Week 7

4.4 Matrix Transformations II

Projection of a Vector ~v on a Line Through the Origin with Direction Vector ~d = [a b c]T .

We will let ~v = [x y z]T .

Recall that proj~d~v =~v ~d~d2

~d =ax+ by + cz

a2 + b2 + c2

abc

.w1 =

ax+ by + cz

a2 + b2 + c2a =

a2x+ aby + acz

a2 + b2 + c2

w2 =ax+ by + cz

a2 + b2 + c2c =

acx+ bcy + c2z

a2 + b2 + c2

w3 =ax+ by + cz

a2 + b2 + c2c =

acx+ bcy + c2z

a2 + b2 + c2

Then the standard matrix for this transformation is

1

a2 + b2 + c2

a2 ab acab b2 bcac bc c2

Projection of a Vector ~v onto a Plane Through the Origin with Normal Vector ~n = [a b c]T .

Here we want the vector ~v proj~n~v or if we let ~v = [x y z]T , then we have

~v proj~n~v =xyz

1a2 + b2 + c2

a2 ab acab b2 bcac bc c2

xyz

=

1 0 00 1 00 0 1

1a2 + b2 + c2

a2 ab acab b2 bcac bc c2

xyz

=

1

a2 + b2 + c2

b2 + c2 ab acab a2 + c2 bcac bc a2 + b2

xyz

Reflection of a Vector ~v in a Line Through the Origin with Direction Vector ~d = [a b c]T

Standard Matrix is1

a2 + b2 + c2

a2 b2 c2 2ab 2ac2ab b2 a2 c2 2bc2ac 2bc c2 a2 b2

.Reflection of a Vector ~v in a Plane Through the Origin with Normal Vector ~n = [a b c]T

Standard Matrix is1

a2 + b2 + c2

b2 + c2 a2 2ab 2ac2ab a2 + c2 b2 2bc2ac 2bc a2 + b2 c2

.1

DarrenReplace

DarrenReplaceb

DarrenReplace

DarrenReplace(abx+(b^2)y+bcz)/(a^2+b^2+c^2)

Projection of a vector ~v onto a Line Through the Origin with Direction Vector ~d = [a b]T

If we let ~v = [x y]T , then we have

T (~v) = proj~d~v

T

([xy

])=

[x y]T [a b]T[a b]T2

[ab

]=

1

a2 + b2(ax+ by)

[ab

]=

1

a2 + b2

[a2x+ bayabx+ b2y

]=

1

a2 + b2

[a2 abab b2

] [xy

]

Reflection of a vector ~v about a Line Through the Origin with Direction Vector ~d = [a b]T

If we let ~v = [x y]T , then we have

T (~v) = ~v + 2(proj~d~v ~v)= 2proj~d~v ~v

T

([xy

])=

(2 1

a2 + b2

[a2 abab b2

][1 00 1

])[xy

]=

1

a2 + b2

[a2 b2 2ab2ab b2 a2

] [xy

]Example:

Find the matrix for the reflection through the plane with normal vector n =

123

followed by aprojection onto the line x = 2t, y = 3t, z = 5t.Answer:

[P R]

=1

22 + (3)2 + 52

22 (2)(3) (2)(5)(2)(3) (3)2 (5)(5)(2)(5) (3)(5) 52

112 + 22 + 32

22 + 32 12 2(1)(2) 2(1)(3)2(1)(2) 12 + 32 22 2(2)(3)2(1)(3) 2(2)(3) 12 + 22 32

=

1

494

12 172 248 258 330 430 5

2

5.1 Subspaces

We will be looking at the set of all vectors with n real components or Rn. This is an example of avector space. We will study other vector spaces in the next chapter.

Definition:

We say a subset of vectors W in Rn is a subspace of Rn if

1. ~0 W2. If ~x W and ~y W , then ~x+ ~y W .3. If ~x W and a is a scalar, then a~x W .

Examples

1. Rn is a subspace of Rn

2. {~0} is a subspace of Rn

3. Every plane P through the origin in R3 is a subspace of R3.Proof: Let ~n be the normal vector of the plane P .Then P = {~v R3 : ~v ~n = 0}.1. ~0 P since ~0 ~n = 0.2. If ~u,~v P , then ~u ~n = 0 and ~v ~n = 0.

Also,

~u+ ~v ~n = ~u ~n+ ~v ~n= 0 + 0

= 0

3. If ~u P and a R, then a~u ~n = a(~u ~n) = a(0) = 0.Thus, P is a subspace of R3.

4. Lines L through the origin are subspaces of R3.Proof: Let ~d be the direction vector of L.Then L = {t~d : t R}.1. ~0 L since when t = 0 0~d = ~0.2. If ~u,~v L, then ~u = t~d and ~v = s~d for some scalars s and t.

Then ~u+ ~v = t~d+ s~d = (t+ s)~d which belongs to L.

3. If a R and ~u L, then ~u = k~d for some k R.Then a~u = a(k~d) = (ak)~d which belongs to L.

3

Example 5

Is W =

{[xy

]: x 0

}a subspace of R2?

Theorem 1:

R has no subspaces other than R and {~0}.

Proof:

Suppose R does have a subspace H, where H 6= {~0} and H 6= R.Since H is non-empty, there exists some non-zero real number H.Since H is a subspace, any scalar multiple of is also in H.

In particular1

= 1 H.

Since 1 H, any scalar multiple of 1 is also in H and so every real number is in H so H = R.This is a contradiction since we assumed H 6= R.

