1. Announcements
Exam 2 on Thurs, Feb 25 in class.
Practice Exam will be uploaded by 2 pm today.
I will do some misc. topics (sec 5.5 and some
applications)tomorrow. These WILL NOT be covered on the
exam but are useful for MA 3521.
Review on Wednesday in class. I will have oce hours on Wed
from 1-4 pm.
Reminder that MA 3521 Di. Eq starts next Monday same
time and place (Prof Todd King)
Please collect your graded exams on
Friday between 7 am and 6 pm in Fisher 214. I would like to
settle any grading issues and submit the grades ocially on
Saturday.
2. Last Week: Orthogonal Projection
The orthogonal projection of y onto any line L through u and 0 is
given by
y u
y = projL y =
ˆ u
u u
The orthogonal projection is a vector (not a number).
The quantity y − y gives the distance between y and the line L.
ˆ
3. Section 6.3 Orthogonal Projections
1. The above formula for orthogonal projection is for R2
2. We can extend this formula to any vector in Rn
4. Section 6.3 Orthogonal Projections
1. The above formula for orthogonal projection is for R2
2. We can extend this formula to any vector in Rn
Consider a vector y and a subspace W in Rn . We can nd the
following
5. Section 6.3 Orthogonal Projections
1. The above formula for orthogonal projection is for R2
2. We can extend this formula to any vector in Rn
Consider a vector y and a subspace W in Rn . We can nd the
following
1. A unique vector y which is a linear combination of vectors in
ˆ
W.
6. Section 6.3 Orthogonal Projections
1. The above formula for orthogonal projection is for R2
2. We can extend this formula to any vector in Rn
Consider a vector y and a subspace W in Rn . We can nd the
following
1. A unique vector y which is a linear combination of vectors in
ˆ
W.
2. The vector y − y which is orthogonal to W (orthogonal to each
ˆ
vector in W ). We can also say that y − y is in W ⊥
ˆ
7. Section 6.3 Orthogonal Projections
1. The above formula for orthogonal projection is for R2
2. We can extend this formula to any vector in Rn
Consider a vector y and a subspace W in Rn . We can nd the
following
1. A unique vector y which is a linear combination of vectors in
ˆ
W.
2. The vector y − y which is orthogonal to W (orthogonal to each
ˆ
vector in W ). We can also say that y − y is in W ⊥
ˆ
3. This vector y is the unique vector in W closest to y
ˆ
8. Decomposition of a vector
Let {u1 , u2 , . . . , un } be an orthogonal basis in Rn . Then any vector y
in Rn can be written as a sum of 2 vectors
y = z1 + z2 ,
where z1 is formed by a linear combination of few of these n
vectors and z2 contains the rest of the vectors not contained in z1 .
9. Decomposition of a vector
Let {u1 , u2 , . . . , un } be an orthogonal basis in Rn . Then any vector y
in Rn can be written as a sum of 2 vectors
y = z1 + z2 ,
where z1 is formed by a linear combination of few of these n
vectors and z2 contains the rest of the vectors not contained in z1 .
Example
Let {u1 , u2 , u3 , u4 , u5 , u6 } be an orthogonal basis in R6 . Then any
vector y in R6 can be written as
10. Decomposition of a vector
Let {u1 , u2 , . . . , un } be an orthogonal basis in Rn . Then any vector y
in Rn can be written as a sum of 2 vectors
y = z1 + z2 ,
where z1 is formed by a linear combination of few of these n
vectors and z2 contains the rest of the vectors not contained in z1 .
Example
Let {u1 , u2 , u3 , u4 , u5 , u6 } be an orthogonal basis in R6 . Then any
vector y in R6 can be written as
y = c1 u1 + c2 u2 + c3 u3 + c4 u4 + c5 u5 + c6 u6
z1 z2
11. Decomposition of a vector
Let {u1 , u2 , . . . , un } be an orthogonal basis in Rn . Then any vector y
in Rn can be written as a sum of 2 vectors
y = z1 + z2 ,
where z1 is formed by a linear combination of few of these n
vectors and z2 contains the rest of the vectors not contained in z1 .
