[PPT] - Diagonalisations and ON-Bases Artem Los (arteml@kth.se) February PowerPoint Presentation

SLIDE 1

Diagonalisations and ON-Bases

Artem Los (arteml@kth.se) February 17th, 2017

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 1 / 14

SLIDE 2

Overview

1

What is Diagonalization?

2

Theorems

3

Orthonormal Bases

4

Exam Question

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 2 / 14

SLIDE 3

What is Diagonalization?

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 3 / 14

SLIDE 4

Definition

Diagonalization of a matrix

Consider matrix A. Let e1 . . . en be a basis of its eigenvectors associated with eigenvalues λ1 . . . λn. Then, AP = P    λ1 . . . . . . ... . . . . . . λn    where P is P =   | |

e1

. . .

en

| |  

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 4 / 14

SLIDE 5

Definition

Diagonalization of a matrix

Consider matrix A. Let e1 . . . en be a basis of its eigenvectors associated with eigenvalues λ1 . . . λn. Then, AP = P    λ1 . . . . . . ... . . . . . . λn    where P is P =   | |

e1

. . .

en

| |   We can use it to: Identify shape such as an ellipsoid even if was rotated. Compute A1000.

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 4 / 14

SLIDE 6

Example from previous lesson (part B)

In Eigenvectors (slide 13), we got that T matrix had eigenvalues λ0 and λ0.5, with eigenvectors shown below: 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) Diagonalisations and ON-Bases February 17th, 2017 5 / 14

SLIDE 7

Example from previous lesson (part B)

In Eigenvectors (slide 13), we got that T matrix had eigenvalues λ0 and λ0.5, with eigenvectors shown below: 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)} Problem. Can we now express T as T = PDP−1?

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 5 / 14

SLIDE 8

Theorems

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 6 / 14

SLIDE 9

Theorems

If we can write A as A = PDP−1, then we know that: det A = det D Both A and D have the same eigenvalues rank(A) = rank(D)

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 7 / 14

SLIDE 10

Theorems

If we can write A as A = PDP−1, then we know that: det A = det D Both A and D have the same eigenvalues rank(A) = rank(D) A matrix A is only diagonalizable if it has n unique eigenvectors. Furthermore, the eigenvectors should span up Rn.

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 7 / 14

SLIDE 11

Theorems

If we can write A as A = PDP−1, then we know that: det A = det D Both A and D have the same eigenvalues rank(A) = rank(D) A matrix A is only diagonalizable if it has n unique eigenvectors. Furthermore, the eigenvectors should span up Rn. We can also compute high powers much easier: Ak = PDkP−1

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 7 / 14

SLIDE 12

Example

Problem. Given P, can A be diagonalized? A = 11 6 9 −4

,

P = 2 −1 1 3

Artem Los (arteml@kth.se)

Diagonalisations and ON-Bases February 17th, 2017 8 / 14

SLIDE 13

Example

Problem. Given P, can A be diagonalized? A = 11 6 9 −4

,

P = 2 −1 1 3

Step 1: Use the definition A

x = λ x, i.e. compute A x and check that it’s a multiple of x.

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 8 / 14

SLIDE 14

Example

Problem. Given P, can A be diagonalized? A = 11 6 9 −4

,

P = 2 −1 1 3

Step 1: Use the definition A

x = λ x, i.e. compute A x and check that it’s a multiple of x. 11 6 9 −4 2 1

Artem Los (arteml@kth.se)

Diagonalisations and ON-Bases February 17th, 2017 8 / 14

SLIDE 15

Example

Problem. Given P, can A be diagonalized? A = 11 6 9 −4

,

P = 2 −1 1 3

Step 1: Use the definition A

x = λ x, i.e. compute A x and check that it’s a multiple of x. 11 6 9 −4 2 1

=

28 14

Artem Los (arteml@kth.se)

Diagonalisations and ON-Bases February 17th, 2017 8 / 14

SLIDE 16

Example

Problem. Given P, can A be diagonalized? A = 11 6 9 −4

,

P = 2 −1 1 3

Step 1: Use the definition A

x = λ x, i.e. compute A x and check that it’s a multiple of x. 11 6 9 −4 2 1

=

28 14

= 14

2 1

=

⇒ yes

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 8 / 14

SLIDE 17

Example

Problem. Given P, can A be diagonalized? A = 11 6 9 −4

,

P = 2 −1 1 3

Step 1: Use the definition A

x = λ x, i.e. compute A x and check that it’s a multiple of x. 11 6 9 −4 2 1

=

28 14

= 14

2 1

=

⇒ yes 11 6 9 −4 −1 3

=

7 −21

= −7

−1 3

=

⇒ yes

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 8 / 14

SLIDE 18

Example

Problem. Given P, can A be diagonalized? A = 11 6 9 −4

,

P = 2 −1 1 3

Step 1: Use the definition A

x = λ x, i.e. compute A x and check that it’s a multiple of x. 11 6 9 −4 2 1

=

28 14

= 14

2 1

=

⇒ yes 11 6 9 −4 −1 3

=

7 −21

= −7

−1 3

=

⇒ yes Step 2: Find D (i.e. eigenvalues in the diagonal).

