From A First Course in Linear Algebra
Version 1.04
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/
Here comes a key definition.
Definition LT
Linear Transformation
A linear transformation, ,
is a function that carries elements of the vector space
(called the domain)
to the vector space
(called the codomain), and which has two additional properties
(This definition contains Notation LT.)
The two defining conditions in the definition of a linear transformation should “feel linear,” whatever that means. Conversely, these two conditions could be taken as a exactly what it means to be linear. As every vector space property derives from vector addition and scalar multiplication, so too, every property of a linear transformation derives from these two defining properties. While these conditions may be reminiscent of how we test subspaces, they really are quite different, so do not confuse the two.
Here are two diagrams that convey the essence of the two defining properties of a linear transformation. In each case, begin in the upper left-hand corner, and follow the arrows around the rectangle to the lower-right hand corner, taking two different routes and doing the indicated operations labeled on the arrows. There are two results there. For a linear transformation these two expressions are always equal.
2006/11/15: Commutative diagrams not available in XML version. See PDF versions.
A couple of words about notation. is the name of the linear transformation, and should be used when we want to discuss the function as a whole. is how we talk about the output of the function, it is a vector in the vector space . When we write , the plus sign on the left is the operation of vector addition in the vector space , since and are elements of . The plus sign on the right is the operation of vector addition in the vector space , since and are elements of the vector space . These two instances of vector addition might be wildly different.
Let’s examine several examples and begin to form a catalog of known linear transformations to work with.
Example ALT
A linear transformation
Define
by describing the output of the function for a generic input with the formula
and check the two defining properties.
So by Definition LT, is a linear transformation.
It can be just as instructive to look at functions that are not linear transformations. Since the defining conditions must be true for all vectors and scalars, it is enough to find just one situation where the properties fail.
Example NLT
Not a linear transformation
Define
by
This function “looks” linear, but consider
So the second required property fails for the choice of and and by Definition LT, is not a linear transformation. It is just about as easy to find an example where the first defining property fails (try it!). Notice that it is the “-2” in the third component of the definition of that prevents the function from being a linear transformation.
Example LTPM
Linear transformation, polynomials to matrices
Define a linear transformation
by
We verify the two defining conditions of a linear transformations.
So by Definition LT, is a linear transformation.
Example LTPP
Linear transformation, polynomials to polynomials
Define a function
by
Then
So by Definition LT, is a linear transformation.
Linear transformations have many amazing properties, which we will investigate through the next few sections. However, as a taste of things to come, here is a theorem we can prove now and put to use immediately.
Theorem LTTZZ
Linear Transformations Take Zero to Zero
Suppose is a linear
transformation. Then .
Proof The two zero vectors in the conclusion of the theorem are different. The first is from while the second is from . We will subscript the zero vectors in this proof to highlight the distinction. Think about your objects. (This proof is contributed by Mark Shoemaker).
Return to Example NLT and compute to quickly see again that is not a linear transformation, while in Example LTPM and compute as an example of Theorem LTTZZ at work.
If you give me a matrix, then I can quickly build you a linear transformation. Always. First a motivating example and then the theorem.
Example LTM
Linear transformation from a matrix
Let
and define a function by
So we are using an old friend, the matrix-vector product (Definition MVP) as a way to convert a vector with 4 components into a vector with 3 components. Applying Definition MVP allows us to write the defining formula for in a slightly different form,
So we recognize the action of the function as using the components of the vector () as scalars to form the output of as a linear combination of the four columns of the matrix , which are all members of , so the result is a vector in . We can rearrange this expression further, using our definitions of operations in (Section VO).
You might recognize this final expression as being similar in style to some previous examples (Example ALT) and some linear transformations defined in the archetypes (Archetype M through Archetype R). But the expression that says the output of this linear transformation is a linear combination of the columns of is probably the most powerful way of thinking about examples of this type.
Almost forgot — we should verify that is indeed a linear transformation. This is easy with two matrix properties from Section MM.
So by Definition LT, is a linear transformation.
So the multiplication of a vector by a matrix “transforms” the input vector into an output vector, possibly of a different size, by performing a linear combination. And this transformation happens in a “linear” fashion. This “functional” view of the matrix-vector product is the most important shift you can make right now in how you think about linear algebra. Here’s the theorem, whose proof is very nearly an exact copy of the verification in the last example.
Theorem MBLT
Matrices Build Linear Transformations
Suppose that
is an matrix.
Define a function
by . Then
is a linear
transformation.
Proof
So by Definition LT, is a linear transformation.
