Math 3191 Applied Linear Algebramath.ucdenver.edu/~billups/courses/ma3191/lectures/lec6.pdf ·...
Transcript of Math 3191 Applied Linear Algebramath.ucdenver.edu/~billups/courses/ma3191/lectures/lec6.pdf ·...
Math 3191Applied Linear AlgebraLecture 6: Linear Transformations
Stephen Billups
University of Colorado at Denver
Math 3191Applied Linear Algebra – p.1/30
Announcements
I am giving a talk tomorrow–you are invited to attend.Title: Analysis of Temporal Gene Expression DataTime: 11-12 (free Pizza)Location: CU-Denver Building, Room 505
HWK correction: Sec 1.6, # 11, reverse the arrow labeled80.
Math 3191Applied Linear Algebra – p.2/30
Comments on Homework
Don’t copy answers from back of book–it makes megrumpy.
For T/F questions, justify your answers.
Answer the question–don’t just plug and chug.
Theorem 4: part d says A has a pivot position in everyrow (not column)–they are different.
A vs. augmented matrix. If a problem specifies a matrix,be sure you know whether it is an augmented matrix, orif it is the A matrix for the equation Ax = b.
Math 3191Applied Linear Algebra – p.3/30
Outline
Revisit Theorem 7 (from Sec. 1.7)
Sec. 1.8–Linear transformations.
Start Sec. 1.9–Matrix of linear transformation.
Math 3191Applied Linear Algebra – p.4/30
Theorem 7
An indexed set S = {v1,v2, . . . ,vp} of two or more vectors is linearly dependent ifand only if at least one of the vectors in S is a linear combination of the others. Infact, if S is linearly dependent, and v1 6= 0, then some vector vj (j ≥ 2) is a linearcombination of the preceding vectors v1, . . . ,vj−1.
Example: S = {v1,v2,v3,v4} =
8
>
>
<
>
>
:
2
6
6
4
1
0
0
3
7
7
5
,
2
6
6
4
0
1
0
3
7
7
5
,
2
6
6
4
1
1
0
3
7
7
5
,
2
6
6
4
0
1
1
3
7
7
5
9
>
>
=
>
>
;
.
v3 is a linear combination of v1 and v2, so S is linearly .
NOTE: v4 is not a linear combination of v1,v2,v3. That doesn’t change thefact that S is linearly independent.
Math 3191Applied Linear Algebra – p.5/30
Section 1.8–Linear Transformations
Key Concepts:
Matrix Transformations.
Solving transformation equations.
Geometry of matrix transformations.
Linear Transformations.
Math 3191Applied Linear Algebra – p.6/30
Matrix-vector multiplication
Another way to view Ax = b:
Matrix A is an object acting on x by multiplication to produce a new vector Ax or b.
EXAMPLE:2
6
6
4
2 −4
3 −6
1 −2
3
7
7
5
2
4
2
3
3
5 =
2
6
6
4
−8
−12
−4
3
7
7
5
2
6
6
4
2 −4
3 −6
1 −2
3
7
7
5
2
4
2
1
3
5 =
2
6
6
4
0
0
0
3
7
7
5
Math 3191Applied Linear Algebra – p.7/30
Transformation Machine
x ∈ IRn
multiply by A (m × n)
b ∈ IRm
Suppose A is m × n. Solving Ax = b amounts to finding all____ in Rn that are transformed into vector b in Rm throughmultiplication by A.
Math 3191Applied Linear Algebra – p.8/30
Matrix Transformations
A transformation T from Rn to Rm is a rule that assigns toeach vector x in Rn a vector T (x) in Rm.
x ∈ IRn
T : Rn −→ Rm
T (x) ∈ IRm
Math 3191Applied Linear Algebra – p.9/30
Terminology
Rn: domain of T
Rm: codomain of T
T (x) in Rm is the image of x under the transformation T
Set of all images T (x) is the range of T
Math 3191Applied Linear Algebra – p.10/30
EXAMPLE
Let A =
2
6
6
4
1 0
2 1
0 1
3
7
7
5
. Define a transformation T : R2−→ R3 by T (x) = Ax.
Then if x =
2
4
2
1
3
5 , T (x) = Ax =
2
6
6
4
1 0
2 1
0 1
3
7
7
5
2
4
2
1
3
5 =
2
6
6
4
2
5
1
3
7
7
5
.
