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CHAPTER 7�EIGENVALUES AND EIGENVECTORS

Elementary Linear Algebra

R. Larson (8 Edition)

7.1 Eigenvalues and Eigenvectors

7.2 Diagonalization

7.3 Symmetric Matrices and Orthogonal Diagonalization

7.4 Applications of Eigenvalues and Eigenvectors

投影片設計製作者

淡江大學 電機系 翁慶昌 教授

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CH 7 Linear Algebra Applied

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Diffusion (p.354) Relative Maxima and Minima (p.375)

Genetics (p.365)

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Population of Rabbits (p.379) Architecture (p.388)

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7.1 Eigenvalues and Eigenvectors

  • Eigenvalue problem:

If A is an n×n matrix, do there exist nonzero vectors x in Rn such that Ax is a scalar multiple of x？

  • Eigenvalue and eigenvector:

A：an n×n matrix

λ：a scalar

x： a nonzero vector in Rn

Eigenvalue

Eigenvector

  • Geometrical Interpretation

Elementary Linear Algebra: Section 7.1, p.348

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  • Ex 1: (Verifying eigenvalues and eigenvectors)

Eigenvalue

Eigenvalue

Eigenvector

Eigenvector

Elementary Linear Algebra: Section 7.1, p.349

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  • Thm 7.1: (The eigenspace of A corresponding to λ)

If A is an n×n matrix with an eigenvalue λ, then the set of all eigenvectors of λ together with the zero vector is a subspace of Rn. This subspace is called the eigenspace of λ .

Pf:

x1 and x2 are eigenvectors corresponding to λ

Elementary Linear Algebra: Section 7.1, p.350

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  • Ex 3: (An example of eigenspaces in the plane)

Find the eigenvalues and corresponding eigenspaces of

If

For a vector on the x-axis

Eigenvalue

Sol:

Elementary Linear Algebra: Section 7.1, p.350

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For a vector on the y-axis

Eigenvalue

Geometrically, multiplying a vector (x, y) in R2 by the matrix A corresponds to a reflection in the y-axis.

The eigenspace corresponding to is the x-axis.

The eigenspace corresponding to is the y-axis.

Elementary Linear Algebra: Section 7.1, p.350

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  • Thm 7.2: (Finding eigenvalues and eigenvectors of a matrix A∈Mn×n )

(1) An eigenvalue of A is a scalar λ such that .

(2) The eigenvectors of A corresponding to λ are the nonzero

solutions of .

  • Characteristic polynomial of A∈Mn×n:
  • Characteristic equation of A:

Let A is an n×n matrix.

If has nonzero solutions iff .

  • Note:

(homogeneous system)

Elementary Linear Algebra: Section 7.1, p.351

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  • Ex 4: (Finding eigenvalues and eigenvectors)

Sol: Characteristic equation:

Elementary Linear Algebra: Section 7.1, p.351

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Elementary Linear Algebra: Section 7.1, p.351

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  • Ex 5: (Finding eigenvalues and eigenvectors)

Find the eigenvalues and corresponding eigenvectors for the matrix A. What is the dimension of the eigenspace of each eigenvalue?

Sol: Characteristic equation:

​

Eigenvalue:

Elementary Linear Algebra: Section 7.1, p.352

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The eigenspace of A corresponding to :

Thus, the dimension of its eigenspace is 2.

Elementary Linear Algebra: Section 7.1, p.352

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  • Notes:

(1) If an eigenvalue λ1 occurs as a multiple root (k times) for the characteristic polynominal, then λ1 has multiplicity k.

(2) The multiplicity of an eigenvalue is greater than or equal to the dimension of its eigenspace.

Elementary Linear Algebra: Section 7.1, p.353

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  • Ex 6：Find the eigenvalues of the matrix A and find a basis

for each of the corresponding eigenspaces.

Sol: Characteristic equation:

Elementary Linear Algebra: Section 7.1, p.353

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is a basis for the eigenspace of A corresponding to

Elementary Linear Algebra: Section 7.1, p.353

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is a basis for the eigenspace of A corresponding to

Elementary Linear Algebra: Section 7.1, p.353

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is a basis for the eigenspace of A corresponding to

Elementary Linear Algebra: Section 7.1, p.353

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  • Thm 7.3: (Eigenvalues of triangular matrices)

If A is an n×n triangular matrix, then its eigenvalues are the entries on its main diagonal.

