1. Announcements
Quiz 4 will be on Thurs Feb 18 on sec 3.3, 5.1 and 5.2.
"Curved" quiz grades (3 point curve as I mentioned in the
email) will be posted by today evening.
I have updated the homework set (again!!). I decided to skip
5.5 and start chap 6 after 5.3.
2. Last Week
1. Dened eigenvalues and eigenvectors of a square matrix
3. Last Week
1. Dened eigenvalues and eigenvectors of a square matrix
2. How to test whether a given vector is an eigenvector of a given
matrix (multiply and check whether we get a scalar multiple of
the original vector)
4. Last Week
1. Dened eigenvalues and eigenvectors of a square matrix
2. How to test whether a given vector is an eigenvector of a given
matrix (multiply and check whether we get a scalar multiple of
the original vector)
3. How to test whether a given number λ is an eigenvalue (check
whether the matrix A − λI has linearly dependent columns)
5. Last Week
1. Dened eigenvalues and eigenvectors of a square matrix
2. How to test whether a given vector is an eigenvector of a given
matrix (multiply and check whether we get a scalar multiple of
the original vector)
3. How to test whether a given number λ is an eigenvalue (check
whether the matrix A − λI has linearly dependent columns)
4. How to nd the eigenvectors of a given eigenvalue (row reduce
A − λI and solve for basic in terms of free variables).
6. Last Week
1. Dened eigenvalues and eigenvectors of a square matrix
2. How to test whether a given vector is an eigenvector of a given
matrix (multiply and check whether we get a scalar multiple of
the original vector)
3. How to test whether a given number λ is an eigenvalue (check
whether the matrix A − λI has linearly dependent columns)
4. How to nd the eigenvectors of a given eigenvalue (row reduce
A − λI and solve for basic in terms of free variables).
5. Eigenvalues of any triangular matrix are the entries of the
main diagonal
7. Last Week
1. Dened eigenvalues and eigenvectors of a square matrix
2. How to test whether a given vector is an eigenvector of a given
matrix (multiply and check whether we get a scalar multiple of
the original vector)
3. How to test whether a given number λ is an eigenvalue (check
whether the matrix A − λI has linearly dependent columns)
4. How to nd the eigenvectors of a given eigenvalue (row reduce
A − λI and solve for basic in terms of free variables).
5. Eigenvalues of any triangular matrix are the entries of the
main diagonal
6. Zero is an eigenvalue of A if and only if A is not invertible.
8. Last Week
1. Dened eigenvalues and eigenvectors of a square matrix
2. How to test whether a given vector is an eigenvector of a given
matrix (multiply and check whether we get a scalar multiple of
the original vector)
3. How to test whether a given number λ is an eigenvalue (check
whether the matrix A − λI has linearly dependent columns)
4. How to nd the eigenvectors of a given eigenvalue (row reduce
A − λI and solve for basic in terms of free variables).
5. Eigenvalues of any triangular matrix are the entries of the
main diagonal
6. Zero is an eigenvalue of A if and only if A is not invertible.
7. We did not see how to nd eigenvalues of a general square
matrix.
9. How to nd eigenvalues of any square matrix A?
Idea: An eigenvalue λ is a scalar such that the equation
(A − λI )x = 0 has free variables. This means
1. The matrix A − λI is not invertible or
10. How to nd eigenvalues of any square matrix A?
Idea: An eigenvalue λ is a scalar such that the equation
(A − λI )x = 0 has free variables. This means
1. The matrix A − λI is not invertible or
2. The determinant of the matrix A − λI is zero
11. How to nd eigenvalues of any square matrix A?
Idea: An eigenvalue λ is a scalar such that the equation
(A − λI )x = 0 has free variables. This means
1. The matrix A − λI is not invertible or
2. The determinant of the matrix A − λI is zero
3. Solve the equation det(A − λI ) = 0 for λ
12. How to nd eigenvalues of any square matrix A?
Idea: An eigenvalue λ is a scalar such that the equation
(A − λI )x = 0 has free variables. This means
1. The matrix A − λI is not invertible or
2. The determinant of the matrix A − λI is zero
3. Solve the equation det(A − λI ) = 0 for λ
4. For a 2 × 2 matrix it is easy, we get a quadratic equation, so
factorize and nd λ.
13. How to nd eigenvalues of any square matrix A?
Idea: An eigenvalue λ is a scalar such that the equation
(A − λI )x = 0 has free variables. This means
1. The matrix A − λI is not invertible or
2. The determinant of the matrix A − λI is zero
3. Solve the equation det(A − λI ) = 0 for λ
4. For a 2 × 2 matrix it is easy, we get a quadratic equation, so
factorize and nd λ.