1 2 3 4x1
1
2x2
0 1 2 3
x1
0 2 4 6 8
x2
0
1
2
x3
0
1
2
x3
Math 3191Applied Linear Algebra – p.11/30
EXAMPLE
Let A =
2
4
1 −2 3
−5 10 −15
3
5, u =
2
6
6
4
2
3
1
3
7
7
5
, b =
2
4
2
−10
3
5 and c =
2
4
3
0
3
5.
Define transformation T : R3→ R2 by T (x) = Ax.
(a) Find an x in R3 whose image under T is b.
(b) Is there more than one x under T whose image is b. (uniqueness problem)
(c) Determine if c is in the range of the transformation T . (existence problem)
Math 3191Applied Linear Algebra – p.12/30
Solution
(a) Solve _______=_____ for x. I.e., solve _______=_____ or
"
1 −2 3
−5 10 −15
#
2
6
6
4
x1
x2
x3
3
7
7
5
=
"
2
−10
#
Augmented matrix:
"
1 −2 3 2
−5 10 −15 −10
#
∼
"
1 −2 3 2
0 0 0 0
# x1 = 2x2 − 3x3 + 2
x2 is free
x3 is free
Let x2 = _____ and x3 = _____. Then x1 = _____.
So x =
2
6
6
4
3
7
7
5
Math 3191Applied Linear Algebra – p.13/30
Solution(cont).
(b) Is there another x for which T (x) = b?
Free variables exist =⇒ There is more than one x for which T (x) = b
(c) Is there an x for which T (x) = c? This is another way of asking if the equation Ax = c
is _______________.
Augmented matrix:
2
4
1 −2 3 3
−5 10 −15 0
3
5 ∼
2
4
1 −2 3 0
0 0 0 1
3
5
c is not in the _______________ of T .
Math 3191Applied Linear Algebra – p.14/30
Applications
Matrix transformations have many applications - including computer graphics.
EXAMPLE: Let A =
2
4
.5 0
0 .5
3
5. The transformation T : R2→ R2 defined by
T (x) = Ax is an example of a contraction transformation.
u =
2
4
8
6
3
5 T (u) =
2
4
.5 0
0 .5
3
5
2
4
8
6
3
5 =
2
4
4
3
3
5
2 4 6 8 10 12
-4
-2
2
4
6
2 4 6 8 10 12
-4
-2
2
4
6
Math 3191Applied Linear Algebra – p.15/30
Linear Transformations
If A is m × n, then the transformation T (x) = Ax has thefollowing properties:
T (u + v) = A (u + v) = _______ + _______
= ______ + ______
andT (cu) = A (cu) = _____Au =_____T (u)
for all u,v in Rn and all scalars c.
Math 3191Applied Linear Algebra – p.16/30
Def. of Linear Transformation
A transformation T is linear if:
T (u + v) = T (u) +T (v) for all u,v in the domain of T .
T (cu) =cT (u) for all u in the domain of T and all scalarsc.
Every matrix transformation is a linear transformation.
Math 3191Applied Linear Algebra – p.17/30
RESULT
If T is a linear transformation, thenT (0) = 0 and T (cu + dv) =cT (u) +dT (v).
Proof:T (0) = T (0u) = ____T (u) = _____.
T (cu + dv) = T ( ) + T ( ) =_____T ( ) + _____T ( )
Math 3191Applied Linear Algebra – p.18/30
Example
Let e1 =
[
1
0
]
, e2 =
[
0
1
]
, y1 =
1
0
2
and y2 =
0
1
1
.
Suppose T : R2 → R3 is a linear transformation that maps
e1 into y1 and e2 into y2. Find the images of
[
3
2
]
and[
x1
x2
]
.
Math 3191Applied Linear Algebra – p.19/30
Solution
First, note that
T (e1) = ______ and T (e2) = ______.Also
___e1 + ___e2 =
[
3
2
]
Then
T
([
3
2
])
= T (___e1 + ___e2) = ___T (e1) + ___T (e2)
=
Math 3191Applied Linear Algebra – p.20/30
1 2 3x1
1
2x2
01
23
x1
01
2 x2
0
2
4
6
8
x3
01
23
x1
01
2 x2
T (3e1 + 2e2) = 3T (e1) + 2T (e2)
Also
T
0
@
2
4
x1
x2
3
5
1
A = T (_____e1 + _____e2) = ____T (e1) + ____T (e2) = ____
Math 3191Applied Linear Algebra – p.21/30
Counterexample
Define T : R3 → R2 such that T (x1, x2, x3) = (|x1 + x3| , 2 + 5x2). Show that T
is a not a linear transformation.