  • Ex 7: (Finding eigenvalues for diagonal and triangular matrices)

Sol:

Elementary Linear Algebra: Section 7.1, p.354

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  • Eigenvalues and eigenvectors of linear transformations:

Elementary Linear Algebra: Section 7.1, p.355

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  • Ex 8: (Finding eigenvalues and eigenspaces)

Sol:

Elementary Linear Algebra: Section 7.1, p.355

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  • Notes:

Elementary Linear Algebra: Section 7.1, p.355

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Key Learning in Section 7.1

  • Verify eigenvalues and corresponding eigenvectors.
  • Find eigenvalues and corresponding eigenspaces.
  • Use the characteristic equation to find eigenvalues and eigenvectors, and find the eigenvalues and eigenvectors of a triangular matrix.
  • Find the eigenvalues and eigenvectors of a linear transformation.

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Keywords in Section 7.1

  • eigenvalue problem: 特徵值問題
  • eigenvalue: 特徵值
  • eigenvector: 特徵向量
  • characteristic polynomial: 特徵多項式
  • characteristic equation: 特徵方程式
  • eigenspace: 特徵空間
  • multiplicity: 重根數

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7.2 Diagonalization

  • Diagonalization problem:

For a square matrix A, does there exist an invertible matrix P such that P-1AP is diagonal?

  • Diagonalizable matrix:

A square matrix A is called diagonalizable if there exists an invertible matrix P such that P−1AP is a diagonal matrix.

(P diagonalizes A)

  • Notes:

(1) If there exists an invertible matrix P such that , then two square matrices A and B are called similar.

(2) The eigenvalue problem is related closely to the diagonalization problem.

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Elementary Linear Algebra: Section 7.2, p.359

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  • Thm 7.4: (Similar matrices have the same eigenvalues)

If A and B are similar n×n matrices, then they have the same eigenvalues.

Pf:

A and B have the same characteristic polynomial.

Thus A and B have the same eigenvalues.

Elementary Linear Algebra: Section 7.2, p.360

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  • Ex 1: (A diagonalizable matrix)

Sol: Characteristic equation:

Elementary Linear Algebra: Section 7.2, p.359

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  • Notes:

Elementary Linear Algebra: Section 7.2, p.359

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  • Thm 7.5: (Condition for diagonalization)

An n×n matrix A is diagonalizable if and only if

it has n linearly independent eigenvectors.

Pf:

Elementary Linear Algebra: Section 7.2, pp.360-361

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Elementary Linear Algebra: Section 7.2, pp.360-361

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Elementary Linear Algebra: Section 7.2, pp.360-361

Note: If n linearly independent vectors do not exist,

then an n × n matrix A is not diagonalizable.

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  • Ex 4: (A matrix that is not diagonalizable)

Sol: Characteristic equation:

A does not have two (n=2) linearly independent eigenvectors, so A is not diagonalizable.

Elementary Linear Algebra: Section 7.2, p.362

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  • Steps for diagonalizing an n×n square matrix:

Step 2: Let

Step 1: Find n linearly independent eigenvectors

for A with corresponding eigenvalues

Step 3:

Elementary Linear Algebra: Section 7.2, p.362

Note:

The order of the eigenvalues used to form P will determine the order in which the eigenvalues appear on the main diagonal of D.

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  • Ex 5: (Diagonalizing a matrix)

Sol: Characteristic equation:

Elementary Linear Algebra: Section 7.2, p.363

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Elementary Linear Algebra: Section 7.2, p.363

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Elementary Linear Algebra: Section 7.2, p.363

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  • Notes: k is a positive integer

Elementary Linear Algebra: Section 7.2, Addition

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  • Thm 7.6: (Sufficient conditions for diagonalization)

If an n × n matrix A has n distinct eigenvalues, then the corresponding eigenvectors are linearly independent and A is diagonalizable.

Elementary Linear Algebra: Section 7.2, p.364

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  • Ex 7: (Determining whether a matrix is diagonalizable)

Sol: Because A is a triangular matrix,

its eigenvalues are the main diagonal entries.

These three values are distinct, so A is diagonalizable. (Thm.7.6)

Elementary Linear Algebra: Section 7.2, p.364

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  • Ex 8: (Finding a diagonalizing matrix for a linear transformation)

Sol:

Elementary Linear Algebra: Section 7.2, p.365

From Ex. 5, there are three distinct eigenvalues

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so A is diagonalizable. (Thm. 7.6)

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The matrix for T relative to this basis is

Elementary Linear Algebra: Section 7.2, p.365

Thus, the three linearly independent eigenvectors found in Ex. 5

​

can be used to form the basis B. That is

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Key Learning in Section 7.2

  • Find the eigenvalues of similar matrices, determine whether a matrix A is diagonalizable, and find a matrix P such that P‒1 AP is diagonal.
  • Find, for a linear transformation T: V→V a basis B for V such that the matrix T for B relative to is diagonal.

