Download 3. Linear function

Linear function
3.
Linear function
Introduction
Definition and examples
3.1.
Introduction
Image, Preimage, and Kernel
Linear operator
In calculus, a vector in the plane R2 with components 2 and −3 is usually written using
→
notation such as −
v = h2, −3i. For our purposes it turns out to be more convenient to
express such a vector as a 2 × 1 matrix:
2
x=
.
−3
More generally, a vector in Rn is written as an n × 1 matrix. When writing vectors in
text we usually use the matrix transpose notation to avoid unseemly vertical spacing. For
instance, we might write x = [6, −1, 3, 2]T , when we want to say
 
6
−1

x=
 3 .
2
Matrix of a linear function
Composition
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The addition and scalar multiplication defined for matrices (Section 2.1) gives an addition
and scalar multiplication for vectors, which coincides with the calculus definitions.
The idea of a function plays a central role in calculus and the same is true for linear algebra.
For most of the functions in calculus the inputs and outputs are both real numbers, but in
linear algebra, the functions we study have inputs and outputs that are vectors.
For instance, here is a function L from the set R2 to the set R3 :


x1 + 4x2
x1
L
= 3x1 − x2  .
x2
x2
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The notation works just like it did in calculus. For example, if the input vector is [2, 1]T ,
then the output vector is

  
(2) + 4(1)
6
2
L
= 3(2) − (1) = 5 .
1
(1)
1
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
This function satisfies a couple of properties that make it “linear,” meaning that it is
compatible with the addition and scalar multiplication of vectors (the precise definition is
given below). Linear functions are the main functions in linear algebra. We study them in
this section.
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3.2.
Definition and examples
Linear function.
n
A function L : R → R
m
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is linear if
(a) L(x + y) = L(x) + L(y),
(b) L(αx) = αL(x),
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for all x, y ∈ Rn , α ∈ R.
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The notation L : Rn → Rm is used to indicate that the input vectors come from the set
Rn (= domain of L) and the output vectors are in the set Rm (= codomain of L).
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3.2.1
Example
Show that the function L : R2 → R3 given by


x1 + 4x2
L(x) = 3x1 − x2 
x2
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
is linear.
Composition
2
2
Solution First, the input vector x is an element of R (according to the notation L : R →
R3 ), so it is of the form x = [x1 , x2 ]T . This is the meaning of x1 and x2 in the formula.
We need to verify that L satisfies the two properties in the definition of linear function.
For any x, y ∈ R2 , we have
x1 + y1
L(x + y) = L
x2 + y2


(x1 + y1 ) + 4(x2 + y2 )
(In the formula, x1 + y1 plays the role of x1
= 3(x1 + y1 ) − (x2 + y2 )
and x2 + y2 plays the role of x2 .)
(x2 + y2 )


(x1 + 4x2 ) + (y1 + 4y2 )
= (3x1 − x2 ) + (3y1 − y2 )
(x2 ) + (y2 )

 

y1 + 4y2
x1 + 4x2
= 3x1 − x2  + 3y1 − y2 
y2
x2
= L(x) + L(y),
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so property (a) holds. Next, for any x ∈ R2 and α ∈ R, we have


(αx1 ) + 4(αx2 )
αx1
L(αx) = L
= 3(αx1 ) − (αx2 )
αx2
(αx2 )