5. For a 3 × 3 matrix, it is not always easy to solve for λ (except
in some carefully constructed matrices)
14. How to nd eigenvalues of any square matrix A?
Idea: An eigenvalue λ is a scalar such that the equation
(A − λI )x = 0 has free variables. This means
1. The matrix A − λI is not invertible or
2. The determinant of the matrix A − λI is zero
3. Solve the equation det(A − λI ) = 0 for λ
4. For a 2 × 2 matrix it is easy, we get a quadratic equation, so
factorize and nd λ.
5. For a 3 × 3 matrix, it is not always easy to solve for λ (except
in some carefully constructed matrices)
6. For a 4 × 4 and larger matrices, use of
calculator/software/approximate numerical schemes are
prefered.
16. Example 2, section 5.2
Find the eigenvalues of the matrix
5 3
.
3 5
Solution: We have to look at the determinant of the matrix
5 3 1 0 5 3 λ 0 5−λ 3
−λ = − = .
3 5 0 1 3 5 0 λ 3 5−λ
17. Example 2, section 5.2
Find the eigenvalues of the matrix
5 3
.
3 5
Solution: We have to look at the determinant of the matrix
5 3 1 0 5 3 λ 0 5−λ 3
−λ = − = .
3 5 0 1 3 5 0 λ 3 5−λ
In other words, form a matrix where you subtract λ from the
diagonal elements (no change to the o diagonal elements). This is
the case with any square matrix of any size.
18. Example 2, section 5.2
Let us look at the determinant of the new matrix
5−λ 3
= (5 − λ )2 − 9 .
3 5−λ
Simplify this quantity. Some high school algebra will be handy.
19. Example 2, section 5.2
Let us look at the determinant of the new matrix
5−λ 3
= (5 − λ )2 − 9 .
3 5−λ
Simplify this quantity. Some high school algebra will be handy.
(5 − λ)2 = 25 − 10λ + λ2 .
20. Example 2, section 5.2
Let us look at the determinant of the new matrix
5−λ 3
= (5 − λ )2 − 9 .
3 5−λ
Simplify this quantity. Some high school algebra will be handy.
(5 − λ)2 = 25 − 10λ + λ2 .
(5 − λ)2 − 9 = 25 − 10λ + λ2 − 9 = 16 − 10λ + λ2 .
21. Example 2, section 5.2
Let us look at the determinant of the new matrix
5−λ 3
= (5 − λ )2 − 9 .
3 5−λ
Simplify this quantity. Some high school algebra will be handy.
(5 − λ)2 = 25 − 10λ + λ2 .
(5 − λ)2 − 9 = 25 − 10λ + λ2 − 9 = 16 − 10λ + λ2 .
This quantity must be equal to zero. In other words,
λ2 − 10λ + 16 = 0.
22. Example 2, section 5.2
Let us look at the determinant of the new matrix
5−λ 3
= (5 − λ )2 − 9 .
3 5−λ
Simplify this quantity. Some high school algebra will be handy.
(5 − λ)2 = 25 − 10λ + λ2 .
(5 − λ)2 − 9 = 25 − 10λ + λ2 − 9 = 16 − 10λ + λ2 .
This quantity must be equal to zero. In other words,
λ2 − 10λ + 16 = 0.
Factorize this and we get,
(λ − 8)(λ − 2) = 0.
23. Example 2, section 5.2
Let us look at the determinant of the new matrix
5−λ 3
= (5 − λ )2 − 9 .
3 5−λ
Simplify this quantity. Some high school algebra will be handy.
(5 − λ)2 = 25 − 10λ + λ2 .
(5 − λ)2 − 9 = 25 − 10λ + λ2 − 9 = 16 − 10λ + λ2 .
This quantity must be equal to zero. In other words,
λ2 − 10λ + 16 = 0.
Factorize this and we get,
(λ − 8)(λ − 2) = 0.
Thus
λ = 8, λ = 2
are the 2 eigenvalues.
24. Comments
1. The equation det(A − λI ) = 0 is called the characteristic
equation of A.
25. Comments
1. The equation det(A − λI ) = 0 is called the characteristic
equation of A.
2. The polynomial involving λ you get after computing the
determinant (and simplifying) in step 1 is called the
characteristic polynomial of A.
26. Comments
1. The equation det(A − λI ) = 0 is called the characteristic
equation of A.