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Keywords in Section 7.2

  • diagonalization problem: 對角化問題
  • diagonalization: 對角化
  • diagonalizable matrix: 可對角化矩陣

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7.3 Symmetric Matrices and Orthogonal Diagonalization

  • Symmetric matrix:

A square matrix A is symmetric if it is equal to its transpose:

  • Ex 1: (Symmetric matrices and nonsymetric matrices)

(symmetric)

(symmetric)

(nonsymmetric)

Elementary Linear Algebra: Section 7.3, p.368

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  • Thm 7.7: (Eigenvalues of symmetric matrices)

If A is an n×n symmetric matrix, then the following properties are true.

(1) A is diagonalizable.

(2) All eigenvalues of A are real.

(3) If λ is an eigenvalue of A with multiplicity k, then λ has k linearly independent eigenvectors. That is, the eigenspace of λ has dimension k.

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Elementary Linear Algebra: Section 7.3, p.368

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  • Ex 2:

Prove that a symmetric matrix is diagonalizable.

Pf: Characteristic equation:

As a quadratic in λ, this polynomial has a discriminant of

Elementary Linear Algebra: Section 7.3, p.369

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The characteristic polynomial of A has two distinct real roots, which implies that A has two distinct real eigenvalues. Thus, A is diagonalizable.

Elementary Linear Algebra: Section 7.3, p.369

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  • Orthogonal matrix:

A square matrix P is called orthogonal if it is invertible and

Elementary Linear Algebra: Section 7.3, p.370

  • Ex 4: (Orthogonal matrices)

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  • Thm 7.8: (Properties of orthogonal matrices)

An n×n matrix P is orthogonal if and only if

its column vectors form an orthogonal set.

Elementary Linear Algebra: Section 7.3, p.370

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  • Ex 5: (An orthogonal matrix)

Sol: If P is a orthogonal matrix, then

Elementary Linear Algebra: Section 7.3, p.371

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Elementary Linear Algebra: Section 7.3, p.371

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  • Thm 7.9: (Properties of symmetric matrices)

Let A be an n×n symmetric matrix. If λ1 and λ2 are distinct eigenvalues of A, then their corresponding eigenvectors x1 and x2 are orthogonal.

Elementary Linear Algebra: Section 7.3, p.372

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  • Sol: Characteristic function

Elementary Linear Algebra: Section 7.3, p.372

  • Ex 6: (Eigenvectors of a symmetric matrix)

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  • Thm 7.10: (Fundamental theorem of symmetric matrices)

Let A be an n×n matrix. Then A is orthogonally diagonalizable and has real eigenvalue if and only if A is symmetric.

  • Orthogonal diagonalization of a symmetric matrix:

Let A be an n×n symmetric matrix.

(1) Find all eigenvalues of A and determine the multiplicity of each.

(2) For each eigenvalue of multiplicity 1, choose a unit eigenvector.

(3) For each eigenvalue of multiplicity k≥2, find a set of k linearly independent eigenvectors. If this set is not orthonormal, apply Gram-Schmidt orthonormalization process.

(4) The composite of steps 2 and 3 produces an orthonormal set of n eigenvectors. Use these eigenvectors to form the columns of P. The matrix will be diagonal.

​

Elementary Linear Algebra: Section 7.3, p.373

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  • Ex 7: (Determining whether a matrix is orthogonally diagonalizable)

Orthogonally

diagonalizable

Symmetric

matrix

Elementary Linear Algebra: Section 7.3, p.373

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  • Ex 9: (Orthogonal diagonalization)

Sol:

Linear Independent

Elementary Linear Algebra: Section 7.3, p.375

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Gram-Schmidt Process:

Elementary Linear Algebra: Section 7.3, p.375

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Key Learning in Section 7.3

  • Recognize, and apply properties of, symmetric matrices.
  • Recognize, and apply properties of, orthogonal matrices.
  • Find an orthogonal matrix P that orthogonally diagonalizes a symmetric matrix A.