x1 + 4x2
α(x1 + 4x2 )
= α(3x1 − x2 ) = α 3x1 − x2 
x2
α(x2 )
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
= αL(x),
so property (b) holds. Therefore, L is linear.
3.2.2
Example
Show that the function L : R1 → R2 given by
2x1
L(x) =
−x1
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is linear.
Solution
1
For any x, y ∈ R , we have
2(x1 + y1 )
L(x + y) = L([x1 + y1 ]) =
−(x1 + y1 )
(2x1 ) + (2y1 )
=
(−x1 ) + (−y1 )
2y1
2x1
+
=
−y1
−x1
= L(x) + L(y),
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so property (a) holds. Next, for any x ∈ R1 and α ∈ R, we have
2(αx1 )
L(αx) = L([αx1 ]) =
−(αx1 )
α(2x1 )
2x1
=
=α
−x1
α(−x1 )
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
= αL(x),
so property (b) holds. Therefore, L is linear.
If a is any number, then the function f : R → R given by f (x) = ax has as its graph a
straight line (through the origin with slope a). In fact, this function is linear in the sense of
the above definition (regarding R as the same thing as R1 ). The next theorem generalizes
this statement with the number a being replaced by a matrix A.
Theorem. Let A be an m × n matrix. The function L : R
defined by
L(x) = Ax
n
→ R
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m
is linear.
The function L in the theorem is called the linear function corresponding to the
matrix A.
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Proof. It should be checked that L makes sense as a function from Rn to Rm . If x is an
input vector, then it is an element of Rn , and is therefore an n × 1 matrix. Since A is
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m × n, the product Ax is defined and equals an m × 1 matrix, which is an element of Rm ,
as desired.
Introduction
Definition and examples
We now check that L satisfies the two properties of a linear function. For any x, y ∈ Rn ,
we have
L(x + y) = A(x + y) = Ax + Ay = L(x) + L(y),
where the second equality is due to the distributive property of matrix multiplication
(property (d) in Section 2.3). This verifies property (a). Next, for any x ∈ Rn and α ∈ R,
we have
L(αx) = A(αx) = α(Ax) = αL(x)
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
where the second equality is due to a property of matrix and scalar multiplication (property
(i) in Section 2.3). This verifies property (b) and finishes the proof that L is linear.
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This gives us another way to check whether a given function is linear:
3.2.3
by
Example
Use the last theorem to show that the function L : R2 → R3 given


x1 + 4x2
L(x) = 3x1 − x2 
x2
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is linear.
Solution
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We have
 
1
x1 + 4x2
L(x) = 3x1 − x2  = 3
x2
0


4 x
−1 1 = Ax,
x2
1
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where


1 4
A = 3 −1 .
0 1
Therefore, L is linear by the preceding result.
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
The zero vector in Rn is the vector
0 = [0, 0, . . . , 0]T .
Theorem. Let L : Rn → Rm be a function. If L is linear, then
L(0) = 0.
Proof. Assume that L is linear. We have
L(0) + L(0) = L(0 + 0) = L(0),
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where the first equality is due to property (a) of a linear function. Subtracting L(0) from
both sides of this equation gives L(0) = 0, as desired.
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Put another way, the theorem says that if L does not send 0 to 0, then it cannot be linear.
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3.2.4
Example
linear? Explain.
1
2
Is the function F : R → R , given by
2x1 + 1
,
F (x) =
−x1
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Solution
Note that
2(0) + 1
1
0
F (0) =
=
6=
=0
−(0)
0
0
(the string says that F (0) 6= 0), so F is not linear according to the preceding theorem.
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
3.2.5
Example
Is the function F : R2 → R2 , given by
x x
F (x) = 1 2 ,
x1
Composition
linear? Explain.
Solution If we can show that the function does not send 0 to 0, then we can quickly
conclude that it is not linear (as in the preceding example). However,
(0)(0)
0
F (0) =
=
= 0,
(0)
0
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so all we know is that F has a chance of being linear.
We see if we can verify property (a) of a linear function. Let x, y ∈ R2 . We have
x1 + y1
(x1 + y1 )(x2 + y2 )
F (x + y) = F
=
(x1 + y1 )
x2 + y2
x1 x2 + x1 y2 + y1 x2 + y1 y2
.
=
x1 + y1
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We are trying to show that this equals
y1 y2
x1 x2
+
F (x) + F (y) =
y1
x1
x1 x2 + y1 y2
=
.
x1 + y1
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
Since the first components (in red) do not match up, we suspect that F is not linear. We
cannot write F (x + y) 6= F (x) + F (y), though, since there are choices for x and y that
actually give equality (for instance, x = 0 and y = 0).
However, in order to show that F fails property (a) it is enough to give a single counterexample. Using inspection, we see that if x1 , x2 , y1 , y2 are all equal to 1, for instance, then
the first components are not equal, so this should give our counterexample.
Everything we have done up to this point can be considered scratch work. It was done just
to come up with an idea for a counterexample. To solve the problem, all we really need to
write is this:
If x = [1, 1]T and y = [1, 1]T , then
2
4
2
1
1
F (x + y) = F
=
6=
=
+
= F (x) + F (y),
2
2
2
1
1
so F is not linear.
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3.3.
Image, Preimage, and Kernel
Introduction
Definition and examples
Image, Preimage, and Kernel
Definition of image.
Let L : Rn → Rm be a function.
ˆ Let x be a vector in R . The image of x under L is L(x).
n
Linear operator
Matrix of a linear function
Composition
ˆ The image of L (denoted im L) is the set of all images L(x) as x
ranges through Rn . In symbols,
im L = {L(x) | x ∈ Rn }.
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In other words, given an input vector x, its image is the corresponding output vector. And
the image of L is the set of all actual output vectors.
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3.3.1
Example
Let L : R3 → R2 be given by
x1 − 3x2 + 2x3
L(x) =
−2x1 + 6x2 − x3
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
T
(a) Find the image of [4, 1, −7] under L.
Matrix of a linear function
Composition
(b) Is [−5, 7]T in im L? Explain.
Solution
(a) The image of [4, 1, −7]T under L is
 