2. The polynomial involving λ you get after computing the
determinant (and simplifying) in step 1 is called the
characteristic polynomial of A.
3. Thus the characteristic polynomial of A is a quadratic equation
if A is a 2 × 2 matrix, a cubic equation if A is a 3 × 3 matrix etc.
27. Comments
1. The equation det(A − λI ) = 0 is called the characteristic
equation of A.
2. The polynomial involving λ you get after computing the
determinant (and simplifying) in step 1 is called the
characteristic polynomial of A.
3. Thus the characteristic polynomial of A is a quadratic equation
if A is a 2 × 2 matrix, a cubic equation if A is a 3 × 3 matrix etc.
4. The characteristic polynomial may not be nicely factorizable in
all cases. In that case you may need your high school
quadratic formula.
28. Back to High School
If you have a quadratic equation
2
ax + bx + c = 0,
the 2 roots (values of x that solve the above equation) are given by
29. Back to High School
If you have a quadratic equation
2
ax + bx + c = 0,
the 2 roots (values of x that solve the above equation) are given by
−b + b 2 − 4ac
2a
and
−b − b 2 − 4ac
2a
This works for any quadratic equation. If you cannot gure out the
factorization immediately, this is a safe option. (provided you won't
make mistakes of course).
30. Trace of a Matrix
Denition
Trace of any square matrix A is the sum of the diagonal elements
of A. It is denoted by tr A.
31. Trace of a Matrix
Denition
Trace of any square matrix A is the sum of the diagonal elements
of A. It is denoted by tr A.
Why is the trace useful? The trace of any square matrix is equal to
the sum of the eigenvalues of A (irrespective of the size of A).
Thus, it is a good check to your eigenvalue calculations.
32. Trace of a Matrix
Denition
Trace of any square matrix A is the sum of the diagonal elements
of A. It is denoted by tr A.
Why is the trace useful? The trace of any square matrix is equal to
the sum of the eigenvalues of A (irrespective of the size of A).
Thus, it is a good check to your eigenvalue calculations.
The product of the eigenvalues of A = det A.
33. How many eigenvalues?
A quadratic equation has exactly 2 roots, if complex roots are
included. (could be the same root repeated)
34. How many eigenvalues?
A quadratic equation has exactly 2 roots, if complex roots are
included. (could be the same root repeated)
A cubic equation has exactly 3 roots, if complex roots are included
(one or more roots could be repeated)
35. How many eigenvalues?
A quadratic equation has exactly 2 roots, if complex roots are
included. (could be the same root repeated)
A cubic equation has exactly 3 roots, if complex roots are included
(one or more roots could be repeated)
In general, a square matrix of size n × n will have exactly n
eigenvalues.
36. Example
Find the characteristic equation, characteristic polynomial and the
eigenvalues of the matrix
5 3 7 9
0 5 9 1
A= .
0 0 2 8
0 0 0 2
37. Example
Find the characteristic equation, characteristic polynomial and the
eigenvalues of the matrix
5 3 7 9
0 5 9 1
A= .
0 0 2 8
0 0 0 2
Observe that this is a triangular matrix (and hence convenient to
work with). The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
38. Example
Find the characteristic equation, characteristic polynomial and the
eigenvalues of the matrix
5 3 7 9
0 5 9 1
A= .
0 0 2 8
0 0 0 2
Observe that this is a triangular matrix (and hence convenient to
work with). The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
5−λ 3 7 9
0 5−λ 9 1
= 0.
0 0 2−λ 8
0 0 0 2−λ
39. Example Contd.
Since A − λI is a triangular matrix, we have
(5 − λ)2 (2 − λ)2 = 0
40. Example Contd.
Since A − λI is a triangular matrix, we have
(5 − λ)2 (2 − λ)2 = 0
This is the characteristic equation. To nd the char. polynomial,
we have to expand both terms
(25 − 10λ + λ2 )(4 − 4λ + λ2 )
41. Example Contd.
Since A − λI is a triangular matrix, we have
(5 − λ)2 (2 − λ)2 = 0
This is the characteristic equation. To nd the char. polynomial,
we have to expand both terms
(25 − 10λ + λ2 )(4 − 4λ + λ2 )
=⇒ 100 − 100λ + 25λ2 − 40λ + 40λ2 −10λ3 + 4λ2 −4λ3 + λ4
42. Example Contd.
Since A − λI is a triangular matrix, we have
(5 − λ)2 (2 − λ)2 = 0
This is the characteristic equation. To nd the char. polynomial,
we have to expand both terms
(25 − 10λ + λ2 )(4 − 4λ + λ2 )
=⇒ 100 − 100λ + 25λ2 − 40λ + 40λ2 −10λ3 + 4λ2 −4λ3 + λ4
=⇒ 100 − 140λ + 69λ2 −14λ3 + λ4
43. Example Contd.
What about the eigenvalues of A? Since A is triangular, the
diagonal entries are its eigenvalues.