Example
Let {u1 , u2 , u3 , u4 , u5 , u6 } be an orthogonal basis in R6 . Then any
vector y in R6 can be written as
y = c1 u1 + c2 u2 + c3 u3 + c4 u4 + c5 u5 + c6 u6
z1 z2
Here the subspace W =Span{u1 , u2 , u3 , u4 }, z1 is in W and z2 is in
W ⊥ (since the dot product of any vector in W with u5 or u6 is zero)
12. Example 2, section 6.3
Let {u1 , u2 , u3 , u4 } be an orthogonal basis for R4 .
1 −2 1 −1 4
2 1 1 1 5
u1 = , u2 =
, u3 =
, u4 =
,v =
1 −1 −2 1 −3
1 1 −1 −2 3
Write v as the sum of 2 vectors, one in the Span{u1 } and the other
in Span{u2 , u3 , u4 }
13. Example 2, section 6.3
Let {u1 , u2 , u3 , u4 } be an orthogonal basis for R4 .
1 −2 1 −1 4
2 1 1 1 5
u1 = , u2 =
, u3 =
, u4 =
,v =
1 −1 −2 1 −3
1 1 −1 −2 3
Write v as the sum of 2 vectors, one in the Span{u1 } and the other
in Span{u2 , u3 , u4 }
Solution: The vector in the Span{u1 } will be
v u1
z1 = c 1 u 1 = u
u1 u1 1
v u1 = 4 + 10 − 3 + 3 = 14, u1 u1 = 1 + 4 + 1 + 1 = 7
14
z1 = u1 = 2u1
7
14. Example 2, section 6.3
The vector in the Span{u2 , u3 , u4 } will be
v u2 v u3 v u4
z2 = c2 u2 + c3 u3 + c4 u4 = u2 + u3 + u
u2 u2 u3 u3 u4 u4 4
15. Example 2, section 6.3
The vector in the Span{u2 , u3 , u4 } will be
v u2 v u3 v u4
z2 = c2 u2 + c3 u3 + c4 u4 = u2 + u3 + u
u2 u2 u3 u3 u4 u4 4
v u2 = −8 + 5 + 3 + 3 = 3, u2 u2 = 4 + 1 + 1 + 1 = 7
16. Example 2, section 6.3
The vector in the Span{u2 , u3 , u4 } will be
v u2 v u3 v u4
z2 = c2 u2 + c3 u3 + c4 u4 = u2 + u3 + u
u2 u2 u3 u3 u4 u4 4
v u2 = −8 + 5 + 3 + 3 = 3, u2 u2 = 4 + 1 + 1 + 1 = 7
v u3 = 4 + 5 + 6 − 3 = 12, u3 u3 = 1 + 1 + 4 + 1 = 7
17. Example 2, section 6.3
The vector in the Span{u2 , u3 , u4 } will be
v u2 v u3 v u4
z2 = c2 u2 + c3 u3 + c4 u4 = u2 + u3 + u
u2 u2 u3 u3 u4 u4 4
v u2 = −8 + 5 + 3 + 3 = 3, u2 u2 = 4 + 1 + 1 + 1 = 7
v u3 = 4 + 5 + 6 − 3 = 12, u3 u3 = 1 + 1 + 4 + 1 = 7
v u4 = −4 + 5 − 3 − 6 = −8, u4 u4 = 4 + 1 + 1 + 1 = 7
18. Example 2, section 6.3
The vector in the Span{u2 , u3 , u4 } will be
v u2 v u3 v u4
z2 = c2 u2 + c3 u3 + c4 u4 = u2 + u3 + u
u2 u2 u3 u3 u4 u4 4
v u2 = −8 + 5 + 3 + 3 = 3, u2 u2 = 4 + 1 + 1 + 1 = 7
v u3 = 4 + 5 + 6 − 3 = 12, u3 u3 = 1 + 1 + 4 + 1 = 7
v u4 = −4 + 5 − 3 − 6 = −8, u4 u4 = 4 + 1 + 1 + 1 = 7
3 12 8
z2 = u 2 + u3 − u4
7 7 7
19. Example 2, section 6.3
The vector in the Span{u2 , u3 , u4 } will be
v u2 v u3 v u4
z2 = c2 u2 + c3 u3 + c4 u4 = u2 + u3 + u
u2 u2 u3 u3 u4 u4 4
v u2 = −8 + 5 + 3 + 3 = 3, u2 u2 = 4 + 1 + 1 + 1 = 7
v u3 = 4 + 5 + 6 − 3 = 12, u3 u3 = 1 + 1 + 4 + 1 = 7
v u4 = −4 + 5 − 3 − 6 = −8, u4 u4 = 4 + 1 + 1 + 1 = 7
3 12 8
z2 = u 2 + u3 − u4
7 7 7
Thus,
3 12 8
y = z1 + z2 = 2u1 + u2 + u3 − u4
7 7 7
20. Orthogonal Decomposition Theorem
Theorem
