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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)
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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
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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.
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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.
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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)=
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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.
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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
.
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Solution
Example 5:
Do ~v1, ~v2, ~v3 span R3 where ~v1 =
143
~v2 = 322
~v3 = 467
?Solution
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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
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