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Keywords in Section 7.3

  • symmetric matrix: 對稱矩陣
  • orthogonal matrix: 正交矩陣
  • orthonormal set: 單範正交集
  • orthogonal diagonalization: 正交對角化

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7.4 Applications of Eigenvalues and Eigenvectors

  • Population growth:

The age distribution vector x represents the number of population members in each age class, where

Number in first age class

Number in second age class

Number in nth age class

Multiplying the age transition matrix by the age distribution vector for a specific time period produces the age distribution vector for the next time period. That is,

Lxj = xj+1

Elementary Linear Algebra: Section 7.4, p.378

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  • Ex 1: (A Population Growth Model)

The current age distribution vector is

0 ≤ age < 1

1 ≤ age < 2

2 ≤ age ≤ 3

and the age transition matrix is

After 1 year, the age distribution vector will be

0 ≤ age < 1

1 ≤ age < 2

2 ≤ age ≤ 3

Elementary Linear Algebra: Section 7.4, p.379

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  • Ex 2: (Finding a Stable Age Distribution Vector)

To solve this problem, find an eigenvalue and a corresponding eigenvector x such that Lx = x. The characteristic polynomial of L is

(check this), which implies that the eigenvalues are −1 and 2. Choosing the positive value, let λ=2. Verify that the corresponding eigenvectors are of the form

Elementary Linear Algebra: Section 7.4, p.379

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  • Ex 2: (Finding a Stable Age Distribution Vector)

For example, if t = 2, then the initial age distribution vector is

and the age distribution vector for the next year is

0 ≤ age < 1

1 ≤ age < 2

2 ≤ age ≤ 3

0 ≤ age < 1

1 ≤ age < 2

2 ≤ age ≤ 3

Notice that the ratio of the three age classes is still 16 : 4 : 1, and so the percent of the population in each age class remains the same.

Elementary Linear Algebra: Section 7.4, p.379

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A system of first-order linear differential equations

  • Systems of Linear Differential Equations (Calculus)

​

y1' = a11y1 + a12y2 + . . . + a1nyn

y2' = a21y1 + a22y2 + . . . + a2nyn

​

yn' = an1y1 + an2y2 + . . . + annyn

where each yi is a function of t and . If you let

then the system can be written in matrix form as y' = Ay.

Elementary Linear Algebra: Section 7.4, p.380

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  • Ex 3: (Solving a System of Linear Differential Equations)

Solve the system of linear differential equations.

y1' = 4y1

y2' = −y2

y3' = 2y3

Sol:

From calculus, you know that the solution of the differential equation y' = ky is

y = Cekt

So, the solution of the system is

y1 = C1e4t

y2 = C2e−t

y3 = C3e2t

Elementary Linear Algebra: Section 7.4, p.380

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  • Ex 4: (Solving a System of Linear Differential Equations)

Solve the system of linear differential equations.

y1' = 3y1 + 2y2

y2' = 6y1 − y2

first find a matrix P that diagonalizes .

Verify that the eigenvalues of A are 1 = −3 and 2 = 5, and that the corresponding eigenvectors are p1 = [1 −3]T and p2 = [1 1]T. Diagonalize A using the matrix P whose columns consist of p1 and p2 to obtain

Elementary Linear Algebra: Section 7.4, p.381

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  • Quadratic Forms

Quadratic equation

Quadratic form

Elementary Linear Algebra: Section 7.4, p.382

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  • Ex 5: (Finding the Matrix of the Quadratic Form)
  • 4x2 + 9y2 − 36 = 0

(b) 13x2 − 10xy + 13y2 − 72 = 0

Elementary Linear Algebra: Section 7.4, p.382

Sol:

(a) a = 4, b = 0, and c = 9, so the matrix is

Diagonal matrix (no xy-term)

(b) a = 13, b = −10, and c = 13, so the matrix is

Nondiagonal matrix (xy-term)

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Elementary Linear Algebra: Section 7.4, p.382

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Elementary Linear Algebra: Section 7.4, p.382

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  • Principal Axes Theorem

Elementary Linear Algebra: Section 7.4, p.383

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  • Ex 6: (Rotation of a Conic)

Sol:

eigenvector

eigenvalue

Orthogonal matrix

Elementary Linear Algebra: Section 7.4, p.383

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x1

x2

x1

x2

x1

x2

Elementary Linear Algebra: Section 7.4, p.384

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  • Ex 7: (Rotation of a Conic)

Sol:

eigenvalue

eigenvector

orthogonal matrix

Elementary Linear Algebra: Section 7.4, p.385

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Elementary Linear Algebra: Section 7.4, p.385

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Quadratic equation

Quadratic form

Elementary Linear Algebra: Section 7.4, p.388

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  • Ellipsoid

Elementary Linear Algebra: Section 7.4, p.386

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  • Hyperboloid of One Sheet