44. Example Contd.
What about the eigenvalues of A? Since A is triangular, the
diagonal entries are its eigenvalues.
Here the eigenvalues are 5, 5, 2 and 2.
45. Example Contd.
What about the eigenvalues of A? Since A is triangular, the
diagonal entries are its eigenvalues.
Here the eigenvalues are 5, 5, 2 and 2.
We say that 5 has multiplicity 2 and 2 has multiplicity 2. In other
words, we have repeated eigenvalues.
46. Example
The characteristic polynomial of a 5 × 5 matrix is
λ5 − 4λ4 − 12λ3 .
Find the eigenvalues and their multiplicities.
47. Example
The characteristic polynomial of a 5 × 5 matrix is
λ5 − 4λ4 − 12λ3 .
Find the eigenvalues and their multiplicities.
Solution: This polynomial (in spite of being of degree 5) can be
factorized easily as follows.
λ3 (λ2 − 4λ − 12) = λ3 (λ − 6)(λ + 2).
48. Example
The characteristic polynomial of a 5 × 5 matrix is
λ5 − 4λ4 − 12λ3 .
Find the eigenvalues and their multiplicities.
Solution: This polynomial (in spite of being of degree 5) can be
factorized easily as follows.
λ3 (λ2 − 4λ − 12) = λ3 (λ − 6)(λ + 2).
Thus we have λ3 = 0 =⇒ λ = 0, λ = 6 and λ = −2.
49. Example
The characteristic polynomial of a 5 × 5 matrix is
λ5 − 4λ4 − 12λ3 .
Find the eigenvalues and their multiplicities.
Solution: This polynomial (in spite of being of degree 5) can be
factorized easily as follows.
λ3 (λ2 − 4λ − 12) = λ3 (λ − 6)(λ + 2).
Thus we have λ3 = 0 =⇒ λ = 0, λ = 6 and λ = −2.
Zero has multiplicity 3 (repeated roots), others have multiplicity 1
each.
50. Example 4, section 5.2
Find the char. polynomial and eigenvalues of the matrix
5 −3
.
−4 3
51. Example 4, section 5.2
Find the char. polynomial and eigenvalues of the matrix
5 −3
.
−4 3
Solution: The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
5 − λ −3
= 0.
−4 3 − λ
52. Example 4, section 5.2
Find the char. polynomial and eigenvalues of the matrix
5 −3
.
−4 3
Solution: The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
5 − λ −3
= 0.
−4 3 − λ
(5 − λ)(3 − λ) − 12 = 0
53. Example 4, section 5.2
Find the char. polynomial and eigenvalues of the matrix
5 −3
.
−4 3
Solution: The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
5 − λ −3
= 0.
−4 3 − λ
(5 − λ)(3 − λ) − 12 = 0
15 − 8λ + λ2 − 12 = 0
3 − 8λ + λ2 = 0 =⇒ λ2 − 8λ + 3 = 0
54. Example 4, section 5.2
Find the char. polynomial and eigenvalues of the matrix
5 −3
.
−4 3
Solution: The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
5 − λ −3
= 0.
−4 3 − λ
(5 − λ)(3 − λ) − 12 = 0
15 − 8λ + λ2 − 12 = 0
3 − 8λ + λ2 = 0 =⇒ λ2 − 8λ + 3 = 0
λ2 − 8λ + 3 is the char.polynomial.
55. Example 4, section 5.2
Use the quadratic formula (factorization will not work here)
56. Example 4, section 5.2
Use the quadratic formula (factorization will not work here)
8 ± 82 − 4(1)(3) 8 ± 52
λ= =
2(1) 2
Since 52=4× 13 we have
8 ± 4.13 8 ± 2 13
λ= = = 4 ± 13
2 2
57. Example 4, section 5.2
Use the quadratic formula (factorization will not work here)
8 ± 82 − 4(1)(3) 8 ± 52
λ= =
2(1) 2
Since 52=4× 13 we have
8 ± 4.13 8 ± 2 13
λ= = = 4 ± 13
2 2
The sum of these eigenvalues is
4 + 13 + 4 − 13 = 8
and the trace of the given matrix is 5+3=8.
