From A First Course in Linear Algebra
Version 2.10
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/
In this section we will conclude our introduction to linear transformations by
bringing together the twin properties of injectivity and surjectivity and consider
linear transformations with both of these properties.
One preliminary definition, and then we will have our main definition for this section.
Definition IDLT
Identity Linear Transformation
The identity linear transformation on the vector space
is
defined as
Informally, is the “do-nothing” function. You should check that is really a linear transformation, as claimed, and then compute its kernel and range to see that it is both injective and surjective. All of these facts should be straightforward to verify (Exercise IVLT.T05). With this in hand we can make our main definition.
Definition IVLT
Invertible Linear Transformations
Suppose that
is a linear transformation. If there is a function
such
that
then is invertible. In this case, we call the inverse of and write .
Informally, a linear transformation is invertible if there is a companion linear transformation, , which “undoes” the action of . When the two linear transformations are applied consecutively (composition), in either order, the result is to have no real effect. It is entirely analogous to squaring a positive number and then taking its (positive) square root.
Here is an example of a linear transformation that is invertible. As usual at the beginning of a section, do not be concerned with where came from, just understand how it illustrates Definition IVLT.
Example AIVLT
An invertible linear transformation
Archetype V is the linear transformation
Define the function defined by
Then
For now, understand why these computations show that is invertible, and that . Maybe even be amazed by how works so perfectly in concert with ! We will see later just how to arrive at the correct form of (when it is possible).
It can be as instructive to study a linear transformation that is not invertible.
Example ANILT
A non-invertible linear transformation
Consider the linear transformation
defined by
Suppose we were to search for an inverse function .
First verify that the matrix is not in the range of . This will amount to finding an input to , , such that
As this system of equations is inconsistent, there is no input column vector, and . How should we define ? Note that
So any definition we would provide for must then be a column vector that sends to and we would have , contrary to the definition of . This is enough to see that there is no function that will allow us to conclude that is invertible, since we cannot provide a consistent definition for if we assume is invertible.
Even though we now know that is not invertible, let’s not leave this example just yet. Check that
How would we define ?
Which definition should we provide for ? Both are necessary. But then is not a function. So we have a second reason to know that there is no function that will allow us to conclude that is invertible. It happens that there are infinitely many column vectors that would have to take to . Construct the kernel of ,
Now choose either of the two inputs used above for and add to it a scalar multiple of the basis vector for the kernel of . For example,
then verify that . Practice creating a few more inputs for that would be sent to , and see why it is hopeless to think that we could ever provide a reasonable definition for ! There is a “whole subspace’s worth” of values that would have to take on.
In Example ANILT you may have noticed that is not surjective, since the matrix was not in the range of . And is not injective since there are two different input column vectors that sends to the matrix . Linear transformations that are not surjective lead to putative inverse functions that are undefined on inputs outside of the range of . Linear transformations that are not injective lead to putative inverse functions that are multiply-defined on each of their inputs. We will formalize these ideas in Theorem ILTIS.
But first notice in Definition IVLT that we only require the inverse (when it exists) to be a function. When it does exist, it too is a linear transformation.
Theorem ILTLT
Inverse of a Linear Transformation is a Linear Transformation
Suppose that
is an invertible linear transformation. Then the function
is a linear
transformation.
Proof We work through verifying Definition LT for , using the fact that is a linear transformation to obtain the second equality in each half of the proof. To this end, suppose and .
So fulfills the requirements of Definition LT and is therefore a linear transformation. So when has an inverse, is also a linear transformation. Additionally, is invertible and its inverse is what you might expect.
Theorem IILT
Inverse of an Invertible Linear Transformation
Suppose that
is an invertible linear transformation. Then
is an invertible linear
transformation and .
Proof Because is invertible, Definition IVLT tells us there is a function such that
Additionally, Theorem ILTLT tells us that is more than just a function, it is a linear transformation. Now view these two statements as properties of the linear transformation . In light of Definition IVLT, they together say that is invertible (let play the role of in the statement of the definition). Furthermore, the inverse of is then , i.e. .
We now know what an inverse linear transformation is, but just which linear transformations have inverses? Here is a theorem we have been preparing for all chapter long.
Theorem ILTIS
Invertible Linear Transformations are Injective and Surjective
Suppose is a linear
transformation. Then is
invertible if and only if is
injective and surjective.
Proof () Since is presumed invertible, we can employ its inverse, (Definition IVLT). To see that is injective, suppose and assume that ,
So there is an element from , when used as an input to (namely ) that produces the desired output, , and hence is surjective by Definition SLT.