So Theorem MBLT gives us a rapid way to construct linear transformations. Grab an matrix , define and Theorem MBLT tells us that is a linear transformation from to , without any further checking.
We can turn Theorem MBLT around. You give me a linear transformation and I will give you a matrix.
Example MFLT
Matrix from a linear transformation
Define the function
by
You could verify that is a linear transformation by applying the definition, but we will instead massage the expression defining a typical output until we recognize the form of a known class of linear transformations.
So if we define the matrix
then . By Theorem MBLT, we can easily recognize as a linear transformation since it has the form described in the hypothesis of the theorem.
Example MFLT was not accident. Consider any one of the archetypes where both the domain and codomain are sets of column vectors (Archetype M through Archetype R) and you should be able to mimic the previous example. Here’s the theorem, which is notable since it is our first occasion to use the full power of the defining properties of a linear transformation when our hypothesis includes a linear transformation.
Theorem MLTCV
Matrix of a Linear Transformation, Column Vectors
Suppose that
is a linear transformation. Then there is an
matrix
such
that .
Proof The conclusion says a certain matrix exists. What better way to prove something exists than to actually build it? So our proof will be constructive, and the procedure that we will use abstractly in the proof can be used concretely in specific examples.
Let be the columns of the identity matrix of size , (Definition SUV). Evaluate the linear transformation with each of these standard unit vectors as an input, and record the result. In other words, define vectors in , , by
Then package up these vectors as the columns of a matrix
Does have the desired properties? First, is clearly an matrix. Then
as desired.
So if we were to restrict our study of linear transformations to those where the domain and codomain are both vector spaces of column vectors (Definition VSCV), every matrix leads to a linear transformation of this type (Theorem MBLT), while every such linear transformation leads to a matrix (Theorem MLTCV). So matrices and linear transformations are fundamentally the same. We call the matrix of Theorem MLTCV the matrix representation of .
We have defined linear transformations for more general vector spaces than just , can we extend this correspondence between linear transformations and matrices to more general linear transformations (more general domains and codomains)? Yes, and this is the main theme of Chapter R. Stay tuned. For now, let’s illustrate Theorem MLTCV with an example.
Example MOLT
Matrix of a linear transformation
Suppose
is defined by
Then
so define
and Theorem MLTCV guarantees that .
As an illuminating exercise, let and compute two different ways. First, return to the definition of and evaluate directly. Then do the matrix-vector product . In both cases you should obtain the vector .
It is the interaction between linear transformations and linear combinations that lies at the heart of many of the important theorems of linear algebra. The next theorem distills the essence of this. The proof is not deep, the result is hardly startling, but it will be referenced frequently. We have already passed by one occasion to employ it, in the proof of Theorem MLTCV. Paraphrasing, this theorem says that we can “push” linear transformations “down into” linear combinations, or “pull” linear transformations “up out” of linear combinations. We’ll have opportunities to both push and pull.
Theorem LTLC
Linear Transformations and Linear Combinations
Suppose that is a
linear transformation,
are vectors from
and are
scalars from .
Then
Proof
Our next theorem says, informally, that it is enough to know how a linear transformation behaves for inputs from a basis of the domain, and all other outputs are described by a linear combination of these values. Again, the theorem and its proof are not remarkable, but the insight that goes along with it is fundamental.
Theorem LTDB
Linear Transformation Defined on a Basis
Suppose that is a
linear transformation,
is a basis for
and is a
vector from .
Let be the
scalars from
such that
Then
Proof For any , Theorem VRRB says there are (unique) scalars such that is a linear combination of the basis vectors in . The result then follows from a straightforward application of Theorem LTLC to the linear combination.
Example LTDB1
Linear transformation defined on a basis
Suppose you are told that
is a linear transformation and given the three values,
Because
is a basis for (Theorem SUVB), Theorem LTDB says we can compute any output of with just this information. For example, consider,
so
Doing it again,
so
Any other value of could be computed in a similar manner. So rather than being given a formula for the outputs of , the requirement that behave as a linear transformation, along with its values on a handful of vectors (the basis), are just as sufficient as a formula for computing any value of the function. You might notice some parallels between this example and Example MOLT or Theorem MLTCV.
Example LTDB2
Linear transformation defined on a basis
Suppose you are told that
is a linear transformation and given the three values,
You can check that
is a basis for (make the vectors the columns of a square matrix and check that the matrix is nonsingular, Theorem CNMB). By Theorem LTDB we can compute any output of with just this information. However, we have to work just a bit harder to take an input vector and express it as a linear combination of the vectors in . For example, consider,
Then we must first write as a linear combination of the vectors in and solve for the unknown scalars, to arrive at
Then Theorem LTDB gives us
Any other value of could be computed in a similar manner.