Let W be a subspace of Rn . Then each y in Rn can be uniquely
written as
y = y + z,
ˆ
where y is in W
ˆ and z is in W ⊥ . If u1 , u2 , . . . , up is an orthogonal
basis for W,
y u1 y u2 y up
y=
ˆ u1 + u2 + . . . + u
u1 u1 u2 u2 up up p
21. Orthogonal Decomposition Theorem
Theorem
Let W be a subspace of Rn . Then each y in Rn can be uniquely
written as
y = y + z,
ˆ
where y is in W
ˆ and z is in W ⊥ . If u1 , u2 , . . . , up is an orthogonal
basis for W,
y u1 y u2 y up
y=
ˆ u1 + u2 + . . . + u
u1 u1 u2 u2 up up p
and
z = y − y.
ˆ
22. Orthogonal Decomposition Theorem
1. As in the case of R2 , the new vector y is the orthogonal
ˆ
projection of y onto W . (In section 6.2, the orthogonal
projection was onto a line through u and the origin)
23. Orthogonal Decomposition Theorem
1. As in the case of R2 , the new vector y is the orthogonal
ˆ
projection of y onto W . (In section 6.2, the orthogonal
projection was onto a line through u and the origin)
2. This is nothing but an extension of the orthogonal projection
formula we had for R2 in section 6.2 to accomodate the
remaining vectors.
24. Orthogonal Decomposition Theorem
1. As in the case of R2 , the new vector y is the orthogonal
ˆ
projection of y onto W . (In section 6.2, the orthogonal
projection was onto a line through u and the origin)
2. This is nothing but an extension of the orthogonal projection
formula we had for R2 in section 6.2 to accomodate the
remaining vectors.
3. If W is one dimensional, we get the formula for y from sec 6.2
ˆ
25. Example 4, section 6.3
Let {u1 , u2 } be an orthogonal set.
6 3
−4
y = 3 , u1 = 4 , u2 = 3
−2 0 0
Find the orthogonal projection of y onto Span{u1 , u2 }
26. Example 4, section 6.3
Let {u1 , u2 } be an orthogonal set.
6 3
−4
y = 3 , u1 = 4 , u2 = 3
−2 0 0
Find the orthogonal projection of y onto Span{u1 , u2 }
Solution: The desired orthogonal projection is
y u1 y u2
y = c1 u1 + c2 u2 =
ˆ u1 + u
u1 u1 u2 u2 2
27. Example 4, section 6.3
Let {u1 , u2 } be an orthogonal set.
6 3
−4
y = 3 , u1 = 4 , u2 = 3
−2 0 0
Find the orthogonal projection of y onto Span{u1 , u2 }
Solution: The desired orthogonal projection is
y u1 y u2
y = c1 u1 + c2 u2 =
ˆ u1 + u
u1 u1 u2 u2 2
y u1 = 18 + 12 = 30, u1 u1 = 9 + 16 = 25
28. Example 4, section 6.3
Let {u1 , u2 } be an orthogonal set.
6 3
−4
y = 3 , u1 = 4 , u2 = 3
−2 0 0
Find the orthogonal projection of y onto Span{u1 , u2 }
Solution: The desired orthogonal projection is
y u1 y u2
y = c1 u1 + c2 u2 =
ˆ u1 + u
u1 u1 u2 u2 2
y u1 = 18 + 12 = 30, u1 u1 = 9 + 16 = 25
y u2 = −24 + 9 = −15, u2 u2 = 16 + 9 = 25
31. Example 8, section 6.3
Let W be the subspace spanned by the u's, then write y as the sum
of a vector in W and a vector orthogonal to W .
