\lambda_1 = 1$ (with eigenvector $(3, 4)^T$ after scaling) and $\lambda_2 = 0.3$ .

Eigenvalues under matrix operations

TheoremEigenvalue relationships

Let $A$ have eigenvalue $\lambda$ with eigenvector $v$ . Then:

$A^k$ has eigenvalue $\lambda^k$ with eigenvector $v$ (for any $k \geq 1$ ).
$A + cI$ has eigenvalue $\lambda + c$ with eigenvector $v$ .
If $A$ is invertible, $A^{-1}$ has eigenvalue $\lambda^{-1}$ with eigenvector $v$ .
$p(A)$ has eigenvalue $p(\lambda)$ for any polynomial $p$ .

ExampleEigenvalue of a matrix polynomial

Let $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ with eigenvalues $2, 3$ .

For $p(x) = x^2 - 5x + 6 = (x-2)(x-3)$ : $p(A)$ has eigenvalues $p(2) = 0$ and $p(3) = 0$ . Indeed, $p(A) = A^2 - 5A + 6I = 0$ (this is the Cayley--Hamilton theorem in action).

For $q(x) = x^2 + 1$ : $q(A)$ has eigenvalues $q(2) = 5$ and $q(3) = 10$ .

Geometric and algebraic multiplicity

Definition5.4Algebraic and geometric multiplicity

For an eigenvalue $\lambda$ of $T$ :

The algebraic multiplicity $m_a(\lambda)$ is the multiplicity of $\lambda$ as a root of the characteristic polynomial.
The geometric multiplicity $m_g(\lambda) = \dim E_\lambda = \dim \ker(T - \lambda I)$ .

Always: $1 \leq m_g(\lambda) \leq m_a(\lambda)$ .

ExampleA defective matrix (m_g less than m_a)

$A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ has characteristic polynomial $(\lambda - 2)^2$ , so $\lambda = 2$ with $m_a = 2$ .

But $A - 2I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ , so $\ker(A - 2I) = \operatorname{span}\{(1, 0)\}$ , giving $m_g = 1 < 2 = m_a$ .

This matrix is called defective: it cannot be diagonalized because there are not enough linearly independent eigenvectors.

ExampleEqual multiplicities

$A = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = 3I$ has $\lambda = 3$ with $m_a = 2$ and $m_g = 2$ (every nonzero vector is an eigenvector). This matrix is diagonalizable (it is already diagonal).

Summary

RemarkEigenvalues as fundamental invariants

Eigenvalues encode essential information about a linear transformation:

The trace is the sum of eigenvalues, the determinant is the product.
Invertibility: $A$ is invertible iff $0 \notin \sigma(A)$ .
Nilpotence: $A$ is nilpotent iff $\sigma(A) = \{0\}$ .
The gap between geometric and algebraic multiplicity measures the "defect" of the transformation -- the failure to be diagonalizable.
Eigenvectors for distinct eigenvalues are always linearly independent, which is the foundation for diagonalization.