Here is a third example of a linear transformation defined by its action on a basis, only with more abstract vector spaces involved.
Example LTDB3
Linear transformation defined on a basis
The set
is a subspace of the vector space of polynomials
. This
subspace has
as a basis (check this!). Suppose we define a linear transformation
by
the values
To illustrate a sample computation of , consider . Verify that is an element of (does it have roots at and ?), then find the scalars needed to write it as a linear combination of the basis vectors in . Because
Theorem LTDB gives us
And all the other outputs of could be computed in the same manner. Every output of will have a zero in the second row, second column. Can you see why this is so?
The definition of a function requires that for each input in the domain there is exactly one output in the codomain. However, the correspondence does not have to behave the other way around. A member of the codomain might have many inputs from the domain that create it, or it may have none at all. To formalize our discussion of this aspect of linear transformations, we define the pre-image.
Definition PI
Pre-Image
Suppose that is a linear
transformation. For each ,
define the pre-image of
to be the subset of
given by
In other words, is the set of all those vectors in the domain that get “sent” to the vector .
TODO: All preimages form a partition of , an equivalence relation is about. Maybe to exercises.
Example SPIAS
Sample pre-images, Archetype S
Archetype S is the linear transformation defined by
We could compute a pre-image for every element of the codomain . However, even in a free textbook, we do not have the room to do that, so we will compute just two.
Choose
for no particular reason. What is ? Suppose . That becomes
Using matrix equality (Definition ME), we arrive at a system of four equations in the three unknowns with an augmented matrix that we can row-reduce in the hunt for solutions,
We recognize this system as having infinitely many solutions described by the single free variable . Eventually obtaining the vector form of the solutions (Theorem VFSLS), we can describe the preimage precisely as,
This last line is merely a suggestive way of describing the set on the previous line. You might create three or four vectors in the preimage, and evaluate with each. Was the result what you expected? For a hint of things to come, you might try evaluating with just the lone vector in the spanning set above. What was the result? Now take a look back at Theorem PSPHS. Hmmmm.
OK, let’s compute another preimage, but with a different outcome this time. Choose
What is ? Suppose . That becomes
Using matrix equality (Definition ME), we arrive at a system of four equations in the three unknowns with an augmented matrix that we can row-reduce in the hunt for solutions,
By Theorem RCLS we recognize this system as inconsistent. So no vector is a member of and so
The preimage is just a set, it is almost never a subspace of (you might think about just when is a subspace, see Exercise ILT.T10). We will describe its properties going forward, and it will be central to the main ideas of this chapter.
We can combine linear transformations in natural ways to create new linear transformations. So we will define these combinations and then prove that the results really are still linear transformations. First the sum of two linear transformations.
Definition LTA
Linear Transformation Addition
Suppose that
and are
two linear transformations with the same domain and codomain. Then their sum is the
function
whose outputs are defined by
Notice that the first plus sign in the definition is the operation being defined, while the second one is the vector addition in . (Vector addition in will appear just now in the proof that is a linear transformation.) Definition LTA only provides a function. It would be nice to know that when the constituents (, ) are linear transformations, then so too is .
Theorem SLTLT
Sum of Linear Transformations is a Linear Transformation
Suppose that
and
are two linear transformations with the same domain and codomain. Then
is a linear
transformation.
Proof We simply check the defining properties of a linear transformation (Definition LT). This is a good place to consistently ask yourself which objects are being combined with which operations.
Example STLT
Sum of two linear transformations
Suppose that
and
are defined by
Then by Definition LTA, we have
and by Theorem SLTLT we know is also a linear transformation from to .
Definition LTSM
Linear Transformation Scalar Multiplication
Suppose that is a linear
transformation and .
Then the scalar multiple is the function
whose outputs are defined by
Given that is a linear transformation, it would be nice to know that is also a linear transformation.
Theorem MLTLT
Multiple of a Linear Transformation is a Linear Transformation
Suppose that is a linear
transformation and .
Then is a linear
transformation.
Proof We simply check the defining properties of a linear transformation (Definition LT). This is another good place to consistently ask yourself which objects are being combined with which operations.
Example SMLT
Scalar multiple of a linear transformation
Suppose that
is defined by
For the sake of an example, choose , so by Definition LTSM, we have
and by Theorem MLTLT we know is also a linear transformation from to .