1
−1 −1
y = 4 , u1 = 1 , u2 = 3
3 1 −2
32. Example 8, section 6.3
Let W be the subspace spanned by the u's, then write y as the sum
of a vector in W and a vector orthogonal to W .
1
−1 −1
y = 4 , u1 = 1 , u2 = 3
3 1 −2
Solution: A vector in W is the orthogonal projection of y onto W
which is y u1 y u2
y = c1 u1 + c2 u2 =
ˆ u + u
u1 u1 1 u2 u2 2
33. Example 8, section 6.3
Let W be the subspace spanned by the u's, then write y as the sum
of a vector in W and a vector orthogonal to W .
1
−1 −1
y = 4 , u1 = 1 , u2 = 3
3 1 −2
Solution: A vector in W is the orthogonal projection of y onto W
which is y u1 y u2
y = c1 u1 + c2 u2 =
ˆ u + u
u1 u1 1 u2 u2 2
y u1 = −1 + 4 + 3 = 6, u1 u1 = 1 + 1 + 1 = 3
34. Example 8, section 6.3
Let W be the subspace spanned by the u's, then write y as the sum
of a vector in W and a vector orthogonal to W .
1
−1 −1
y = 4 , u1 = 1 , u2 = 3
3 1 −2
Solution: A vector in W is the orthogonal projection of y onto W
which is y u1 y u2
y = c1 u1 + c2 u2 =
ˆ u + u
u1 u1 1 u2 u2 2
y u1 = −1 + 4 + 3 = 6, u1 u1 = 1 + 1 + 1 = 3
y u2 = 1 + 12 − 6 = 7, u2 u2 = 1 + 9 + 4 = 14
37. Example 4, section 6.3
6 7
y = u1 +
ˆ u2 .
3 14
1
−1
1
= 2 1 + 3
2
1 −2
1 3
2
−2 2
= 2 + 3 = 7
2 2
2 −1 1
A vector orthogonal to W will be z = y − y which is
ˆ
3 5
−1 2 −2
4 − 7 = 1
2 2
3 1 2
38. Example 10, section 6.3
Let W be the subspace spanned by the u's, then write y as the sum
of a vector in W and a vector orthogonal to W .
3 1 1 0
4 1 0 −1
y= , u1 =
, u2 =
, u3 =
5 0 1 1
6 −1 1 −1
39. Example 10, section 6.3
Let W be the subspace spanned by the u's, then write y as the sum
of a vector in W and a vector orthogonal to W .
3 1 1 0
4 1 0 −1
y= , u1 =
, u2 =
, u3 =
5 0 1 1
6 −1 1 −1
Solution: A vector in W is the orthogonal projection of y onto W
which is
y u1 y u2 y u3
y = c1 u1 + c2 u2 + c3 u3 =
ˆ u1 + u2 + u
u1 u1 u2 u2 u3 u3 3
40. Example 10, section 6.3
Let W be the subspace spanned by the u's, then write y as the sum
of a vector in W and a vector orthogonal to W .
3 1 1 0
4 1 0 −1
y= , u1 =
, u2 =
, u3 =
5 0 1 1
6 −1 1 −1
Solution: A vector in W is the orthogonal projection of y onto W
which is
y u1 y u2 y u3
y = c1 u1 + c2 u2 + c3 u3 =
ˆ u1 + u2 + u
u1 u1 u2 u2 u3 u3 3
y u1 = 3 + 4 − 6 = 1, u1 u1 = 1 + 1 + 1 = 3
41. Example 10, section 6.3
Let W be the subspace spanned by the u's, then write y as the sum
of a vector in W and a vector orthogonal to W .