4
(4) − 3(1) + 2(−7)
−13




1
L
=
=
.
−2(4) + 6(1) − (−7)
5
−7
(b) The question amounts to asking if there is a vector x in R3 such that L(x) = [−5, 7]T ,
that is,
−5
x1 − 3x2 + 2x3
=
.
7
−2x1 + 6x2 − x3
This equality of vectors holds if and only if the vectors’ components are the same, so
this leads to a system of equations with corresponding augmented matrix
−5
1
−3
2
,
−2
6
−1
7
which has row echelon form
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1
0
−3
0
2
3
−5
−3
.
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There is no pivot in the augmented column, so a solution x = [x1 , x2 , x3 ]T exists.
Therefore, [−5, 7]T is in im L.
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Definition of preimage.
Composition
Let L : Rn → Rm be a function. Let y be a vector in Rm . The
preimage of y under L (denoted L−1 (y)) is the set of all x in Rn that
have image under L equal to y. In symbols:
L−1 (y) = {x ∈ Rn | L(x) = y}.
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Definition of kernel. Let L : Rn → Rm be a function. The kernel
of L (denoted ker L) is the preimage of 0 under L. In other words, ker L
is the set of all vectors in Rn that have image under L equal to 0. In
symbols:
ker L = L−1 (0) = {x ∈ Rn | L(x) = 0}.
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
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3.3.2
Example
3
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2
Let L : R → R be given by
x1 − 3x2 + 2x3
L(x) =
−2x1 + 6x2 − x3
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T
T
(a) Determine whether the vector [2, 0, −3] is in the preimage of [−4, 8] under L.
(b) Find L−1 ([−5, 7]T ).
(c) Find ker L.
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Solution
(a) Asking whether the vector [2, 0, −3]T is in the preimage of [−4, 8]T under
L is asking whether L([2, 0, −3]T ) = [−4, 8]T . Since
 