() Now assume that is both injective and surjective. We will build a function that will establish that is invertible. To this end, choose any . Since is surjective, Theorem RSLT says , so we have . Theorem RPI says that the pre-image of , , is nonempty. So we can choose a vector from the pre-image of , say . In other words, there exists .
Since is non-empty, Theorem KPI then says that
However, because is injective, by Theorem KILT the kernel is trivial, . So the pre-image is a set with just one element, . Now we can define by . This is the key to this half of this proof. Normally the preimage of a vector from the codomain might be an empty set, or an infinite set. But surjectivity requires that the preimage not be empty, and then injectivity limits the preimage to a singleton. Since our choice of was arbitrary, we know that every pre-image for is a set with a single element. This allows us to construct as a function. Now that it is defined, verifying that it is the inverse of will be easy. Here we go.
Choose . Define . Then , so that and,
and since our choice of was arbitrary we have function equality, .
Now choose . Define to be the single vector in the set , in other words, . Then , so
and since our choice of was arbitrary we have function equality, .
When a linear transformation is both injective and surjective, the pre-image of any element of the codomain is a set of size one (a “singleton”). This fact allowed us to construct the inverse linear transformation in one half of the proof of Theorem ILTIS (see Technique C). We can follow this approach to construct the inverse of a specific linear transformation, as the next example shows.
Example CIVLT
Computing the Inverse of a Linear Transformations
Consider the linear transformation
defined by
is invertible, which you are able to verify, perhaps by determining that the kernel of is empty and the range of is all of . This will be easier once we have Theorem RPNDD, which appears later in this section.
By Theorem ILTIS we know exists, and it will be critical shortly to realize that is automatically known to be a linear transformation as well (Theorem ILTLT). To determine the complete behavior of we can simply determine its action on a basis for the domain, . This is the substance of Theorem LTDB, and an excellent example of its application. Choose any basis of , the simpler the better, such as . Values of for these three basis elements will be the single elements of their preimages. In turn, we have
Theorem LTDB says, informally, “it is enough to know what a linear transformation does to a basis.” Formally, we have the outputs of for a basis, so by Theorem LTDB there is a unique linear transformation with these outputs. So we put this information to work. The key step here is that we can convert any element of into a linear combination of the elements of the basis (Theorem VRRB). We are after a “formula” for the value of on a generic element of , say .
Notice how a linear combination in the domain of has been translated into a linear combination in the codomain of since we know is a linear transformation by Theorem ILTLT.
Also, notice how the augmented matrices used to determine the three pre-images could be combined into one calculation of a matrix in extended echelon form, reminiscent of a procedure we know for computing the inverse of a matrix (see Example CMI). Hmmmm.
We will make frequent use of the characterization of invertible linear transformations provided by Theorem ILTIS. The next theorem is a good example of this, and we will use it often, too.
Theorem CIVLT
Composition of Invertible Linear Transformations
Suppose that
and
are invertible linear transformations. Then the composition,
is an invertible linear
transformation.
Proof Since and are both linear transformations, is also a linear transformation by Theorem CLTLT. Since and are both invertible, Theorem ILTIS says that and are both injective and surjective. Then Theorem CILTI says is injective, and Theorem CSLTS says is surjective. Now apply the “other half” of Theorem ILTIS and conclude that is invertible.
When a composition is invertible, the inverse is easy to construct.
Theorem ICLT
Inverse of a Composition of Linear Transformations
Suppose that and
are invertible linear
transformations. Then
is invertible and .
Proof Compute, for all
so . By Definition IVLT, is invertible and .
Notice that this theorem not only establishes what the inverse of is, it also duplicates the conclusion of Theorem CIVLT and also establishes the invertibility of . But somehow, the proof of Theorem CIVLT is nicer way to get this property.
Does Theorem ICLT remind you of the flavor of any theorem we have seen about matrices? (Hint: Think about getting dressed.) Hmmmm.
A vector space is defined (Definition VS) as a set of objects (“vectors”) endowed with a definition of vector addition () and a definition of scalar multiplication (written with juxtaposition). Many of our definitions about vector spaces involve linear combinations (Definition LC), such as the span of a set (Definition SS) and linear independence (Definition LI). Other definitions are built up from these ideas, such as bases (Definition B) and dimension (Definition D). The defining properties of a linear transformation require that a function “respect” the operations of the two vector spaces that are the domain and the codomain (Definition LT). Finally, an invertible linear transformation is one that can be “undone” — it has a companion that reverses its effect. In this subsection we are going to begin to roll all these ideas into one.