58. Example 8, section 5.2
Find the char. polynomial and eigenvalues of the matrix
7 −2
.
2 3
59. Example 8, section 5.2
Find the char. polynomial and eigenvalues of the matrix
7 −2
.
2 3
Solution: The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
7 − λ −2
= 0.
2 3−λ
60. Example 8, section 5.2
Find the char. polynomial and eigenvalues of the matrix
7 −2
.
2 3
Solution: The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
7 − λ −2
= 0.
2 3−λ
(7 − λ)(3 − λ) + 4 = 0
21 − 10λ + λ2 + 4 = 0
61. Example 8, section 5.2
Find the char. polynomial and eigenvalues of the matrix
7 −2
.
2 3
Solution: The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
7 − λ −2
= 0.
2 3−λ
(7 − λ)(3 − λ) + 4 = 0
21 − 10λ + λ2 + 4 = 0
25 − 10λ + λ2 = 0 =⇒ λ2 − 10λ + 25 = 0
λ2 − 10λ + 25 is the char.polynomial. What about eigenvalues? Try
it yourself!!!!
62. Example 10, section 5.2
Find the char. polynomial of the matrix
0 3 1
3 0 2 .
1 2 0
63. Example 10, section 5.2
Find the char. polynomial of the matrix
0 3 1
3 0 2 .
1 2 0
For a 3 × 3 matrix, nding the char equation (and the char
polynomial) is more involved.
You could do a cofactor expansion or the Sarru's mnemonic rule
(where you repeat the rst 2 rows and strike through) of A − λI .
64. Example 10, section 5.2
Find the char. polynomial of the matrix
0 3 1
3 0 2 .
1 2 0
For a 3 × 3 matrix, nding the char equation (and the char
polynomial) is more involved.
You could do a cofactor expansion or the Sarru's mnemonic rule
(where you repeat the rst 2 rows and strike through) of A − λI .
NEVER do a reduction to echeleon form. The eigenvalues of the
echelon form are totally dierent from that of the original matrix.
65. Example 10, section 5.2
Solution: The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
0−λ 3 1
3 0−λ 2 = 0.
1 2 0−λ
66. Example 10, section 5.2
Solution: The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
0−λ 3 1
3 0−λ 2 = 0.
1 2 0−λ
−λ 2 3 2 3 −λ
−λ −3 +1
2 −λ 1 −λ 1 2
λ2 −4 −3λ−2 6+λ
67. Example 10, section 5.2
Solution: The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
0−λ 3 1
3 0−λ 2 = 0.
1 2 0−λ
−λ 2 3 2 3 −λ
−λ −3 +1
2 −λ 1 −λ 1 2
λ2 −4 −3λ−2 6+λ
−λ(λ2 − 4) − 3(−3λ − 2) + 6 + λ = 0
68. Example 10, section 5.2
Solution: The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
0−λ 3 1
3 0−λ 2 = 0.
1 2 0−λ
−λ 2 3 2 3 −λ
−λ −3 +1
2 −λ 1 −λ 1 2
λ2 −4 −3λ−2 6+λ
−λ(λ2 − 4) − 3(−3λ − 2) + 6 + λ = 0
−λ3 + 4λ + 9λ + 6 + 6 + λ = 0 =⇒ −λ3 + 14λ + 12 = 0
−λ3 + 14λ + 12 is the char.polynomial.
69. Example 14, section 5.2
Find the char. polynomial of the matrix
5 −2 3
0 1 0 .
6 7 −2
Solution: The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
5 − λ −2 3
0 1−λ 0 = 0.
6 7 −2 − λ
70. Example 14, section 5.2
Find the char. polynomial of the matrix
5 −2 3
0 1 0 .
6 7 −2
Solution: The characteristic equation is found by solving the
equation det (A − λI ) = 0 which is
5 − λ −2 3
0 1−λ 0 = 0.
6 7 −2 − λ
Expand along second row, we get
5−λ 3
(1 − λ) =0
6 −2 − λ
(5−λ)(−2−λ)−18=−10−3λ+λ2 −18=λ2 −3λ−28
74. Similarity
Denition
Let A and B be 2 square matrices. We say that A is similar to B if
we can nd an invertible matrix P such that
−1
P AP = B
Theorem
If n × n matrices A and B are similar, they have the same char
polynomial and hence the same eigenvalues (with same
multiplicities).