3 1 1 0
4 1 0 −1
y= , u1 =
, u2 =
, u3 =
5 0 1 1
6 −1 1 −1
Solution: A vector in W is the orthogonal projection of y onto W
which is
y u1 y u2 y u3
y = c1 u1 + c2 u2 + c3 u3 =
ˆ u1 + u2 + u
u1 u1 u2 u2 u3 u3 3
y u1 = 3 + 4 − 6 = 1, u1 u1 = 1 + 1 + 1 = 3
y u2 = 3 + 5 + 6 = 14, u2 u2 = 1 + 1 + 1 = 3
42. Example 10, section 6.3
Let W be the subspace spanned by the u's, then write y as the sum
of a vector in W and a vector orthogonal to W .
3 1 1 0
4 1 0 −1
y= , u1 =
, u2 =
, u3 =
5 0 1 1
6 −1 1 −1
Solution: A vector in W is the orthogonal projection of y onto W
which is
y u1 y u2 y u3
y = c1 u1 + c2 u2 + c3 u3 =
ˆ u1 + u2 + u
u1 u1 u2 u2 u3 u3 3
y u1 = 3 + 4 − 6 = 1, u1 u1 = 1 + 1 + 1 = 3
y u2 = 3 + 5 + 6 = 14, u2 u2 = 1 + 1 + 1 = 3
y u3 = −4 + 5 − 6 = −5, u3 u3 = 1 + 1 + 1 = 3
46. Applications
1. Let W = Span u1 , u2 , . . . , up (an orthogonal basis). Let y be
any vector in Rn . Then the orthogonal projection y gives the
ˆ
closest point in W to y.
47. Applications
1. Let W = Span u1 , u2 , . . . , up (an orthogonal basis). Let y be
any vector in Rn . Then the orthogonal projection y gives the
ˆ
closest point in W to y.
2. The orthogonal projection y is also the best approximation to
ˆ
y by the elements of W (this means we can replace y by a
vector in some xed subspace W and this replacement comes
with the least error if we choose y).ˆ
48. Example 12, section 6.3
Find the closest point to y in the subspace spanned by v1 and v2 .
3 1 −4
−1 −2 1
y= , v1 =
, v2 =
1 −1 0
13 2 3
49. Example 12, section 6.3
Find the closest point to y in the subspace spanned by v1 and v2 .
3 1 −4
−1 −2 1
y= , v1 =
, v2 =
1 −1 0
13 2 3
Solution: The closest point to y is the orthogonal projection of y
onto the span of v1 and v2 which is
50. Example 12, section 6.3
Find the closest point to y in the subspace spanned by v1 and v2 .
3 1 −4
−1 −2 1
y= , v1 =
, v2 =
1 −1 0
13 2 3
Solution: The closest point to y is the orthogonal projection of y
onto the span of v1 and v2 which is
y v1 y v2
y = c1 v1 + c2 v2 =
ˆ v + v
v1 v1 1 v2 v2 2
51. Example 12, section 6.3
Find the closest point to y in the subspace spanned by v1 and v2 .
3 1 −4
−1 −2 1
y= , v1 =
, v2 =
1 −1 0
13 2 3
Solution: The closest point to y is the orthogonal projection of y
onto the span of v1 and v2 which is
y v1 y v2
y = c1 v1 + c2 v2 =
ˆ v + v
v1 v1 1 v2 v2 2
y v1 = 3 + 2 − 1 + 26 = 30, v1 v1 = 1 + 4 + 1 + 4 = 10
52. Example 12, section 6.3
Find the closest point to y in the subspace spanned by v1 and v2 .
3 1 −4
−1 −2 1
y= , v1 =
, v2 =
1 −1 0
13 2 3
Solution: The closest point to y is the orthogonal projection of y
onto the span of v1 and v2 which is
y v1 y v2
y = c1 v1 + c2 v2 =
ˆ v + v
v1 v1 1 v2 v2 2
y v1 = 3 + 2 − 1 + 26 = 30, v1 v1 = 1 + 4 + 1 + 4 = 10
y v2 = −12 − 1 + 39 = 26, v2 v2 = 16 + 1 + 9 = 26
53. Example 12, section 6.3
Find the closest point to y in the subspace spanned by v1 and v2 .