2
2 − 3(0) + 2(−3)
−4
−4
L  0  =
=
6=
,
−2(2) + 6(0) − (−3)
−1
8
−3
[2, 0, −3]T is not in the preimage of [−4, 8]T .
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
(b) We seek the set of all vectors x in R3 for which L(x) = [−5, 7]T , that is,
x1 − 3x2 + 2x3
−5
=
.
7
−2x1 + 6x2 − x2
This equality of vectors holds if and only if the vectors’ components are the same, so
this leads to a system of equations with corresponding augmented matrix
−5
1
−3
2
,
7
−2
6
−1
which has reduced row echelon form (RREF)
1 −3 0 −3
.
0
0
1 −1
The preimage of [−5, 7]T is the solution set of the corresponding system, which is
{[3t − 3, t, −1]T | t ∈ R}.
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(c) The kernel of L is the preimage of the zero vector, so the solution is just like the
solution to (b) except with [0, 0]T in place of [−5, 7]T . The augmented column in
the augmented matrix now consists of 0’s and, since row operations never change a
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column of all 0’s, we can immediately write down the reduced row echelon form of
the system:
1 −3 0 0
.
0
0
1 0
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
Therefore, ker L = {[3t, t, 0]T | t ∈ R}.
Matrix of a linear function
Composition
3.4.
Linear operator
A special name is given to a linear function L : Rn → Rm in the case m = n, that is, when
the domain and the codomain of L are the same:
Linear operator.
n
n
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n
A linear operator on R is a linear function from R to R .
Let L be a linear operator on R2 (the plane). Since L is a linear function from the plane
to itself, we can think of it as simply moving vectors in the plane: an input vector gets
moved to the corresponding output vector. (A similar statement can be made for a linear
operator on Rn for any n.)
3.4.1
Example
Let L : R2 → R2 be “projection onto the x1 -axis.”
(a) Find the image of [2, 3]T under L geometrically.
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(b) Find the kernel of L geometrically.
(c) Find a general formula for L(x).
(d) Use the general formula found in part (c) to redo parts (a) and (b) analytically.
Solution
(a) The image of [2, 3]T under L is L([2, 3]T ), which is [2, 0]T :
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
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(b) The kernel of L is the set of all vectors x such that L(x) = 0. This set is the x2 -axis,
so ker L = {[0, t]T | t ∈ R}:
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Linear operator
Matrix of a linear function
Composition
(c) A general formula for L(x) is L(x) = [x1 , 0]T (keep the first component the same,
but change the second component to 0).
T
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T
(d) Redoing part (a) using the formula, we have L([2, 3] ) = [2, 0] .
For part (b), we seek the set of all x for which L(x) = 0, that is, [x1 , 0]T = [0, 0]T . This
last equation forces x1 = 0 but places no restriction on x2 , so ker L = {[0, x2 ]T | x2 ∈
R} (which is the same as the set obtained in (b) since x2 acts as a dummy variable,
meaning that renaming it has no effect).
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3.4.2
Example
Let L : R2 → R2 be “90◦ counterclockwise rotation.”
(a) Find the image of [3, 1]T under L geometrically.
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(b) Find the preimage of [2, −3]T under L geometrically.
(c) Find a general formula for L(x).
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(d) Use the general formula found in part (c) to redo parts (a) and (b) analytically.
Solution
T
T
T
(a) The image of [3, 1] under L is L([3, 1] ), which is [−1, 3] :
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
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(b) The preimage of [2, −3]T under L is the set of all those vectors that L moves to
[2, −3]T . There is only one such vector, namely [−3, −2]T , so L−1 ([2, −3]T ) =
{[−3, −2]T }:
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Linear operator
Matrix of a linear function
Composition
(c) The general formula for L is L(x) = [−x2 , x1 ]T (switch components and then negate
the first). (Part (a) shows that this formula works for x in the first quadrant and one
can check that it works in the other three quadrants as well.)
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(d) Redoing part (a) using the formula, we have L([3, 1]T ) = [−1, 3]T .
For part (b), we seek the set of all x for which L(x) = [2, −3]T , that is, [−x2 , x1 ]T =
[2, −3]T . This equation forces x1 = −3 and x2 = −2, so L−1 ([2, −3]T ) = {[−3, −2]T }.
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The functions given in the last two examples are linear (as can be checked by using the
general formulas). The following functions from R2 to itself are all linear:
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ˆ projection onto line through origin,
ˆ rotation about origin,
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ˆ reflection across line through origin,
ˆ dilation (= multiplication by scalar > 1),
ˆ contraction (= multiplication by scalar between 0 and 1).
However, translation by a vector t (which sends x to x + t) is not linear if t is nonzero
(since, for instance, it does not send 0 to 0).
3.5.
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
Matrix of a linear function
We have seen that if A is an m × n matrix, then we get a linear function L : Rn → Rm by
defining
L(x) = Ax.
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Here we turn things around and show that if we start with a linear function L : Rn → Rm ,
then we can use it to build a matrix A so that the above equation holds.
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The construction requires the following notation:
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1
0
, e2 =
;
0
1
 
