MARGINALIZATION (Different learners in Marginalized Group
Eigenvalues - Contd
1. Announcements
Quiz 3 after lecture.
Grades will be updated online (including quiz 3 grades) over
the weekend. Please let me know if you spot any mistakes.
Exam 2 will be on Feb 25 Thurs in class. Details later.
Make-up exams will be given only if there is an excused
absence from the Dean of Students or a Doctor's note about
sudden serious illness. No exceptions on this. Travel
plans/broken alarm clock are unacceptable excuses.
2. Last Class...
Denition
An eigenvector of an n × n matrix A is a NON-ZERO vector x
such that Ax = λx for some scalar λ.
A scalar λ is called an eigenvalue of A if there is a nontrivial (or
nonzero) solution x to Ax = λx; such an x is called an eigenvector
corresponding to λ.
3. Triangular Matrices
Theorem
The eigenvalues of a triangular matrix are the entries on its main
diagonal.
4. Triangular Matrices
Example
1. Let
5 1 9
A = 0 2 3
0 0 6
The eigenvalues of A are 5, 2 and 6.
5. Triangular Matrices
Example
1. Let
5 1 9
A = 0 2 3
0 0 6
The eigenvalues of A are 5, 2 and 6.
2. Let
4 1 9
A= 0 0 3
0 0 6
The eigenvalues of A are 4, 0 and 6.
6. Zero Eigenvalue??
Zero eigenvalue means
1. The equation Ax = 0x has a nontrivial or nonzero solution
7. Zero Eigenvalue??
Zero eigenvalue means
1. The equation Ax = 0x has a nontrivial or nonzero solution
2. This means Ax = 0 has a nontrivial solution.
8. Zero Eigenvalue??
Zero eigenvalue means
1. The equation Ax = 0x has a nontrivial or nonzero solution
2. This means Ax = 0 has a nontrivial solution.
3. This means A is not invertible (or det A = 0) (by invertible
matrix theorem)
9. Zero Eigenvalue??
Zero eigenvalue means
1. The equation Ax = 0x has a nontrivial or nonzero solution
2. This means Ax = 0 has a nontrivial solution.
3. This means A is not invertible (or det A = 0) (by invertible
matrix theorem)
Zero is an eigenvalue of A if and only if A is not invertible
10. Important
If λ is an eigenvalue of a square matrix A, prove that λ2 is an
eigenvalue of A2 .
For any problem of this type, start with the equation that denes
eigenvalue and take it from there.
11. Important
If λ is an eigenvalue of a square matrix A, prove that λ2 is an
eigenvalue of A2 .
For any problem of this type, start with the equation that denes
eigenvalue and take it from there.
Solution: If λ is an eigenvalue of a square matrix A, we have
Ax = λx.
12. Important
If λ is an eigenvalue of a square matrix A, prove that λ2 is an
eigenvalue of A2 .
For any problem of this type, start with the equation that denes
eigenvalue and take it from there.
Solution: If λ is an eigenvalue of a square matrix A, we have
Ax = λx.
Multiply both sides by A. We get A2 x = A(λx).
13. Important
If λ is an eigenvalue of a square matrix A, prove that λ2 is an
eigenvalue of A2 .
For any problem of this type, start with the equation that denes
eigenvalue and take it from there.
Solution: If λ is an eigenvalue of a square matrix A, we have
Ax = λx.
Multiply both sides by A. We get A2 x = A(λx).
This is same as writing A2 x = λ (Ax) since λ is a scalar.
14. Important
If λ is an eigenvalue of a square matrix A, prove that λ2 is an
eigenvalue of A2 .
For any problem of this type, start with the equation that denes
eigenvalue and take it from there.
Solution: If λ is an eigenvalue of a square matrix A, we have
Ax = λx.
Multiply both sides by A. We get A2 x = A(λx).
This is same as writing A2 x = λ (Ax) since λ is a scalar.
Again Ax = λx. So, A2 x = λ (λx) = λ2 x.
15. Important
If λ is an eigenvalue of a square matrix A, prove that λ2 is an
eigenvalue of A2 .
For any problem of this type, start with the equation that denes
eigenvalue and take it from there.
Solution: If λ is an eigenvalue of a square matrix A, we have
Ax = λx.
Multiply both sides by A. We get A2 x = A(λx).
This is same as writing A2 x = λ (Ax) since λ is a scalar.
Again Ax = λx. So, A2 x = λ (λx) = λ2 x. This equation means that
λ2 is an eigenvalue of A2 .
16. Important, see prob 25 sec 5.1
If λ is an eigenvalue of an invertible matrix A, prove that λ−1 is an
eigenvalue of A−1 .
17. Important, see prob 25 sec 5.1
If λ is an eigenvalue of an invertible matrix A, prove that λ−1 is an
eigenvalue of A−1 .
Solution: If λ is an eigenvalue of a square matrix A, we have
Ax = λx.
18. Important, see prob 25 sec 5.1
If λ is an eigenvalue of an invertible matrix A, prove that λ−1 is an
eigenvalue of A−1 .
Solution: If λ is an eigenvalue of a square matrix A, we have
Ax = λx.
Since A is invertible, multiply both sides by A−1 . We get
−1
A A x = A−1 (λx)
I
19. Important, see prob 25 sec 5.1
If λ is an eigenvalue of an invertible matrix A, prove that λ−1 is an
eigenvalue of A−1 .