3 1 −4
−1 −2 1
y= , v1 =
, v2 =
1 −1 0
13 2 3
Solution: The closest point to y is the orthogonal projection of y
onto the span of v1 and v2 which is
y v1 y v2
y = c1 v1 + c2 v2 =
ˆ v + v
v1 v1 1 v2 v2 2
y v1 = 3 + 2 − 1 + 26 = 30, v1 v1 = 1 + 4 + 1 + 4 = 10
y v2 = −12 − 1 + 39 = 26, v2 v2 = 16 + 1 + 9 = 26
1 −4 −1
30 26 −2 1 −5
y=
ˆ v1 + v2 = 3 +
=
−1 0 −3
10 26
2 3 9
54. Example 14, section 6.3
Find the best approximation z by v1 and v2 of the form c1 v1 + c2 v2 .
2 2 5
4 0 −2
z= , v1 =
, v2 =
0 −1 4
−1 −3 2
55. Example 14, section 6.3
Find the best approximation z by v1 and v2 of the form c1 v1 + c2 v2 .
2 2 5
4 0 −2
z= , v1 =
, v2 =
0 −1 4
−1 −3 2
Solution: The best approximation to z is the orthogonal projection
of z onto the span of v1 and v2 which is
56. Example 14, section 6.3
Find the best approximation z by v1 and v2 of the form c1 v1 + c2 v2 .
2 2 5
4 0 −2
z= , v1 =
, v2 =
0 −1 4
−1 −3 2
Solution: The best approximation to z is the orthogonal projection
of z onto the span of v1 and v2 which is
z v1 z v2
z = c1 v1 + c2 v2 =
ˆ v + v
v1 v1 1 v2 v2 2
57. Example 14, section 6.3
Find the best approximation z by v1 and v2 of the form c1 v1 + c2 v2 .
2 2 5
4 0 −2
z= , v1 =
, v2 =
0 −1 4
−1 −3 2
Solution: The best approximation to z is the orthogonal projection
of z onto the span of v1 and v2 which is
z v1 z v2
z = c1 v1 + c2 v2 =
ˆ v + v
v1 v1 1 v2 v2 2
z v1 = 4 + 3 = 7, v1 v1 = 4 + 1 + 9 = 14
58. Example 14, section 6.3
Find the best approximation z by v1 and v2 of the form c1 v1 + c2 v2 .
2 2 5
4 0 −2
z= , v1 =
, v2 =
0 −1 4
−1 −3 2
Solution: The best approximation to z is the orthogonal projection
of z onto the span of v1 and v2 which is
z v1 z v2
z = c1 v1 + c2 v2 =
ˆ v + v
v1 v1 1 v2 v2 2
z v1 = 4 + 3 = 7, v1 v1 = 4 + 1 + 9 = 14
z v2 = 10 − 8 − 2 = 0, v2 v2 = 25 + 4 + 16 + 4 = 49
59. Example 14, section 6.3
Find the best approximation z by v1 and v2 of the form c1 v1 + c2 v2 .
2 2 5
4 0 −2
z= , v1 =
, v2 =
0 −1 4
−1 −3 2
Solution: The best approximation to z is the orthogonal projection
of z onto the span of v1 and v2 which is
z v1 z v2
z = c1 v1 + c2 v2 =
ˆ v + v
v1 v1 1 v2 v2 2
z v1 = 4 + 3 = 7, v1 v1 = 4 + 1 + 9 = 14
z v2 = 10 − 8 − 2 = 0, v2 v2 = 25 + 4 + 16 + 4 = 49
2 1
7 0 0 0
z=
ˆ v1 + v2 = 0.5 =
−1 −0.5
14 49
=0 −3 −1.5
60. Topics to learn for Test 2
1. Evaluation of determinants, use of determinants in Cramer's
rule, nding adjugate of a matrix and nding areas of
parallelograms.
2. Finding char. equation, eigenvalues, eigenvectors of a square
matrix (including triangular matrix) and use these in
diagonalizing a matrix and computing higher exponents of the
matrix. (you should know the condition under which
diagonalization may fail).
3. Finding length of a vector, unit vector, orthogonal vectors and
orthogonal basis, orthogonal projection onto a subspace
spanned by any number of vectors, closest point/best
approximation of a vector
Good luck. Feel free to email me if you have any questions. Please
come prepared for Wednesday's review. Bring any problems you
want me to go over.