Now, let’s imagine we have two vector spaces, and , and we collect every possible linear transformation from to into one big set, and call it . Definition LTA and Definition LTSM tell us how we can “add” and “scalar multiply” two elements of . Theorem SLTLT and Theorem MLTLT tell us that if we do these operations, then the resulting functions are linear transformations that are also in . Hmmmm, sounds like a vector space to me! A set of objects, an addition and a scalar multiplication. Why not?
Theorem VSLT
Vector Space of Linear Transformations
Suppose that
and
are vector spaces. Then the set of all linear transformations from
to
,
is a vector
space when the operations are those given in Definition LTA and Definition LTSM.
Proof Theorem SLTLT and Theorem MLTLT provide two of the ten properties in Definition VS. However, we still need to verify the remaining eight properties. By and large, the proofs are straightforward and rely on concocting the obvious object, or by reducing the question to the same vector space property in the vector space .
The zero vector is of some interest, though. What linear transformation would we add to any other linear transformation, so as to keep the second one unchanged? The answer is defined by for every . Notice how we do not need to know any specifics about and to make this definition.
Definition LTC
Linear Transformation Composition
Suppose that
and
are linear transformations. Then the composition of
and
is the
function
whose outputs are defined by
Given that and are linear transformations, it would be nice to know that is also a linear transformation.
Theorem CLTLT
Composition of Linear Transformations is a Linear Transformation
Suppose that
and are linear
transformations. Then is
a linear transformation.
Proof We simply check the defining properties of a linear transformation (Definition LT).
Example CTLT
Composition of two linear transformations
Suppose that
and
are defined by
Then by Definition LTC
and by Theorem CLTLT is a linear transformation from to .
Here is an interesting exercise that will presage an important result later. In Example STLT compute (via Theorem MLTCV) the matrix of , and . Do you see a relationship between these three matrices?
In Example SMLT compute (via Theorem MLTCV) the matrix of and . Do you see a relationship between these two matrices?
Here’s the tough one. In Example CTLT compute (via Theorem MLTCV) the matrix of , and . Do you see a relationship between these three matrices???
C15 The archetypes below are all linear transformations whose domains and
codomains are vector spaces of column vectors (Definition VSCV). For
each one, compute the matrix representation described in the proof of
Theorem MLTCV.
Archetype M
Archetype N
Archetype O
Archetype P
Archetype Q
Archetype R
Contributed by Robert Beezer
C20 Let . Referring to
Example MOLT, compute
two different ways. First use the definition of
, then compute the
matrix-vector product
(Definition MVP).
Contributed by Robert Beezer Solution [1306]
C25 Define the linear transformation
Verify that
is a linear transformation.
Contributed by Robert Beezer Solution [1306]
C26 Verify that the function below is a linear transformation.
Contributed by Robert Beezer Solution [1306]
C30 Define the linear transformation
Compute the preimages,
and .
Contributed by Robert Beezer Solution [1307]
C31 For the linear transformation compute the pre-images.
M10 Define two linear transformations, and by
Using the proof of Theorem MLTCV compute the matrix representations of the three linear
transformations ,
and
.
Discover and comment on the relationship between these three matrices.
Contributed by Robert Beezer Solution [1312]
C20 Contributed by Robert Beezer Statement [1302]
In both cases the result will be .
C25 Contributed by Robert Beezer Statement [1302]
We can rewrite
as follows:
and Theorem MBLT tell us that any function of this form is a linear transformation.
C26 Contributed by Robert Beezer Statement [1303]
Check the two conditions of Definition LT.
So is indeed a linear transformation.
C30 Contributed by Robert Beezer Statement [1303]
For the first pre-image, we want
such that .
This becomes,
Vector equality gives a system of two linear equations in three variables, represented by the augmented matrix
so the system is inconsistent and the pre-image is the empty set. For the second pre-image the same procedure leads to an augmented matrix with a different vector of constants
This system is consistent and has infinitely many solutions, as we can see from the presence of the two free variables ( and ) both to zero. We apply Theorem VFSLS to obtain
C31 Contributed by Robert Beezer Statement [1304]
We work from the definition of the pre-image, Definition PI. Setting
we arrive at a system of three equations in three variables, with an augmented matrix that we row-reduce in a search for solutions,
With a leading 1 in the last column, this system is inconsistent (Theorem RCLS), and there are no values of , and that will create an element of the pre-image. So the preimage is the empty set.
We work from the definition of the pre-image, Definition PI. Setting
we arrive at a system of three equations in three variables, with an augmented matrix that we row-reduce in a search for solutions,
The solution set to this system, which is also the desired pre-image, can be expressed using the vector form of the solutions (Theorem VFSLS)
Does the final expression for this set remind you of Theorem KPI?
M10 Contributed by Robert Beezer Statement [1304]