Solution: If λ is an eigenvalue of a square matrix A, we have
Ax = λx.
Since A is invertible, multiply both sides by A−1 . We get
−1
A A x = A−1 (λx) =⇒ A−1 (λx) = x
I
20. Important, see prob 25 sec 5.1
If λ is an eigenvalue of an invertible matrix A, prove that λ−1 is an
eigenvalue of A−1 .
Solution: If λ is an eigenvalue of a square matrix A, we have
Ax = λx.
Since A is invertible, multiply both sides by A−1 . We get
−1
A A x = A−1 (λx) =⇒ A−1 (λx) = x
I
This is same as writing λ(A−1 x) = x since λ is a scalar.
21. Important, see prob 25 sec 5.1
If λ is an eigenvalue of an invertible matrix A, prove that λ−1 is an
eigenvalue of A−1 .
Solution: If λ is an eigenvalue of a square matrix A, we have
Ax = λx.
Since A is invertible, multiply both sides by A−1 . We get
−1
A A x = A−1 (λx) =⇒ A−1 (λx) = x
I
This is same as writing λ(A−1 x) = x since λ is a scalar.
Since λ = 0 (Why?) we can divide both sides by λ and we get
−1 1
A x = λ x.
22. Important, see prob 25 sec 5.1
If λ is an eigenvalue of an invertible matrix A, prove that λ−1 is an
eigenvalue of A−1 .
Solution: If λ is an eigenvalue of a square matrix A, we have
Ax = λx.
Since A is invertible, multiply both sides by A−1 . We get
−1
A A x = A−1 (λx) =⇒ A−1 (λx) = x
I
This is same as writing λ(A−1 x) = x since λ is a scalar.
Since λ = 0 (Why?) we can divide both sides by λ and we get
−1 1
A x = λ x.
1
Thus λ or λ−1 is an eigenvalue of A−1 .
23. Example 20 section 5.1
Without calculation, nd one eigenvalue and 2 linearly independent
5 5 5
eigenvectors of A = 5 5 5 . Justify your answer.
5 5 5
Solution: What is special about this matrix? Invertible/Not
Invertible?
24. Example 20 section 5.1
Without calculation, nd one eigenvalue and 2 linearly independent
5 5 5
eigenvectors of A = 5 5 5 . Justify your answer.
5 5 5
Solution: What is special about this matrix? Invertible/Not
Invertible?
Clearly not invertible (same rows, columns).
What is an eigenvalue of A?
25. Example 20 section 5.1
Without calculation, nd one eigenvalue and 2 linearly independent
5 5 5
eigenvectors of A = 5 5 5 . Justify your answer.
5 5 5
Solution: What is special about this matrix? Invertible/Not
Invertible?
Clearly not invertible (same rows, columns).
What is an eigenvalue of A?0!!
To nd eigenvectors for this eigenvalue, we look at A − 0I and row
reduce. Or row reduce A.
26. Example 20 section 5.1
We get (do the row reductions yourself)
1 1 1 0
0 0 0 0
0 0 0 0
.
27. Example 20 section 5.1
We get (do the row reductions yourself)
1 1 1 0
0 0 0 0
0 0 0 0
. Thus x1 + x2 + x3 = 0 where x2 and x3 are free. So x1 = −x2 − x3 .
28. Example 20 section 5.1
We get (do the row reductions yourself)
1 1 1 0
0 0 0 0
0 0 0 0
. Thus x1 + x2 + x3 = 0 where x2 and x3 are free. So x1 = −x2 − x3 .
We have
−1 −1
x1 −x2 − x3
x2 = x2 = x2 1 + x3 0
x3 x3 0 1
29. Example 20 section 5.1
We get (do the row reductions yourself)
1 1 1 0
0 0 0 0
0 0 0 0
. Thus x1 + x2 + x3 = 0 where x2 and x3 are free. So x1 = −x2 − x3 .
We have
−1 −1
x1 −x2 − x3
x2 = x2 = x2 1 + x3 0
x3 x3 0 1
Choose x2 = 1 and x3 = 1 (or anything nonzero) and two linearly
independent eigenvectors are
−1 −1
1 , 0
0 1
30. Observations
1. The eigenvalue λ = 0 has 2 linearly independent eigenvectors
31. Observations
1. The eigenvalue λ = 0 has 2 linearly independent eigenvectors
2. We say that this eigenspace is a two-dimensional subspace of
R3 .
32. Observations
1. The eigenvalue λ = 0 has 2 linearly independent eigenvectors
2. We say that this eigenspace is a two-dimensional subspace of
R3 .
3. Examples with 2 linearly independent eigenvectors are very
important for section 5.3 when we do diagonalization and in
dierential equations where eigenvalues repeat.
39. Theorem
Theorem
Eigenvectors corresponding to distinct eigenvalues of an n ×n
matrix A are linearly independent.
40. Next week...
1. How to nd eigenvalues of a 2 × 2 and 3 × 3 matrix?
2. The process of diagonalization (uses eigenvalues and
eigenvectors)
3. Finding complex eigenvalues
4. Quick look at eigenvalues and eigenvectors being used to learn
long-term behavior of a dynamical system.
5. Quiz 4 (last quiz) will be on thurs Feb 18 based on sections
3.3, 5.1 and 5.2.