Kuzee
Linear Algebra

Basic Concepts of Eigenvalues and Eigenvectors

The Definitions of Eigenvalues and Eigenvectors

The Definition of an Eigenvalue

Eigenvalue

Let AA be an n-th order square matrix. If there exist a nonzero vector x⃗\vec{x} and a number λ\lambda such that: Ax⃗=λx⃗A\vec{x} = \lambda\vec{x}

then λ\lambda is called an eigenvalue of the matrix AA, and x⃗\vec{x} is called an eigenvector corresponding to the eigenvalue λ\lambda.

The Definition of an Eigenvector

Eigenvector

A nonzero vector x⃗\vec{x} satisfying Ax⃗=λx⃗A\vec{x} = \lambda\vec{x} is called an eigenvector of the matrix AA corresponding to the eigenvalue λ\lambda.

The Eigenspace

Eigenspace

The subspace formed by all the eigenvectors corresponding to the eigenvalue λ\lambda together with the zero vector is called the eigenspace, written EλE_\lambda.

The Characteristic Equation

The Definition of the Characteristic Polynomial

Characteristic Polynomial

The characteristic polynomial of the matrix AA is defined as: fA(λ)=det⁡(λI−A)f_A(\lambda) = \det(\lambda I - A)

where II is the identity matrix, λ\lambda is the variable, and det⁡\det denotes the determinant.

The Characteristic Equation

Characteristic Equation

The equation ∣A−λI∣=0\lvert A - \lambda I\rvert = 0 is called the characteristic equation of the matrix AA.

How to Find the Eigenvalues

Steps:

  1. Write down the characteristic equation ∣A−λI∣=0\lvert A - \lambda I\rvert = 0
  2. Solve the characteristic equation to obtain the eigenvalues
  3. For each eigenvalue, solve the homogeneous system (A−λI)x⃗=0⃗(A - \lambda I)\vec{x} = \vec{0} to obtain the eigenvectors

Example

Example 1: find the eigenvalues and eigenvectors of the matrix A=(2112)A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}.

Solution:

  1. The characteristic equation: ∣A−λI∣=∣2−λ112−λ∣=0\lvert A - \lambda I\rvert = \begin{vmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{vmatrix} = 0
  2. Expanding: (2−λ)2−1=λ2−4λ+3=0(2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = 0
  3. Solving: λ=1\lambda = 1 or λ=3\lambda = 3
  4. For λ=1\lambda = 1:
    • Solve the system (A−I)x⃗=0⃗(A - I)\vec{x} = \vec{0}
    • (1111)(xy)=(00)\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
    • This gives the eigenvector x⃗=(1,−1)\vec{x} = (1, -1)
  5. For λ=3\lambda = 3:
    • Solve the system (A−3I)x⃗=0⃗(A - 3I)\vec{x} = \vec{0}
    • (−111−1)(xy)=(00)\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
    • This gives the eigenvector x⃗=(1,1)\vec{x} = (1, 1)

Properties of Eigenvalues and Eigenvectors

Basic Properties

Property 1: an n-th order matrix has nn eigenvalues (counting multiplicities).

Property 2: the sum of the eigenvalues equals the trace of the matrix: ∑i=1nλi=tr⁡(A)\sum_{i=1}^n \lambda_i = \operatorname{tr}(A)

Property 3: the product of the eigenvalues equals the determinant of the matrix: ∏i=1nλi=∣A∣\prod_{i=1}^n \lambda_i = \lvert A\rvert

Properties of Eigenvectors

Property 1: eigenvectors corresponding to distinct eigenvalues are linearly independent.

Property 2: a nonzero linear combination of eigenvectors corresponding to the same eigenvalue is still an eigenvector for that eigenvalue.

Property 3: the eigenvector is not unique, but the eigenvalue is.

Properties of the Characteristic Polynomial

Property 1: the degree of the characteristic polynomial equals the order of the matrix.

Property 2: the constant term of the characteristic polynomial equals (−1)n∣A∣(-1)^n\lvert A\rvert.

Property 3: the leading coefficient of the characteristic polynomial is (−1)n(-1)^n.

The Multiplicity of an Eigenvalue

Algebraic Multiplicity

Algebraic Multiplicity

The multiplicity with which the eigenvalue λ\lambda appears as a root of the characteristic polynomial is called the algebraic multiplicity.

Geometric Multiplicity

Geometric Multiplicity

The largest number of linearly independent eigenvectors corresponding to the eigenvalue λ\lambda is called the geometric multiplicity.

