The document summarizes a lesson on diagonalization of matrices. It defines eigenvalues and eigenvectors, provides an example to illustrate the geometric effect of a non-diagonal linear transformation, and outlines the procedure for diagonalizing a matrix by finding its eigenvalues and eigenvectors and arranging them into diagonal and invertible matrices. It also gives an worked example of diagonalizing a specific 2x2 matrix.
You will learn how to evaluate algebraic expressions by substitution.
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You've seen that many quantities are related to each other. However, not all of them are directly related. Now you will explore quantities that vary inversely. In inverse variation, one quantity decreases as the other increases.
You will learn how to evaluate algebraic expressions by substitution.
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Lesson 15: Diagonalization
1. Lesson 15 (S&H, Section 14.5)
Diagonalization
Math 20
October 24, 2007
Announcements
Midterm done. Nice job!
Problem Set 6 assigned today. Due October 31.
OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323)
Prob. Sess.: Sundays 6–7 (SC B-10), Tuesdays 1–2 (SC 116)
2.
3. Outline
Concept Review
Diagonalization
Motivating Example
Procedure
Computations
Uses of Diagonalization
4. Concept Review
Definition
Let A be an n × n matrix. The number λ is called an eigenvalue
of A if there exists a nonzero vector x ∈ Rn such that
Ax = λx. (1)
Every nonzero vector satisfying (1) is called an eigenvector of A
associated with the eigenvalue λ.
5. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by A.
y
x
6. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by A.
y
v1
x
7. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by A.
y
Av1
v1
x
8. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by A.
y
Av1
v1
x
v2
9. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by A.
y
Av1
v2
v1
x
v2
10. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by A.
y
Av1
v2
v1
2e2 x
v2
11. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by A.
y
Av1
v2
v1
2e2 x
v2
12. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by A.
y
Av1
v2
v1
2e2 x
v2
13. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by A.
y
Ae2
Av1
v2
v1
2e2 x
v2
14. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by A.
y
Ae2
Av1
v2
v1
2e2 x
v2
15. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by A.
y
Ae2
Av1
v2
v1
2e2 x
v2
16. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by A.
y
Ae2
Av1
v2
v1
2e2 x
v2
17. Methods
To find the eigenvalues of a matrix A, find the determinant of
A − λI. This will be a polynomial in λ (called the
characteristic polynomial of A, and its roots are the
eigenvalues.
To find the eigenvector(s) of a matrix corresponding to an
eigenvalue λ, do Gaussian Elimination on A − λI.
18. Outline
Concept Review
Diagonalization
Motivating Example
Procedure
Computations
Uses of Diagonalization
19.
20. The fact that we have eigenvectors corresponding to two different
eigenvalues gives us the following:
1 1 1 1 1 1
A = A A = 2 −1
1 −1 1 −1 1 −1
P
1 1 2 0
=
1 −1 0 −1
D
21. The fact that we have eigenvectors corresponding to two different
eigenvalues gives us the following:
1 1 1 1 1 1
A = A A = 2 −1
1 −1 1 −1 1 −1
P
1 1 2 0
=
1 −1 0 −1
D
So we have found a matrix P and a diagonal matrix D such that
AP = PD
22. The fact that we have eigenvectors corresponding to two different
eigenvalues gives us the following:
1 1 1 1 1 1
A = A A = 2 −1
1 −1 1 −1 1 −1
P
1 1 2 0
=
1 −1 0 −1
D
So we have found a matrix P and a diagonal matrix D such that
AP = PD
Since P is invertible (det P = −2), we have
A = PDP−1
23. Diagonalization Procedure
Find the eigenvalues and eigenvectors.
Arrange the eigenvectors in a matrix P and the corresponding
eigenvalues in a diagonal matrix D.
If you have “enough” eigenvectors so that the matrix P is
square and invertible, the original matrix is diagonalizable and
equal to PDP−1 .
24. Outline
Concept Review
Diagonalization
Motivating Example
Procedure
Computations
Uses of Diagonalization
25. Example
Example (Worksheet Problem 1)
Let
0 −2
A= .
−3 1
Find an invertible matrix P and a diagonal matrix D such that
A = PDP−1 .
