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May 25, 2015
Announcements
Ï Quiz 4 after lecture.
Ï Exam 2 on Thurs, Feb 25 in class.
Ï Exam 2 will cover material taught after exam 1 and upto what
is covered on Monday Feb 22.
Ï Practice Exam will be uploaded on Monday after I �nish the
material.
Ï I will do some misc. topics (sec 5.5 and some applications) on
Tuesday. These WILL NOT be covered on the exam but are
useful for MA 3521. Attendance not mandatory.
Ï Review on Wednesday in class. I will have o�ce hours on Wed
from 1-4 pm.
Yesterday
The inner product (or the inner product) of two vectors u and v in
Rn .
uTv= [u1 u2 . . . un
]
v1
v2...
vn
= u1v1 +u2v2 + . . .+un vn
1. Inner product of 2 vectors is a number.
2. Inner product is also called dot product (in Calculus II)
3. Often written as u �v
Yesterday
De�nition
The length (or the norm) of v is the nonnegative scalar ‖v‖ de�ned
by
‖v‖ =pv �v=
√v2
1 + v22 + . . .+ v2
n
De�nition
A vector of length 1 is called a unit vector.
Yesterday
De�nition
For any two vectors u and v in Rn , the distance between u and v
written as dist(u,v) is the length of the vector u-v.
dist(u,v) = ‖u-v‖
De�nition
Two vectors u and v in Rn are orthogonal (to each other) if
u �v= 0
Yesterday
Consider a set of vectors{u1,u2, . . . ,up
}in Rn . If each pair of
distinct vectors from the set is orthogonal (that is u1 �u2 = 0,u1 �u3 = 0, u2 �u3 = 0 etc etc) then the set is called an orthogonal
set.
An orthogonal basis for a subspace W of Rn is a set
1. spans W and
2. is linearly independent and
3. is orthogonal
An Orthogonal ProjectionLet u be a nonzero vector in Rn . Suppose we want to write another
vector y in Rn as the sum of 2 vectors
1. one vector a multiple of u
2. the second vector orthogonal to u
That is, we want to do the following
y= y+z
where y= αu for some scalar α and z is some vector orthogonal to
u.
Thus
z= y−αuIf z is orthogonal to u, we have
z �u= 0
=⇒ (y−αu) �u= 0 =⇒ y �u= α(u �u)
An Orthogonal ProjectionLet u be a nonzero vector in Rn . Suppose we want to write another
vector y in Rn as the sum of 2 vectors
1. one vector a multiple of u
2. the second vector orthogonal to u
That is, we want to do the following
y= y+z
where y= αu for some scalar α and z is some vector orthogonal to
u.
Thus
z= y−αuIf z is orthogonal to u, we have
z �u= 0
=⇒ (y−αu) �u= 0 =⇒ y �u= α(u �u)
An Orthogonal ProjectionLet u be a nonzero vector in Rn . Suppose we want to write another
vector y in Rn as the sum of 2 vectors
1. one vector a multiple of u
2. the second vector orthogonal to u
That is, we want to do the following
y= y+z
where y= αu for some scalar α and z is some vector orthogonal to
u.
Thus
z= y−αuIf z is orthogonal to u, we have
z �u= 0
=⇒ (y−αu) �u= 0 =⇒ y �u= α(u �u)
An Orthogonal ProjectionLet u be a nonzero vector in Rn . Suppose we want to write another
vector y in Rn as the sum of 2 vectors
1. one vector a multiple of u
2. the second vector orthogonal to u
That is, we want to do the following
y= y+z
where y= αu for some scalar α and z is some vector orthogonal to
u.
Thus
z= y−αuIf z is orthogonal to u, we have
z �u= 0
=⇒ (y−αu) �u= 0 =⇒ y �u= α(u �u)
An Orthogonal ProjectionLet u be a nonzero vector in Rn . Suppose we want to write another
vector y in Rn as the sum of 2 vectors
1. one vector a multiple of u
2. the second vector orthogonal to u
That is, we want to do the following
y= y+z
where y= αu for some scalar α and z is some vector orthogonal to
u.
Thus
z= y−αu
If z is orthogonal to u, we have
z �u= 0
=⇒ (y−αu) �u= 0 =⇒ y �u= α(u �u)
An Orthogonal ProjectionLet u be a nonzero vector in Rn . Suppose we want to write another
vector y in Rn as the sum of 2 vectors
1. one vector a multiple of u
2. the second vector orthogonal to u
That is, we want to do the following
y= y+z
where y= αu for some scalar α and z is some vector orthogonal to
u.