Solution: Goal: Find a counterexample to one of the requirements i)T (cu) =cT (u) or ii) T (u + v) = T (u) + T (v) .
T (0) = T
0
B
B
@
2
6
6
4
0
0
0
3
7
7
5
1
C
C
A
=
2
4
3
5 6= _____
which means that T is not linear.
Math 3191Applied Linear Algebra – p.22/30
Another counterexample
Let c = −1 and u =
2
6
6
4
1
1
1
3
7
7
5
. Then
T (cu) = T
0
B
B
@
2
6
6
4
−1
−1
−1
3
7
7
5
1
C
C
A
=
2
4
|−1 + −1|
2 + 5 (−1)
3
5 =
2
4
2
−3
3
5
and
cT (u) = −1T
0
B
B
@
2
6
6
4
1
1
1
3
7
7
5
1
C
C
A
= −1
2
4
3
5 =
2
4
3
5 .
Therefore T (cu) 6= ___T (u) and therefore T is not _________________.
Math 3191Applied Linear Algebra – p.23/30
Sec 1.9: Matrix of a Linear Transformation
Identity Matrix In is an n × n matrix with 1’s on the main(left to right) diagonal and 0’s elsewhere. The ith column ofIn is denoted ei.
EXAMPLE:
I3 =[
e1 e2 e3
]
=
1 0 0
0 1 0
0 0 1
Math 3191Applied Linear Algebra – p.24/30
Note that
I3x =
1 0 0
0 1 0
0 0 1
x1
x2
x3
= ____
+ ____
+ ____
= ____.
In general, for x in Rn,Inx = ___
Math 3191Applied Linear Algebra – p.25/30
From Section 1.8, if T : Rn→ Rm is a linear transformation, then
T (cu + dv) =cT (u)+dT (v) .
Generalized Result:
T (c1v1 + · · · + cpvp) =c1T (v1) + · · · + cpT (vp).
EXAMPLE: Suppose T is a linear transformation from R2 to R3 where
T (e1) =
2
6
6
4
2
−3
4
3
7
7
5
and T (e2) =
2
6
6
4
5
0
1
3
7
7
5
.
where e1 =
2
4
1
0
3
5 and e2 =
2
4
0
1
3
5 .
Compute T (x) for any x =
2
4
x1
x2
3
5.Math 3191Applied Linear Algebra – p.26/30
Solution
A vector x in R2 can be written as[
x1
x2
]
= _____
[
1
0
]
+ _____
[
0
1
]
= _____e1 + _____e2
Then
T (x) = T (x1e1 + x2e2) = _____T (e1) + _____T (e2)
= _____
2
−3
4
+ _____
5
0
1
=
.
Math 3191Applied Linear Algebra – p.27/30
Finding the matrix of a linear transformation
On the previous slide, note that
T (x) =
[
x1
x2
]
.
So
T (x) =[
T (e1) T (e2)]
x = Ax
To get A, replace the identity matrix[
e1 e2
]
with[
T (e2) T (e2)]
.
Math 3191Applied Linear Algebra – p.28/30
Theorem 10
Let T : Rn → Rm be a linear transformation. Then thereexists a unique matrix A such that
T (x) = Ax for all x in Rn.
In fact, A is the m × n matrix whose jth column is the vectorT (ej), where ej is the jth column of the identity matrix in Rn.
A = [T (e1) T (e2) · · · T (en)]
↑
standard matrix for the linear transformation T
Math 3191Applied Linear Algebra – p.29/30
EXAMPLE
T (x) =
2
6
6
4
? ?
? ?
? ?
3
7
7
5
2
4
x1
x2
3
5 =
2
6
6
4
x1 − 2x2
4x1
3x1 + 2x2
3
7
7
5
Solution:
2
6
6
4
? ?
? ?
? ?
3
7
7
5
= standard matrix of the linear transformation T .
2
6
6
4
? ?
? ?
? ?
3
7
7
5
=h
T (e1) T (e2)i
=
2
6
6
4
3
7
7
5
.
Math 3191Applied Linear Algebra – p.30/30