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)

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

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

27. Checking the Solution
1 −2 −2 0
P= D= .
1 3 0 3
Check this: We have
1 3 2
P−1 = .
5 −1 1
1 1 −2 −2 0 3 2
PDP−1 =
5 1 3 0 3 −1 1
1 1 −2 −6 −4
=
5 1 3 −3 3
1 0 −10 0 −2
= = .
5 −15 5 −3 1

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