26. Example
Example (Worksheet Problem 1)
Let
0 −2
A= .
−3 1
Find an invertible matrix P and a diagonal matrix D such that
A = PDP−1 .
Solution
We found that −2 and 3 are the eigenvalues for A. The eigenvalue
1
−2 has an associated eigenvector , and the eigenvalue 3 has
1
−2
eigenvector . Thus
3
1 −2 −2 0
P= D= .
1 3 0 3
28. Example (Worksheet Problem 2)
−7 4
Let B = . Find an invertible matrix P and a diagonal
−9 5
matrix D such that B = PDP−1 .
29. Example (Worksheet Problem 2)
−7 4
Let B = . Find an invertible matrix P and a diagonal
−9 5
matrix D such that B = PDP−1 .
Solution
The characteristic polynomial of B is (λ + 1)2 , which has the
2
double root −1. There is one eigenvector, , but nothing more.
3
So there is no diagonal D which works.
30. Example (Worksheet Problem 3)
0 1
Let B = . Find an invertible matrix P and a diagonal
−1 0
matrix D such that B = PDP−1 .
31. Example (Worksheet Problem 3)
0 1
Let B = . Find an invertible matrix P and a diagonal
−1 0
matrix D such that B = PDP−1 .
Solution
The characteristic polynomial of B is λ2 + 1, which has no real
roots. The eigenvalues are i and −i. We could consider the
complex eigenvectors
i −i
z1 = and z2 =
1 1
but scaling by a complex number is more complicated than it looks.
32. Outline
Concept Review
Diagonalization
Motivating Example
Procedure
Computations
Uses of Diagonalization
33. Geometric effects of powers of matrices
Example
2 0
Let D = . Draw the effect of the linear transformation
0 −1
which is multiplication by Dn .
y
S
x
34. Geometric effects of powers of matrices
Example
2 0
Let D = . Draw the effect of the linear transformation
0 −1
which is multiplication by Dn .
y
S
x
D(S)
35. Geometric effects of powers of matrices
Example
2 0
Let D = . Draw the effect of the linear transformation
0 −1
which is multiplication by Dn .
y
S D2 (S)
x
D(S)
36. Geometric effects of powers of matrices
Example
2 0
Let D = . Draw the effect of the linear transformation
0 −1
which is multiplication by Dn .
y
S D2 (S)
x
D(S) D3 (S)
37. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by An .
y
S
x
38. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by An .
y
S
x
39. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by An .
y
A(S)
S
x
40. Geometric effect of a non-diagonal linear transformation
Example
1/2 3/2
Let A = 3/2 1/2
. Draw the effect of the linear transformation
which is multiplication by An .
y A2 (S)
A(S)
S
x
41. Computing An with diagonalization
Example (Worksheet Problem 4)
1/2 3/2
Let A = 3/2 1/2
. Find A100 .
42. Computing An with diagonalization
Example (Worksheet Problem 4)
1/2 3/2
Let A = 3/2 1/2
. Find A100 .
Solution
1 1 2 0
We know A = PDP−1 , where P = and D = .
1 −1 0 −1
43. Computing An with diagonalization
Example (Worksheet Problem 4)
1/2 3/2
Let A = 3/2 1/2
. Find A100 .
Solution
1 1 2 0
We know A = PDP−1 , where P = and D = .
1 −1 0 −1
Now
An = (PDP−1 )n = (PDP−1 )(PDP−1 ) · · · (PDP−1 )
n
−1 −1 −1
= PD(P P)D(P P) · · · D(P P)DP−1 = PDn P−1
And Dn is easy!
44. Computing An with diagonalization
Example (Worksheet Problem 4)
1/2 3/2
Let A = 3/2 1/2
. Find A100 .
Solution
1 1 2 0
We know A = PDP−1 , where P = and D = .
1 −1 0 −1
Now
An = (PDP−1 )n = (PDP−1 )(PDP−1 ) · · · (PDP−1 )
n
−1 −1 −1
= PD(P P)D(P P) · · · D(P P)DP−1 = PDn P−1
And Dn is easy! So
1 1 1 2100 0 1 1 1 2100 + 1 2100 − 1
A100 = =
2 1 −1 0 1 1 −1 2 2100 − 1 2100 + 1