Eigenvalues and Eigenvectors Artem Los (arteml@kth.se) February - - PowerPoint PPT Presentation

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Eigenvalues and Eigenvectors Artem Los (arteml@kth.se) February - - PowerPoint PPT Presentation

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Eigenvalues and Eigenvectors Artem Los (arteml@kth.se) February 9th, 2017 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 1 / 16 Overview What is an Eigenvalue? 1 Finding eigenvalues and eigenvectors 2 Exam

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Eigenvalues and Eigenvectors

Artem Los (arteml@kth.se) February 9th, 2017

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 1 / 16

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Overview

1

What is an Eigenvalue?

2

Finding eigenvalues and eigenvectors

3

Exam question

4

Martin’s method

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 2 / 16

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What is an Eigenvalue?

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 3 / 16

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Introduction problem

Problem. Given A = 3/4 1/4 1/4 3/4

  • find A1000(2, 1) using A(1, 1) = (1, 1)

and A(1, −1) = 1

2(1, −1).

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 4 / 16

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Introduction problem

Problem. Given A = 3/4 1/4 1/4 3/4

  • find A1000(2, 1) using A(1, 1) = (1, 1)

and A(1, −1) = 1

2(1, −1).

Step 1: Find x1, x2 so that (2, 1) = x1(1, 1) + x2(1, −1). x1 = 3

2 and

x2 = 1

2

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 4 / 16

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Introduction problem

Problem. Given A = 3/4 1/4 1/4 3/4

  • find A1000(2, 1) using A(1, 1) = (1, 1)

and A(1, −1) = 1

2(1, −1).

Step 1: Find x1, x2 so that (2, 1) = x1(1, 1) + x2(1, −1). x1 = 3

2 and

x2 = 1

2

Step 2: A1000(2, 1) = x1A1000(1, 1) + x2A1000(1, −1) =

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 4 / 16

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Introduction problem

Problem. Given A = 3/4 1/4 1/4 3/4

  • find A1000(2, 1) using A(1, 1) = (1, 1)

and A(1, −1) = 1

2(1, −1).

Step 1: Find x1, x2 so that (2, 1) = x1(1, 1) + x2(1, −1). x1 = 3

2 and

x2 = 1

2

Step 2: A1000(2, 1) = x1A1000(1, 1) + x2A1000(1, −1) = = x1(1, 1) + x2 1 2 1000 (1, −1) =

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 4 / 16

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Introduction problem

Problem. Given A = 3/4 1/4 1/4 3/4

  • find A1000(2, 1) using A(1, 1) = (1, 1)

and A(1, −1) = 1

2(1, −1).

Step 1: Find x1, x2 so that (2, 1) = x1(1, 1) + x2(1, −1). x1 = 3

2 and

x2 = 1

2

Step 2: A1000(2, 1) = x1A1000(1, 1) + x2A1000(1, −1) = = x1(1, 1) + x2 1 2 1000 (1, −1) = = x1(1, 1) = (3/2, 3/2)

Based on lecture notes from Olof Heden’s lecture in SF1604 Linear Algebra (in 2014). Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 4 / 16

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Definition

Eigenvalue

The number λ is an eigenvalue of matrix A if A x = λ x for some vector x = 0.

  • x is an eigenvector to A belonging to the eigenvalue λ.

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 5 / 16

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Definition

Eigenvalue

The number λ is an eigenvalue of matrix A if A x = λ x for some vector x = 0.

  • x is an eigenvector to A belonging to the eigenvalue λ.

We can use it to: find A1000 given A. perform diagonalisations

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 5 / 16

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Finding eigenvalues and eigenvectors

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 6 / 16

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Finding eigenvalues and their corresponding eigenvectors

Using the definition, we can deduce the following:

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 7 / 16

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Finding eigenvalues and their corresponding eigenvectors

Using the definition, we can deduce the following: To find eigenvalues. Solve for λ (in the characteristic equation below): det (A − λI) = 0

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 7 / 16

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Finding eigenvalues and their corresponding eigenvectors

Using the definition, we can deduce the following: To find eigenvalues. Solve for λ (in the characteristic equation below): det (A − λI) = 0

  • Eg. we may get (2 − λ)(3 − λ)2 = 0, so the eigenvalues are λ2 and λ3.

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 7 / 16

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Finding eigenvalues and their corresponding eigenvectors

Using the definition, we can deduce the following: To find eigenvalues. Solve for λ (in the characteristic equation below): det (A − λI) = 0

  • Eg. we may get (2 − λ)(3 − λ)2 = 0, so the eigenvalues are λ2 and λ3.

The algebraic multiplicity of λ2 and λ3 are 1 and 2, respectively.

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 7 / 16

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Finding eigenvalues and their corresponding eigenvectors

Using the definition, we can deduce the following: To find eigenvalues. Solve for λ (in the characteristic equation below): det (A − λI) = 0

  • Eg. we may get (2 − λ)(3 − λ)2 = 0, so the eigenvalues are λ2 and λ3.