The Relation between the Multiplicities

Geometric Multiplicity ≤ Algebraic Multiplicity

Geometric multiplicity ≤ algebraic multiplicity

Proof

Proof: the geometric multiplicity equals the dimension of the eigenspace, and the dimension of the eigenspace never exceeds the algebraic multiplicity.

Computing Eigenvalues and Eigenvectors

The Steps

  1. Write down the characteristic equation: ∣A−λI∣=0\lvert A - \lambda I\rvert = 0
  2. Solve for the eigenvalues: solve the characteristic equation to obtain all the eigenvalues
  3. Find the eigenvectors: for each eigenvalue, solve the homogeneous system (A−λI)x⃗=0⃗(A - \lambda I)\vec{x} = \vec{0}
  4. Check the result: verify that Ax⃗=λx⃗A\vec{x} = \lambda\vec{x} holds

Points to Watch

  1. Eigenvectors are nonzero: an eigenvector cannot be the zero vector
  2. Eigenvectors are not unique: a given eigenvalue has infinitely many corresponding eigenvectors
  3. Handling repeated roots: the eigenvectors of a repeated root need special treatment
  4. Complex eigenvalues: the eigenvalues may be complex

Exercises

Exercise 1

Find the eigenvalues and eigenvectors of the matrix A=(3102)A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}.

eigenvalueeigenvectorcharacteristic equationvector

Approach: Follow the steps for finding eigenvalues and eigenvectors.

Steps:

  1. The characteristic equation: ∣A−λI∣=∣3−λ102−λ∣=0\lvert A - \lambda I\rvert = \begin{vmatrix} 3-\lambda & 1 \\ 0 & 2-\lambda \end{vmatrix} = 0
  2. Expanding: (3−λ)(2−λ)=0(3-\lambda)(2-\lambda) = 0
  3. Solving: λ=2\lambda = 2 or λ=3\lambda = 3
  4. For λ=2\lambda = 2:
    • Solve the system (A−2I)x⃗=0⃗(A - 2I)\vec{x} = \vec{0}
    • (1100)(xy)=(00)\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
    • This gives the eigenvector x⃗=(1,−1)\vec{x} = (1, -1)
  5. For λ=3\lambda = 3:
    • Solve the system (A−3I)x⃗=0⃗(A - 3I)\vec{x} = \vec{0}
    • (010−1)(xy)=(00)\begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
    • This gives the eigenvector x⃗=(1,0)\vec{x} = (1, 0)

Answer: the eigenvalues are λ1=2\lambda_1 = 2, λ2=3\lambda_2 = 3; the corresponding eigenvectors are x⃗1=(1,−1)\vec{x}_1 = (1, -1), x⃗2=(1,0)\vec{x}_2 = (1, 0)

Exercise 2

Find the eigenvalues and eigenvectors of the matrix A=(1101)A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.

eigenvalueeigenvectorcharacteristic equationvector

Approach: Follow the steps for finding eigenvalues and eigenvectors.

Steps:

  1. The characteristic equation: ∣A−λI∣=∣1−λ101−λ∣=0\lvert A - \lambda I\rvert = \begin{vmatrix} 1-\lambda & 1 \\ 0 & 1-\lambda \end{vmatrix} = 0
  2. Expanding: (1−λ)2=0(1-\lambda)^2 = 0
  3. Solving: λ=1\lambda = 1 (a double root)
  4. For λ=1\lambda = 1:
    • Solve the system (A−I)x⃗=0⃗(A - I)\vec{x} = \vec{0}
    • (0100)(xy)=(00)\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
    • This gives the eigenvector x⃗=(1,0)\vec{x} = (1, 0)

Answer: the eigenvalue is λ=1\lambda = 1 (a double root); the corresponding eigenvector is x⃗=(1,0)\vec{x} = (1, 0)

Exercise 3

Prove that: eigenvectors corresponding to distinct eigenvalues are linearly independent.

eigenvalueeigenvectorcharacteristic equationvector

Approach: Use a proof by contradiction.

Steps:

  1. Suppose that the eigenvectors x⃗1,x⃗2,…,x⃗k\vec{x}_1, \vec{x}_2, \dots, \vec{x}_k corresponding to the distinct eigenvalues λ1,λ2,…,λk\lambda_1, \lambda_2, \dots, \lambda_k are linearly dependent
  2. Then there exist constants c1,c2,…,ckc_1, c_2, \dots, c_k, not all zero, such that c1x⃗1+c2x⃗2+⋯+ckx⃗k=0⃗c_1\vec{x}_1 + c_2\vec{x}_2 + \dots + c_k\vec{x}_k = \vec{0}
  3. Taking c1≠0c_1 \neq 0, we get x⃗1=−c2c1x⃗2−⋯−ckc1x⃗k\vec{x}_1 = -\frac{c_2}{c_1}\vec{x}_2 - \dots - \frac{c_k}{c_1}\vec{x}_k
  4. Left-multiplying by AA: Ax⃗1=−c2c1Ax⃗2−⋯−ckc1Ax⃗kA\vec{x}_1 = -\frac{c_2}{c_1}A\vec{x}_2 - \dots - \frac{c_k}{c_1}A\vec{x}_k
  5. That is, λ1x⃗1=−c2c1λ2x⃗2−⋯−ckc1λkx⃗k\lambda_1\vec{x}_1 = -\frac{c_2}{c_1}\lambda_2\vec{x}_2 - \dots - \frac{c_k}{c_1}\lambda_k\vec{x}_k
  6. This contradicts the fact that the eigenvalues are distinct

Answer: the proof is complete

Exercise 4

Let the eigenvalues of the matrix AA be 1,2,31, 2, 3. Find ∣A∣\lvert A\rvert and tr⁡(A)\operatorname{tr}(A).

eigenvalueeigenvectorcharacteristic equationdeterminant

Approach: Use the properties of the eigenvalues.

Steps:

  1. The product of the eigenvalues equals the determinant: ∣A∣=1×2×3=6\lvert A\rvert = 1 \times 2 \times 3 = 6
  2. The sum of the eigenvalues equals the trace: tr⁡(A)=1+2+3=6\operatorname{tr}(A) = 1 + 2 + 3 = 6

Answer: ∣A∣=6\lvert A\rvert = 6, tr⁡(A)=6\operatorname{tr}(A) = 6

Exercise 5

Find the eigenvalues and eigenvectors of the matrix A=(0110)A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.

eigenvalueeigenvectorcharacteristic equationvector

Approach: Follow the steps for finding eigenvalues and eigenvectors.

Steps:

  1. The characteristic equation: ∣A−λI∣=∣−λ11−λ∣=0\lvert A - \lambda I\rvert = \begin{vmatrix} -\lambda & 1 \\ 1 & -\lambda \end{vmatrix} = 0
  2. Expanding: λ2−1=0\lambda^2 - 1 = 0
  3. Solving: λ=1\lambda = 1 or λ=−1\lambda = -1
  4. For λ=1\lambda = 1:
    • Solve the system (A−I)x⃗=0⃗(A - I)\vec{x} = \vec{0}
    • (−111−1)(xy)=(00)\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
    • This gives the eigenvector x⃗=(1,1)\vec{x} = (1, 1)
  5. For λ=−1\lambda = -1:
    • Solve the system (A+I)x⃗=0⃗(A + I)\vec{x} = \vec{0}
    • (1111)(xy)=(00)\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
    • This gives the eigenvector x⃗=(1,−1)\vec{x} = (1, -1)

Answer: the eigenvalues are λ1=1\lambda_1 = 1, λ2=−1\lambda_2 = -1; the corresponding eigenvectors are x⃗1=(1,1)\vec{x}_1 = (1, 1), x⃗2=(1,−1)\vec{x}_2 = (1, -1)

Summary

Symbols Used in This Article

SymbolTypePronunciation / NoteMeaning in This Article
λ\lambdaGreek letterLambdaAn eigenvalue of the matrix AA, satisfying Ax⃗=λx⃗A\vec{x} = \lambda\vec{x}
x⃗\vec{x}Mathematical symbolx vectorAn eigenvector corresponding to the eigenvalue (nonzero)
Ax⃗=λx⃗A\vec{x} = \lambda\vec{x}Mathematical symbolA x equals lambda xThe defining relation for eigenvalues and eigenvectors
EλE_\lambdaMathematical symbolE sub lambdaThe eigenspace: all the eigenvectors corresponding to λ\lambda together with the zero vector
fA(λ)=det⁡(λI−A)f_A(\lambda) = \det(\lambda I - A)Mathematical symbolcharacteristic polynomial of AThe characteristic polynomial of the matrix AA
IIMathematical symbolidentity matrixThe identity matrix
∣A−λI∣=0\lvert A - \lambda I\rvert = 0Mathematical symboldeterminant of (A minus lambda I) equals zeroThe characteristic equation
tr⁡(A)\operatorname{tr}(A)Mathematical symboltrace of AThe trace of the matrix, equal to the sum of the eigenvalues ∑i=1nλi\sum_{i=1}^n \lambda_i
∣A∣\lvert A\rvertMathematical symboldeterminant of AThe determinant of the matrix, equal to the product of the eigenvalues ∏i=1nλi\prod_{i=1}^n \lambda_i
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