Thus
z= y−αuIf z is orthogonal to u, we have
z �u= 0
=⇒ (y−αu) �u= 0 =⇒ y �u= α(u �u)
An Orthogonal ProjectionLet u be a nonzero vector in Rn . Suppose we want to write another
vector y in Rn as the sum of 2 vectors
1. one vector a multiple of u
2. the second vector orthogonal to u
That is, we want to do the following
y= y+z
where y= αu for some scalar α and z is some vector orthogonal to
u.
Thus
z= y−αuIf z is orthogonal to u, we have
z �u= 0
=⇒ (y−αu) �u= 0 =⇒ y �u= α(u �u)
=⇒ α= y �uu �u
Thus,
y= y �uu �u
u
x0 u
y
y= αy
z= y− y
=⇒ α= y �uu �u
Thus,
y= y �uu �u
u
x0 u
y
y= αy
z= y− y
=⇒ α= y �uu �u
Thus,
y= y �uu �u
u
x0 u
y
y= αy
z= y− y
=⇒ α= y �uu �u
Thus,
y= y �uu �u
u
x0 u
y
y= αy
z= y− y
1. The new vector y is called the orthogonal projection of y onto
u
2. The vector z is called the complement of y orthogonal to u
The orthogonal projection of y onto any line L through u and 0 is
given by
y= projLy=y �uu �u
u
The orthogonal projection is a vector (not a number).
The quantity ‖y− y‖ gives the distance between y and the line L.
These two formulas are to be used in problems 11, 13 and 15 of
section 6.2.
1. The new vector y is called the orthogonal projection of y onto
u
2. The vector z is called the complement of y orthogonal to u
The orthogonal projection of y onto any line L through u and 0 is
given by
y= projLy=y �uu �u
u
The orthogonal projection is a vector (not a number).
The quantity ‖y− y‖ gives the distance between y and the line L.
These two formulas are to be used in problems 11, 13 and 15 of
section 6.2.
1. The new vector y is called the orthogonal projection of y onto
u
2. The vector z is called the complement of y orthogonal to u
The orthogonal projection of y onto any line L through u and 0 is
given by
y= projLy=y �uu �u
u
The orthogonal projection is a vector (not a number).
The quantity ‖y− y‖ gives the distance between y and the line L.
These two formulas are to be used in problems 11, 13 and 15 of
section 6.2.
Example 12, section 6.2
Compute the orthogonal projection of
[1−1
]onto the line through[ −1
3
]and the origin.
Solution: Here y=[
1−1
]and u=
[ −13
]. So, y �u=−1−3 =−4
and u �u= 1+9 = 10. The orthogonal projection of y onto u is
y= y �uu �u
u= −4
10
[ −13
]=
[0.4−1.2
]
Example 12, section 6.2
Compute the orthogonal projection of
[1−1
]onto the line through[ −1
3
]and the origin.
Solution: Here y=[
1−1
]and u=
[ −13
]. So, y �u=−1−3 =−4
and u �u= 1+9 = 10. The orthogonal projection of y onto u is
y= y �uu �u
u= −4
10
[ −13
]=
[0.4−1.2
]
Example 14, section 6.2
Let y=[
26
]and u=
[71
]. Write y as the sum of 2 orthogonal
vectors, one in Span{u} and one orthogonal to u
Solution: A vector in Span{u} is the orthogonal projection of y onto
the line containing u and the origin.
Here y=[
26
]and u=
[71
]. So, y �u= 14+6 = 20 and
u �u= 49+1 = 50. The orthogonal projection of y onto u is
y= y �uu �u
u= 20
50
[71
]=
[2.80.4
]The vector orthogonal to u will be
z= y− y=[
26
]−
[2.80.4
]=
[ −0.85.6
](Check: z �u= 0. )
Example 14, section 6.2
Let y=[
26
]and u=
[71
]. Write y as the sum of 2 orthogonal
vectors, one in Span{u} and one orthogonal to u
Solution: A vector in Span{u} is the orthogonal projection of y onto
the line containing u and the origin.