Theorem 2:

Let H1 and H2 be subspaces of Rn. Then H1 H2 is a subspace of Rn.

Proof:

1. ~0 H1 H2 since ~0 H1 and ~0 H2.2. Let ~u,~v H1 H2.

Then ~u H1 and ~u H2and ~v H1 and ~v H2.Thus, ~u+ ~v H1 and ~u+ ~v H2 since H1 and H2 are subspaces (closed under addition).Therefore, ~u+ ~v H1 H2.

3. Let ~u H1 H2 and a R.Then ~u H1 and ~u H2.Thus, a~u H1 and a~u H2.Therefore, a~u H1 H2.

4

Example to Illustrate Theorem 2

Let H1 =

xyz

: 2x y z = 0. and H2 =

xyz

: x+ 2y + 3z = 0.

Then H1 H2 is the intersection of two planes through the origin.In this case, by solving the above system of equations

2x y z = 0x+ 2y + 3z = 0

we obtain the solution

xyz

=1/57/5

1

t, which is a line through the origin.We showed in Example 4 that this is also a subspace of R3.

Span

Definition

If ~v1, ~v2, . . . , ~vk Rn, then S = span{~v1, ~v2, . . . , ~vk} is the set of all linear combinations of ~v1, ~v2, . . . , ~vk},and we say {~v1, ~v2, . . . , ~vk} span S.

S = {t1~v1 + t2~v2 + . . .+ tk~vk : ti R}

Examples

1. A line L in R3 = span{~d} or ~d spans L.2. A plane P in R3 = span{~u,~v}, where ~u,~v are non-parallel, non-zero vectors in R3. Equivalently,

we could say ~u and ~v span the plane P .

Row Space, Column Space & Null Space

There are 3 particular subspaces associated with an m n matrix A. The Row Space of A, or Row(A), is the set of all linear combinations of the row vectors of A.

ie. If A =

~r1~r2...~rm

, thenRow(A) =

The Column Space of A, or Col(A), is the set of all linear combinations of the column vectorsof A.ie. If A =

[~c1 ~c2 ~cn

], then

Col(A)=

5

The Null Space of A, Null(A), is the set of all solutions to the matrix equation A~x = ~0.ie. Null(A) =

Exercise

Verify that Row(A), Col(A) and Null(A) are indeed subspaces. What are they subspaces of?

Theorem 1:

Let U = span{~v1, ~v2, . . . , ~vk} Rn.(i) U is a subspace of Rn containing each ~vi.

(ii) If W is a subspace of Rn and each ~vi W , then U W .

Proof:

(i) 1. ~0 U since ~0 = 0~v1 + 0~v2 + . . .+ 0~vk2. If ~x, ~y U , then we can write ~x = t1~v1+ t2~v2+ . . .+ tk~vk and ~y = s1~v1+ s2~v2+ . . .+ sk~vk.

Then ~x+ ~y = (t1 + s1)~v1 + (t2 + s2)~v2 + . . .+ (tk + sk)~vk and thus lies in U .

3. If ~x U , then ~x = t1~v1 + t2~v2 + . . .+ tk~vk.Then a~x = at1~v1 + at2~v2 + . . .+ atk~vk and thus lies in U .

Finally, each ~vi lies in U since ~v1 = 1~v1 + 0~v2 + . . .+ 0~vk etc.

(ii) Let ~u U . We want to show ~u W .Since ~u U , we can write ~u = t1~v1 + t2~v2 + . . .+ tk~vk.

6

Now, since ~v1, ~v2, . . . , ~vk W , t1~v1, t2~v2, . . . tk~vk W also, since a subspace is closed underscalar multiplication.Since W is also closed under addition, we have t1~v1 + t2~v2 + . . .+ tk~vk W and so ~u W .

Example 3:

If ~x, ~y Rn, show span{~x+ ~y, ~x ~y} =span{~x, ~y}.

Solution:

To show two sets are equal we need to show they are each contained inside the other. In other words,to show A = B, we need to show A B and B A.We will use the result of Theorem 3(ii) above.Is ~x+ ~y, ~x ~y span{~x, ~y}?Yes, since ~x+ ~y = 1(~x) + 1(~y) and ~x ~y = 1~x+ (1)~y.Thus, by Theorem 3(ii), span{~x+ ~y, ~x ~y} span{~x, ~y}.Is ~x, ~y span{~x+ ~y, ~x ~y}?Yes, since

~x = (~x+ ~y) + (~x ~y)~y = (~x+ ~y) + (~x ~y)

Thus, span{~x, ~y} span{~x+ ~y, ~x ~y}.

Example 4:

Is ~b span{~u,~v}, where ~b = 743

~u = 125

~v =256

.

7

Solution

Example 5:

Do ~v1, ~v2, ~v3 span R3 where ~v1 =

143

~v2 = 322

~v3 = 467

?Solution

8

Note: The vector equation x1~v1 + x2~v2 + x3~v3 = ~b is equivalent to the matrix equation A~x = ~b. Byplacing the vectors ~v1, ~v2, ~v3 as columns of a matrix A and reducing the augmented matrix [A|~b], wecan see that this system is only consistent for some vectors ~b R3. This tells us that there are somevectors in Rn can not be written as a linear combination of the vectors ~v1, ~v2, ~v3 and so these vectorsdo not span R3.What is significant about the coefficient matrix A in this case?

Example 5:

Does the set of vectors {~v1, ~v2, ~v3, ~v4} span R3, where ~v1 = 121

~v2 = 233

~v3 = 457

~v4 =282

?Solution

9