 
 
1
0
0
ˆ In R3 , e1 = 0, e2 = 1, e3 = 0;
0
0
1
ˆ In R2 , e1 =
and so forth. These are the standard unit vectors.
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Matrix of a linear function.
Introduction
Let L : Rn → Rm be a linear function. There is a unique m × n matrix
A such that
L(x) = Ax
Definition and examples
for all x ∈ Rn . Moreover,
A = L(e1 ) L(e2 ) · · · L(en ) .
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
The matrix A is called the matrix of L.
This is a special case of a theorem that will be presented later, so we postpone the proof
till then.
3.5.1
Example
Let L : R2 → R3 be the linear function given by


x1 + 4x2
L(x) = 3x1 − x2  .
x2
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(a) Find the matrix of L.
(b) Use part (a) to find L([5, −2]T ).
(c) Find L([5, −2]T ) directly from the formula for L and verify that it agrees with the
answer to part (b).
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Solution
(a) We have
Introduction

  
(1) + 4(0)
1
L(e1 ) = L([1, 0]T ) = 3(1) − (0) = 3 ,
(0)
0
and similarly, L(e2 ) = [4, −1, 1]T , so the matrix A of L is


1 4
A = L(e1 ) L(e2 ) = 3 −1 .
0 1
(b) Using the formula L(x) = Ax, we have


 
1 4 −3
5
L([5, −2]T ) = 3 −1
=  17  .
−2
0 1
−2
Definition and examples
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
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(c) The formula for L gives

  
(5) + 4(−2)
−3
L([5, −2]T ) = 3(5) − (−2) =  17  ,
(−2)
−2
in agreement with part (b).
3.5.2
Example
Let L : R2 → R2 be “reflection across the x2 -axis.”
(a) Find the matrix of L.
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(b) Use part (a) to find L([1, 3]T ).
Introduction
(c) Find L([1, 3]T ) geometrically and verify that it agrees with the answer to part (b).
Definition and examples
Image, Preimage, and Kernel
Solution
(a) The matrix A of L is
Linear operator
−1 0
A = L(e1 ) L(e2 ) =
.
0 1
(b) Using the formula L(x) = Ax, we have
−1
T
L([1, 3] ) =
0
Composition
0 1
−1
=
.
1 3
3
(c) Since reflection across the x2 -axis negates the first component of a vector and keeps
the second component the same, we have L([1, 3]T ) = [−1, 3]T , in agreement with
part (b).
3.6.
Matrix of a linear function
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Composition
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The reader is likely familiar with the concept of a composition of real-valued functions. For
instance, if f (x) = 2x + 3 and g(x) = x2 , then the composition of f and g is given by
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(g ◦ f )(x) = g(f (x)) = (f (x))2 = (2x + 3)2 .
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The composition can be described as “doing f first and then g”. In more detail, the
composition takes an input x, uses f to produce the output f (x), and then uses g with
input f (x) to produce the final output g(f (x)).
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We can compose linear functions as well: If L : Rn → Rm and M : Rm → Rl are linear
functions, then the composition of L and M is the function M ◦ L : Rn → Rl given by
Introduction
Definition and examples
(M ◦ L)(x) = M (L(x)).
In order for the composition to make sense, the domain of M must be the same as the
codomain of L (both equal to Rm above), for otherwise the output produced by L could
not be used as an input for M .
3.6.1
Let L : R2 → R3 be the linear function given by