Here y=[
26
]and u=
[71
]. So, y �u= 14+6 = 20 and
u �u= 49+1 = 50. The orthogonal projection of y onto u is
y= y �uu �u
u= 20
50
[71
]=
[2.80.4
]The vector orthogonal to u will be
z= y− y=[
26
]−
[2.80.4
]=
[ −0.85.6
](Check: z �u= 0. )
Example 14, section 6.2
Let y=[
26
]and u=
[71
]. Write y as the sum of 2 orthogonal
vectors, one in Span{u} and one orthogonal to u
Solution: A vector in Span{u} is the orthogonal projection of y onto
the line containing u and the origin.
Here y=[
26
]and u=
[71
]. So, y �u= 14+6 = 20 and
u �u= 49+1 = 50.
The orthogonal projection of y onto u is
y= y �uu �u
u= 20
50
[71
]=
[2.80.4
]The vector orthogonal to u will be
z= y− y=[
26
]−
[2.80.4
]=
[ −0.85.6
](Check: z �u= 0. )
Example 14, section 6.2
Let y=[
26
]and u=
[71
]. Write y as the sum of 2 orthogonal
vectors, one in Span{u} and one orthogonal to u
Solution: A vector in Span{u} is the orthogonal projection of y onto
the line containing u and the origin.
Here y=[
26
]and u=
[71
]. So, y �u= 14+6 = 20 and
u �u= 49+1 = 50. The orthogonal projection of y onto u is
y= y �uu �u
u= 20
50
[71
]=
[2.80.4
]
The vector orthogonal to u will be
z= y− y=[
26
]−
[2.80.4
]=
[ −0.85.6
](Check: z �u= 0. )
Example 14, section 6.2
Let y=[
26
]and u=
[71
]. Write y as the sum of 2 orthogonal
vectors, one in Span{u} and one orthogonal to u
Solution: A vector in Span{u} is the orthogonal projection of y onto
the line containing u and the origin.
Here y=[
26
]and u=
[71
]. So, y �u= 14+6 = 20 and
u �u= 49+1 = 50. The orthogonal projection of y onto u is
y= y �uu �u
u= 20
50
[71
]=
[2.80.4
]The vector orthogonal to u will be
z= y− y=[
26
]−
[2.80.4
]=
[ −0.85.6
](Check: z �u= 0. )
Example 16, section 6.2
Let y=[ −3
9
]and u=
[12
]. Compute the distance from y to the
line through u and the origin.
Solution: We have to compute ‖y− y‖Here y=
[ −39
]and u=
[12
]. So, y �u=−3+18 = 15 and
u �u= 1+4 = 5. The orthogonal projection of y onto u is
y= y �uu �u
u= 15
5
[12
]=
[36
]The distance from y to the line containing u and the origin will be
‖y− y‖y− y=
[ −39
]−
[36
]=
[ −63
]‖y− y‖ =p
36+9 =p45
Example 16, section 6.2
Let y=[ −3
9
]and u=
[12
]. Compute the distance from y to the
line through u and the origin.
Solution: We have to compute ‖y− y‖
Here y=[ −3
9
]and u=
[12
]. So, y �u=−3+18 = 15 and
u �u= 1+4 = 5. The orthogonal projection of y onto u is
y= y �uu �u
u= 15
5
[12
]=
[36
]The distance from y to the line containing u and the origin will be
‖y− y‖y− y=
[ −39
]−
[36
]=
[ −63
]‖y− y‖ =p
36+9 =p45
Example 16, section 6.2
Let y=[ −3
9
]and u=
[12
]. Compute the distance from y to the
line through u and the origin.
Solution: We have to compute ‖y− y‖Here y=
[ −39
]and u=
[12
]. So, y �u=−3+18 = 15 and
u �u= 1+4 = 5. The orthogonal projection of y onto u is
y= y �uu �u
u= 15
5
[12
]=
[36
]
The distance from y to the line containing u and the origin will be
‖y− y‖y− y=
[ −39
]−
[36
]=
[ −63
]‖y− y‖ =p
36+9 =p45
Example 16, section 6.2
Let y=[ −3
9
]and u=
[12
]. Compute the distance from y to the
line through u and the origin.
Solution: We have to compute ‖y− y‖Here y=
[ −39
]and u=
[12
]. So, y �u=−3+18 = 15 and
u �u= 1+4 = 5. The orthogonal projection of y onto u is
y= y �uu �u
u= 15
5
[12
]=
[36
]The distance from y to the line containing u and the origin will be
‖y− y‖y− y=
[ −39
]−
[36
]=
[ −63
]‖y− y‖ =p
36+9 =p45