Orthogonal Projection

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)






y= y+z


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)






y= y+z


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

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

]


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. )


Let y=[

26

]and u=

[71





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


z= y− y=[

26

]−

[2.80.4

]=

[ −0.85.6

](Check: z �u= 0. )


Let y=[

26

]and u=

[71





Here y=[

26

]and u=

[71

]. So, y �u= 14+6 = 20 and


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. )


Let y=[

26

]and u=

[71





Here y=[

26

]and u=

[71

]. So, y �u= 14+6 = 20 and


y= y �uu �u

u= 20

50

[71

]=

[2.80.4


z= y− y=[

26

]−

[2.80.4

]=

[ −0.85.6

](Check: z �u= 0. )


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


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


Let y=[ −3

9

]and u=

[12



Solution: We have to compute ‖y− y‖

Here y=[ −3

9

]and u=

[12

]. So, y �u=−3+18 = 15 and


y= y �uu �u

u= 15

5

[12

]=

[36


‖y− y‖y− y=

[ −39

]−

[36

]=

[ −63

]‖y− y‖ =p

36+9 =p45


Let y=[ −3

9

]and u=

[12




[ −39

]and u=

[12

]. So, y �u=−3+18 = 15 and


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


Let y=[ −3

9

]and u=

[12




[ −39

]and u=

[12

]. So, y �u=−3+18 = 15 and


y= y �uu �u

u= 15

5

[12

]=

[36


‖y− y‖y− y=

[ −39

]−

[36

]=

[ −63

]‖y− y‖ =p

36+9 =p45

Orthogonal Projection

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