Elementary Linear Algebra: Section 7.4, p.386

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  • Hyperboloid of Two Sheet

Elementary Linear Algebra: Section 7.4, p.386

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  • Elliptic Cone

Elementary Linear Algebra: Section 7.4, p.387

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  • Elliptic Paraboloid

Elementary Linear Algebra: Section 7.4, p.387

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  • Hyperbolic Paraboloid

Elementary Linear Algebra: Section 7.4, p.387

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  • Ex 8: (Rotation of a Quadric Surface)

Sol:

Elementary Linear Algebra: Section 7.4, p.388

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Elementary Linear Algebra: Section 7.4, p.388

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Key Learning in Section 7.4

  • Model population growth using an age transition matrix and an age distribution vector, and find a stable age distribution vector.
  • Use a matrix equation to solve a system of first-order linear differential equations.
  • Find the matrix of a quadratic form and use the Principal Axes Theorem to perform a rotation of axes for a conic and a quadric surface.
  • Solve a constrained optimization problem.

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Keywords in Section 7.4

  • population growth: 人口成長
  • age distribution vector: 年齡分佈向量
  • age transition matrix: 年齡轉換矩陣
  • quadratic form: 二次式
  • quadratic equation: 二次方程式
  • principal axes theorem: 主軸定理

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  • Diffusion

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Eigenvalues and eigenvectors are useful for modeling real-life phenomena. For example, consider an experiment to determine the diffusion of a fluid from one flask to another through a permeable membrane and then out of the second flask, researchers determine that the flow rate between flasks is twice the volume of fluid in the first flask and the flow rate out of the second flask is three times the volume of fluid in the second flask, then the system of linear differential equations below, where yi represents the volume of fluid in flask i, models this situation.

y1' = ‒2y1

y2' = 2y1 ‒ 3y2

In Section 7.4, you will use eigenvalues and eigenvectors to solve such systems of linear differential equations. For now, verify that the solution of this system is

y1 = C1e ‒2t

y2 = 2C1e ‒2t + C2e ‒3t .

7.1 Linear Algebra Applied

Elementary Linear Algebra: Section 7.1, p.354

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  • Genetics

​

Genetics is the science of heredity. A mixture of chemistry and biology, genetics attempts to explain hereditary evolution and gene movement between generations based on the deoxyribonucleic acid (DNA) of a species. Research in the area of genetics called population genetics, which focuses on genetic structures of specific populations, is especially popular today. Such research has led to a better understanding of the types of genetic inheritance. For instance, in humans, one type of genetic inheritance is called X–linked inheritance (or sex-linked inheritance), which refers to recessive genes on the X chromosome. Males have one X and one Y chromosome, and females have two X chromosomes. If a male has a defective gene on the X chromosome, then its corresponding trait will be expressed because there is not a normal gene on the Y chromosome to suppress its activity. With females, the trait will not be expressed unless it is present on both X chromosomes, which is rare. This is why inherited diseases or conditions are usually found in males, hence the term sex-linked inheritance. Some of these include hemophilia A, Duchenne muscular dystrophy, red-green color blindness, and hereditary baldness. Matrix eigenvalues and diagonalization can be useful for coming up with mathematical models to describe X-linked inheritance in a given population.

7.2 Linear Algebra Applied

Elementary Linear Algebra: Section 7.2, p.365

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  • Relative Maxima and Minima

​

The Hessian matrix is a symmetric matrix that can be helpful in finding relative maxima and minima of functions of several variables. For a function f of two variables x and y—that is, a surface in R3 —the Hessian matrix has the form

​

​

​

The determinant of this matrix, evaluated at a point for which fx and fy are zero, is the expression used in the Second Partials Test for relative extrema.

​

7.3 Linear Algebra Applied

Elementary Linear Algebra: Section 7.3, p.375

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  • Architecture

​

Some of the world’s most unusual architecture makes use of quadric surfaces. For example, Catedral Metropolitana Nossa Senhora Aparecida, a cathedral located in Brasilia, Brazil, is in the shape of a hyperboloid of one sheet. It was designed by Pritzker Prize winning architect Oscar Niemeyer, and dedicated in 1970. The sixteen identical curved steel columns are intended to represent two hands reaching up to the sky. In the triangular gaps formed by the columns, semitransparent stained glass allows light inside for nearly the entire height of the columns.

7.4 Linear Algebra Applied

Elementary Linear Algebra: Section 7.4, p.388

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