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 8 / 14

SLIDE 19

Example

Problem. Given P, can A be diagonalized? A = 11 6 9 −4

,

P = 2 −1 1 3

Step 1: Use the definition A

x = λ x, i.e. compute A x and check that it’s a multiple of x. 11 6 9 −4 2 1

=

28 14

= 14

2 1

=

⇒ yes 11 6 9 −4 −1 3

=

7 −21

= −7

−1 3

=

⇒ yes Step 2: Find D (i.e. eigenvalues in the diagonal). 14 −7

Artem Los (arteml@kth.se)

Diagonalisations and ON-Bases February 17th, 2017 8 / 14

SLIDE 20

Example

Problem. Given P, can A be diagonalized? A = 11 6 9 −4

,

P = 2 −1 1 3

Step 1: Use the definition A

x = λ x, i.e. compute A x and check that it’s a multiple of x. 11 6 9 −4 2 1

=

28 14

= 14

2 1

=

⇒ yes 11 6 9 −4 −1 3

=

7 −21

= −7

−1 3

=

⇒ yes Step 2: Find D (i.e. eigenvalues in the diagonal). 14 −7

Step 3: Confirm that det A = det D.

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 8 / 14

SLIDE 21

OrthoNormal Bases

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 9 / 14

SLIDE 22

Definition

There are two terms worth distinguishing:

rthogonal - ’perpendicular’, i.e.

a, b are orthogonal if a · b = 0

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 10 / 14

SLIDE 23

Definition

There are two terms worth distinguishing:

rthogonal - ’perpendicular’, i.e.

a, b are orthogonal if a · b = 0

rthonormal - orthogonal and normalized vectors such that

|| a|| = || b|| = 1

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 10 / 14

SLIDE 24

Definition

There are two terms worth distinguishing:

rthogonal - ’perpendicular’, i.e.

a, b are orthogonal if a · b = 0

rthonormal - orthogonal and normalized vectors such that

|| a|| = || b|| = 1 Caution. Orthogonal matrix means that the set of column and row vectors is an orthonormal set.

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 10 / 14

SLIDE 25

Definition

There are two terms worth distinguishing:

rthogonal - ’perpendicular’, i.e.

a, b are orthogonal if a · b = 0

rthonormal - orthogonal and normalized vectors such that

|| a|| = || b|| = 1 Caution. Orthogonal matrix means that the set of column and row vectors is an orthonormal set. Another useful property is that if a matrix is orthogonal, its inverse is the same as its transpose, i.e. A−1 = AT

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 10 / 14

SLIDE 26

Definition

There are two terms worth distinguishing:

rthogonal - ’perpendicular’, i.e.

a, b are orthogonal if a · b = 0

rthonormal - orthogonal and normalized vectors such that

|| a|| = || b|| = 1 Caution. Orthogonal matrix means that the set of column and row vectors is an orthonormal set. Another useful property is that if a matrix is orthogonal, its inverse is the same as its transpose, i.e. A−1 = AT What does this mean in the context of base change and rotation?

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 10 / 14

SLIDE 27

Example

Problem. Convert the vector

x = (4, 3, 5) to base spanned by {1

3(1, −2, 2), 1 3(2, 2, 1), 1 3(−2, 1, 2)}.

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 11 / 14

SLIDE 28

Example

Problem. Convert the vector

x = (4, 3, 5) to base spanned by {1

3(1, −2, 2), 1 3(2, 2, 1), 1 3(−2, 1, 2)}.

Step 1: Use the fact that bi = x · vi

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 11 / 14

SLIDE 29

Example

Problem. Convert the vector

x = (4, 3, 5) to base spanned by {1

3(1, −2, 2), 1 3(2, 2, 1), 1 3(−2, 1, 2)}.

Step 1: Use the fact that bi = x · vi 1 3(4, 3, 5)(1, −2, 2) = 8 3

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 11 / 14

SLIDE 30

Example

Problem. Convert the vector

x = (4, 3, 5) to base spanned by {1

3(1, −2, 2), 1 3(2, 2, 1), 1 3(−2, 1, 2)}.

Step 1: Use the fact that bi = x · vi 1 3(4, 3, 5)(1, −2, 2) = 8 3 1 3(4, 3, 5)(2, 2, 1) = 19 3

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 11 / 14

SLIDE 31

Example

Problem. Convert the vector

x = (4, 3, 5) to base spanned by {1

3(1, −2, 2), 1 3(2, 2, 1), 1 3(−2, 1, 2)}.

Step 1: Use the fact that bi = x · vi 1 3(4, 3, 5)(1, −2, 2) = 8 3 1 3(4, 3, 5)(2, 2, 1) = 19 3 1 3(4, 3, 5)(−2, 1, 2) = 5 3

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 11 / 14

SLIDE 32

Example

Problem. Convert the vector

x = (4, 3, 5) to base spanned by {1

3(1, −2, 2), 1 3(2, 2, 1), 1 3(−2, 1, 2)}.

Step 1: Use the fact that bi = x · vi 1 3(4, 3, 5)(1, −2, 2) = 8 3 1 3(4, 3, 5)(2, 2, 1) = 19 3 1 3(4, 3, 5)(−2, 1, 2) = 5 3 ∴ [ w]B = ( 8

3, 19 3 , 5 3)

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 11 / 14

SLIDE 33

Exam Question

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 12 / 14

SLIDE 34

Exam question

Problem. Show that 1 and −1 are the only eigenvalues of orthogonal matrices.

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 13 / 14

SLIDE 35

Exam question

Problem. Show that 1 and −1 are the only eigenvalues of orthogonal matrices. Prerequisite. xT · x - square of the magnitude of x (expressed using dot product).

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 13 / 14

SLIDE 36

Exam question

Problem. Show that 1 and −1 are the only eigenvalues of orthogonal matrices. Prerequisite. xT · x - square of the magnitude of x (expressed using dot product). If A is orthogonal, then A−1 = AT

Artem Los (arteml@kth.se) Diagonalisations and ON-Bases February 17th, 2017 13 / 14

SLIDE 37

Exam question

Problem. Show that 1 and −1 are the only eigenvalues of orthogonal matrices. Step 1: Show that if let A y = x, then ||A y|| = ||x||