x1 + 2x2
L(x) =  −x1 
3x1 − 4x2
Example
and let M : R3 → R2 be the linear function given by
5x1 + x2 − 7x3
.
M (x) =
−x1 + x3
Find a formula for the composition M ◦ L : R2 → R2 .
Solution
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
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We have


x1 + 2x2
(M ◦ L)(x) = M (L(x)) = M  −x1 
3x1 − 4x2
5(x1 + 2x2 ) + (−x1 ) − 7(3x1 − 4x2 )
=
−(x1 + 2x2 ) + (3x1 − 4x2 )
−17x1 + 38x2
.
=
2x1 − 6x2
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Introduction
Definition and examples
Theorem. Let L : Rn → Rm and M : Rm → Rl be linear functions,
let A be the matrix of L, and let B be the matrix of M .
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
(a) M ◦ L is linear,
Composition
(b) the matrix of M ◦ L is BA.
Proof. (a) For any x, y ∈ Rn , we have
(M ◦ L)(x + y) = M (L(x + y))
= M (L(x) + L(y))
(L is linear)
= M (L(x)) + M (L(y))
(M is linear)
= (M ◦ L)(x) + (M ◦ L)(y),
so M ◦ L satisfies the first property of a linear function. Verification of the second property
is left to the exercises (see Exercise 3 – 10).
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(b) For any x ∈ Rn , we have
(M ◦ L)(x) = M (L(x)) = M (Ax) = B(Ax) = (BA)x,
so the matrix of M ◦ L is BA (by the uniqueness assertion in 3.5).
3.6.2
Example
Let L : R2 → R3 and M : R3 → R2 be as in Example 3.6.1.
(a) Find the matrix A of L and the matrix B of M and use these matrices to find the
matrix C of the composition M ◦ L.
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(b) Use the formula for M ◦ L found in Example 3.6.1 to find the matrix of M ◦ L directly
and compare with the answer to part (a).
Introduction
Definition and examples
Solution
Image, Preimage, and Kernel
(a) We have

and

1
2
A = L(e1 ) L(e2 ) = −1 0 
3 −4
B = M (e1 ) M (e2 ) M (e3 ) =
Linear operator
Matrix of a linear function
Composition
5 1 −7
.
−1 0 1
Therefore, according to the theorem, the matrix C of the composition is


1
2
5 1 −7 
−17 38

−1 0 =
C = BA =
.
−1 0 1
2
−6
3 −4
(b) Using the formula for M ◦ L found in Example 3.6.1, we have
−17 38
C = (M ◦ L)(e1 ) (M ◦ L)(e2 ) =
,
2
−6
in agreement with part (a).
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Page 26 of 32
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2
2
3.6.3 Example Let L : R → R be “90 counterclockwise rotation” and let M :
R2 → R2 be “projection onto the x1 -axis.”
◦
(a) Using geometry, find the matrix A of L and the matrix B of M and then use these
matrices to find the matrix C of the composition M ◦ L.
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(b) Using geometry, find the matrix of M ◦ L directly and compare with the answer to
part (a).
Introduction
Definition and examples
Solution
and
(a) Using geometry to see where L and M send the vectors e1 and e2 , we get
0 −1
A = L(e1 ) L(e2 ) =
1 0
1
B = M (e1 ) M (e2 ) =
0
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
0
.
0
Therefore, according to the theorem, the matrix C of the composition is
1 0 0 −1
0 −1
C = BA =
=
.
0 0 1 0
0 0
(b) The vector e1 = [1, 0]T is sent by L to [0, 1]T , which is sent by M to [0, 0]T , so
(M ◦ L)(e1 ) = [0, 0]T . Similarly, (M ◦ L)(e2 ) = [−1, 0]T . Therefore,
0 −1
C = (M ◦ L)(e1 ) (M ◦ L)(e2 ) =
,
0 0
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Page 27 of 32
in agreement with part (a).
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3 – Exercises
Introduction
Definition and examples
Image, Preimage, and Kernel
3–1
Show directly from the definition that the function L : R3 → R2 given by
2x1 − x2 + 4x3
L(x) =
x1 − 6x3
Linear operator
Matrix of a linear function
Composition
is linear.
3–2
Show directly from the definition that the function L : R2 → R1 given by
L(x) = 5x1 − 8x2
is linear.
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3–3
Let L : R3 → R2 be the linear function corresponding to the matrix
−1 7 2
A=
1 −6 1
Introduction
Definition and examples
Image, Preimage, and Kernel
Linear operator
(see Section 3.2 for what this means).
Matrix of a linear function
(a) Find L([4, 1, −3]T )
Composition
(b) Determine whether [6, 2, −2]T is in L−1 ([4, −8]T ).
(c) Find L−1 ([4, −8]T ) and use it to verify your answer to part (b).
3–4
3–5
In each case, determine whether the function F is linear:
(a) F : R2 → R1 given by F (x) = x21 − x22 ,
x1
.
(b) F : R2 → R2 given by F (x) =
cos x2
Let L : R2 → R3 be given by
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
x1 + 3x2
L(x) = −2x1 − 6x2 
3x1 + 9x2