The algebraic multiplicity of λ2 and λ3 are 1 and 2, respectively. To find corresponding eigenvectors. Solve for v (for λ2 and λ3): (A − λI) v = The dimension of the solution space is the geometric multiplicity.

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 7 / 16

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Example

Problem. Find the eigenvalues and eigenvectors of A = 2 −2 3

  • Artem Los (arteml@kth.se)

Eigenvalues and Eigenvectors February 9th, 2017 8 / 16

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Example

Problem. Find the eigenvalues and eigenvectors of A = 2 −2 3

  • Step 1: Solve for λ in det(A − λI) = 0

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 8 / 16

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Example

Problem. Find the eigenvalues and eigenvectors of A = 2 −2 3

  • Step 1: Solve for λ in det(A − λI) = 0
  • 2 − λ

−2 3 − λ

  • =

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 8 / 16

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Example

Problem. Find the eigenvalues and eigenvectors of A = 2 −2 3

  • Step 1: Solve for λ in det(A − λI) = 0
  • 2 − λ

−2 3 − λ

  • = (2 − λ)(3 − λ) = 0

The roots are λ1 = 2 and λ2 = 3, which are also the eigenvalues

  • f A.

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 8 / 16

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Example

Problem. Find the eigenvalues and eigenvectors of A = 2 −2 3

  • Step 1: Solve for λ in det(A − λI) = 0
  • 2 − λ

−2 3 − λ

  • = (2 − λ)(3 − λ) = 0

The roots are λ1 = 2 and λ2 = 3, which are also the eigenvalues

  • f A. Both of them have algebraic multiplicity of 1, since we

don’t have any double roots, etc.

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 8 / 16

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Example (continued)

Problem. Find the eigenvalues and eigenvectors of A = 2 −2 3

  • Step 2: To find the corresponding eigenvectors, solve (A − λI)

v =

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 9 / 16

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Example (continued)

Problem. Find the eigenvalues and eigenvectors of A = 2 −2 3

  • Step 2: To find the corresponding eigenvectors, solve (A − λI)

v = Case λ = 2 −2 1

  • Artem Los (arteml@kth.se)

Eigenvalues and Eigenvectors February 9th, 2017 9 / 16

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Example (continued)

Problem. Find the eigenvalues and eigenvectors of A = 2 −2 3

  • Step 2: To find the corresponding eigenvectors, solve (A − λI)

v = Case λ = 2 −2 1

  • x2 has to be 0 and x1 can

be any, let’s say 1.

  • v =

1

  • Artem Los (arteml@kth.se)

Eigenvalues and Eigenvectors February 9th, 2017 9 / 16

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Example (continued)

Problem. Find the eigenvalues and eigenvectors of A = 2 −2 3

  • Step 2: To find the corresponding eigenvectors, solve (A − λI)

v = Case λ = 2 −2 1

  • x2 has to be 0 and x1 can

be any, let’s say 1.

  • v =

1

  • Case λ = 3

−1 −2

  • Artem Los (arteml@kth.se)

Eigenvalues and Eigenvectors February 9th, 2017 9 / 16

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Example (continued)

Problem. Find the eigenvalues and eigenvectors of A = 2 −2 3

  • Step 2: To find the corresponding eigenvectors, solve (A − λI)

v = Case λ = 2 −2 1

  • x2 has to be 0 and x1 can

be any, let’s say 1.

  • v =

1

  • Case λ = 3

−1 −2

  • By parametrization of

−x1 − 2x2 = 0, we get:

  • v =

2 −1

  • Artem Los (arteml@kth.se)

Eigenvalues and Eigenvectors February 9th, 2017 9 / 16

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Example (continued)

Problem. Find the eigenvalues and eigenvectors of A = 2 −2 3

  • Step 2: To find the corresponding eigenvectors, solve (A − λI)

v = Case λ = 2 −2 1

  • x2 has to be 0 and x1 can

be any, let’s say 1.

  • v =

1

  • Case λ = 3

−1 −2

  • By parametrization of

−x1 − 2x2 = 0, we get:

  • v =

2 −1

  • Geometric multiplicity is 1 in both cases, since the dimension is 1.

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 9 / 16

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Exam question

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 10 / 16

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Exam question

From https://www.kth.se/social/files/5858ed94f276542bb5877d62/tentor.pdf, p. 2 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 11 / 16

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Exam question

From https://www.kth.se/social/files/5858ed94f276542bb5877d62/tentor.pdf, p. 2

Idea. Step 1: Find the T matrix in the standard basis using Martin’s method, i.e. T(1, 0, 0) = . . . , T(0, 1, 0) = . . . , T(0, 0, 1) = . . . .

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 11 / 16

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Exam question

From https://www.kth.se/social/files/5858ed94f276542bb5877d62/tentor.pdf, p. 2

Idea. Step 1: Find the T matrix in the standard basis using Martin’s method, i.e. T(1, 0, 0) = . . . , T(0, 1, 0) = . . . , T(0, 0, 1) = . . . . This is optional but good practise.

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 11 / 16

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Exam question

From https://www.kth.se/social/files/5858ed94f276542bb5877d62/tentor.pdf, p. 2

Idea. Step 1: Find the T matrix in the standard basis using Martin’s method, i.e. T(1, 0, 0) = . . . , T(0, 1, 0) = . . . , T(0, 0, 1) = . . . . This is optional but good practise. Step 2: Find the eigenvalues using det (A − λI) = 0

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 11 / 16

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Exam question

From https://www.kth.se/social/files/5858ed94f276542bb5877d62/tentor.pdf, p. 2

Idea. Step 1: Find the T matrix in the standard basis using Martin’s method, i.e. T(1, 0, 0) = . . . , T(0, 1, 0) = . . . , T(0, 0, 1) = . . . . This is optional but good practise. Step 2: Find the eigenvalues using det (A − λI) = 0 Step 3: Find the eigenvectors using (A − λI) v = 0 for each unique eigenvalue.

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 11 / 16

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Exam question (continued)

Step 1: The standard matrix T is given by: T =   −2

1 2 3 2

 

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 12 / 16

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Exam question (continued)

Step 1: The standard matrix T is given by: T =   −2

1 2 3 2

  Step 2: Find the eigenvalues using det (A − λI) = 0

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 12 / 16

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Exam question (continued)

Step 1: The standard matrix T is given by: T =   −2

1 2 3 2

  Step 2: Find the eigenvalues using det (A − λI) = 0

  • −λ

−2

1 2 − λ 3 2

−λ

  • =

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 12 / 16

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Exam question (continued)

Step 1: The standard matrix T is given by: T =   −2

1 2 3 2

  Step 2: Find the eigenvalues using det (A − λI) = 0

  • −λ

−2

1 2 − λ 3 2

−λ

  • = − λ
  • 1

2 − λ 3 2

−λ

  • =

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 12 / 16

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Exam question (continued)

Step 1: The standard matrix T is given by: T =   −2

1 2 3 2

  Step 2: Find the eigenvalues using det (A − λI) = 0

  • −λ

−2

1 2 − λ 3 2

−λ

  • = − λ
  • 1

2 − λ 3 2

−λ

  • = λ2(0.5 − λ) = 0

∴ λ = 0, λ = 0.5. Keep in mind the double root for λ0.

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 12 / 16

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Exam question (continued)

Step 1: The standard matrix T is given by: T =   −2

1 2 3 2

  Step 2: Find the eigenvalues using det (A − λI) = 0

  • −λ

−2

1 2 − λ 3 2

−λ

  • = − λ
  • 1

2 − λ 3 2

−λ

  • = λ2(0.5 − λ) = 0

∴ λ = 0, λ = 0.5. Keep in mind the double root for λ0. Step 3: Find the eigenvectors using (A − λI) v = 0 for λ0 and λ0.5.

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 12 / 16

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Exam question (continued)

Step 1: The standard matrix T is given by: T =   −2

1 2 3 2

  Step 2: Find the eigenvalues using det (A − λI) = 0

  • −λ

−2

1 2 − λ 3 2

−λ

  • = − λ
  • 1

2 − λ 3 2

−λ

  • = λ2(0.5 − λ) = 0

∴ λ = 0, λ = 0.5. Keep in mind the double root for λ0. Step 3: Find the eigenvectors using (A − λI) v = 0 for λ0 and λ0.5. λ0 :   −2 0.5 1.5   λ0.5 :   −0.5 −2 1.5 −0.5  

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 12 / 16

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Exam question (continued)

Step 3: Find the eigenvectors using (A − λI) v = 0 for λ0 and λ0.5. λ0 :   −2 −0.5 1.5   span{(−0.25, 1, 0), (0.75, 0, 1)} λ0.5 :   −0.5 −2 1.5 −0.5   span{(0, 1, 0)}

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 13 / 16

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Exam question (reflections)

The eigenvectors span up a space. λ0 had a two-dimensional eigenvector space, whereas λ0.5 had one-dimensional.

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 14 / 16

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Exam question (reflections)

The eigenvectors span up a space. λ0 had a two-dimensional eigenvector space, whereas λ0.5 had one-dimensional. An alternative approach would be to use the definition of eigenvalues i.e. T a = λ

  • a. We would clearly see that

T(1, 1, 1) = 0 × (1, 1, 1) =

T(1,1,1)

(0, 0, 0) T(0, 2, 0) = 0.5 × (0, 2, 0) =

T(0,2,0)

(0, 1, 0) and that there is no c such that T(0, 1, 1) = c × (0, 1, 1) =

T(0,1,1)

(0, 2, 0)

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 14 / 16

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Martin’s method

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 15 / 16

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Martin’s method

  • Idea. Put all the vectors as rows in the matrix. Those being transformed

are on the left (i.e. v) and the result on the right (i.e. T( v)). Use row

  • perations to get the identify matrix on LHS.

  1 1 1 2 1 1 1 2   ∼ el row op ∼   1 −2 1 0.5 1 1.5   If you transpose what you have on RHS, you have the standard matrix for the transformation. So, T(1, 0, 0) = (0, −2, 0), for example.

Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 16 / 16