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(a) Find the image of [−2, 1]T under L.
(b) Is [−2, 4, 6]T in im L? Explain.
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Introduction
Definition and examples
3–6
Find the kernel of the linear function L : R3 → R3 given by L(x) = [0, 2x2 − x3 , −6x2 +
3x3 ]T .
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
3–7
Let L : R2 → R2 be “reflection across the 45◦ line x2 = x1 .” (In calculus, this line is
written y = x.)
(a) Find the image of [2, 1]T under L geometrically.
(b) Find the preimage of [1, −3]T under L geometrically.
(c) Find a general formula for L(x).
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(d) Use the general formula found in part (c) to redo parts (a) and (b) analytically.
3–8
Let L : R3 → R2 be the linear function given by
2x1 − x2 + 5x3
L(x) =
.
7x2 + 4x3
(a) Find the matrix of L.
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T
(b) Use part (a) to find L([3, 2, −1] ).
(c) Find L([3, 2, −1]T ) directly from the formula for L and verify that it agrees with the
answer to part (b).
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3–9
Let L : R2 → R2 be “90◦ clockwise rotation.”
Introduction
(a) Find the matrix of L.
Definition and examples
(b) Use part (a) to find L([2, 1]T ).
Image, Preimage, and Kernel
Linear operator
T
(c) Find L([2, 1] ) geometrically and verify that it agrees with the answer to part (b).
Matrix of a linear function
Composition
3 – 10
Let L : Rn → Rm and M : Rm → Rl be linear functions. Verify that
(M ◦ L)(αx) = α(M ◦ L)(x)
for all x ∈ Rn and α ∈ R. (This is the second part of the verification that M ◦ L is linear.
See the theorem of Section 3.6 and its proof.)
3 – 11
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Let L : R3 → R2 be the linear function given by
−x1 + 4x2 + 2x3
L(x) =
3x1 − x3
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and let M : R2 → R3 be the linear function given by


7x1 + x2
M (x) = x1 + 6x2  .
5x2
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3
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3
Find a formula for the composition M ◦ L : R → R .
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3 – 12
Let L : R2 → R2 be “reflection across the line x2 = −x1 ” and let M : R2 → R2 be
“projection onto the x2 -axis.”
Introduction
Definition and examples
(a) Using geometry, find the matrix A of L and the matrix B of M and then use these
matrices to find the matrix C of the composition M ◦ L.
(b) Using geometry, find the matrix of M ◦ L directly and compare with the answer to
part (a).
Image, Preimage, and Kernel
Linear operator
Matrix of a linear function
Composition
Table of Contents
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