A vector space has “structure” derived from definitions of the two operations and the requirement that these operations interact in ways that satisfy the ten properties of Definition VS. When two different vector spaces have an invertible linear transformation defined between them, then we can translate questions about linear combinations (spans, linear independence, bases, dimension) from the first vector space to the second. The answers obtained in the second vector space can then be translated back, via the inverse linear transformation, and interpreted in the setting of the first vector space. We say that these invertible linear transformations “preserve structure.” And we say that the two vector spaces are “structurally the same.” The precise term is “isomorphic,” from Greek meaning “of the same form.” Let’s begin to try to understand this important concept.
Definition IVS
Isomorphic Vector Spaces
Two vector spaces
and
are isomorphic if there exists an invertible linear transformation
with
domain and
codomain ,
. In this case, we
write , and the linear
transformation is known as
an isomorphism between
and .
A few comments on this definition. First, be careful with your language (Technique L). Two vector spaces are isomorphic, or not. It is a yes/no situation and the term only applies to a pair of vector spaces. Any invertible linear transformation can be called an isomorphism, it is a term that applies to functions. Second, a given pair of vector spaces there might be several different isomorphisms between the two vector spaces. But it only takes the existence of one to call the pair isomorphic. Third, isomorphic to , or isomorphic to ? Doesn’t matter, since the inverse linear transformation will provide the needed isomorphism in the “opposite” direction. Being “isomorphic to” is an equivalence relation on the set of all vector spaces (see Theorem SER for a reminder about equivalence relations).
Example IVSAV
Isomorphic vector spaces, Archetype V
Archetype V is a linear transformation from
to
,
Since it is injective and surjective, Theorem ILTIS tells us that it is an invertible linear transformation. By Definition IVS we say and are isomorphic.
At a basic level, the term “isomorphic” is nothing more than a codeword for the presence of an invertible linear transformation. However, it is also a description of a powerful idea, and this power only becomes apparent in the course of studying examples and related theorems. In this example, we are led to believe that there is nothing “structurally” different about and . In a certain sense they are the same. Not equal, but the same. One is as good as the other. One is just as interesting as the other.
Here is an extremely basic application of this idea. Suppose we want to compute the following linear combination of polynomials in ,
Rather than doing it straight-away (which is very easy), we will apply the transformation to convert into a linear combination of matrices, and then compute in according to the definitions of the vector space operations there (Example VSM),
Now we will translate our answer back to by applying , which we found in Example AIVLT,
We compute,
which is, as expected, exactly what we would have computed for the original linear combination had we just used the definitions of the operations in (Example VSP). Notice this is meant only as an illustration and not a suggested route for doing this particular computation.
Checking the dimensions of two vector spaces can be a quick way to establish that they are not isomorphic. Here’s the theorem.
Theorem IVSED
Isomorphic Vector Spaces have Equal Dimension
Suppose and
are isomorphic
vector spaces. Then .
Proof If and are isomorphic, there is an invertible linear transformation (Definition IVS). is injective by Theorem ILTIS and so by Theorem ILTD, . Similarly, is surjective by Theorem ILTIS and so by Theorem SLTD, . The net effect of these two inequalities is that .
The contrapositive of Theorem IVSED says that if and have different dimensions, then they are not isomorphic. Dimension is the simplest “structural” characteristic that will allow you to distinguish non-isomorphic vector spaces. For example is not isomorphic to since their dimensions (7 and 12, respectively) are not equal. With tools developed in Section VR we will be able to establish that the converse of Theorem IVSED is true. Think about that one for a moment.
Just as a matrix has a rank and a nullity, so too do linear transformations. And just like the rank and nullity of a matrix are related (they sum to the number of columns, Theorem RPNC) the rank and nullity of a linear transformation are related. Here are the definitions and theorems, see the Archetypes (Appendix A) for loads of examples.
Definition ROLT
Rank Of a Linear Transformation
Suppose that
is a linear transformation. Then the rank of
,
, is the dimension
of the range of ,
(This definition contains Notation ROLT.)
Definition NOLT
Nullity Of a Linear Transformation
Suppose that
is a linear transformation. Then the nullity of
,
, is the dimension
of the kernel of ,
(This definition contains Notation NOLT.)
Here are two quick theorems.
Theorem ROSLT
Rank Of a Surjective Linear Transformation
Suppose that
is a linear transformation. Then the rank of
is the
dimension of ,
, if and
only if is
surjective.
Proof By Theorem RSLT, is surjective if and only if . Applying Definition ROLT, if and only if .
Theorem NOILT
Nullity Of an Injective Linear Transformation
Suppose that
is a linear transformation. Then the nullity of
is zero,
, if and
only if is
injective.
Proof By Theorem KILT, is injective if and only if . Applying Definition NOLT, if and only if .
Just as injectivity and surjectivity come together in invertible linear transformations, there is a clear relationship between rank and nullity of a linear transformation. If one is big, the other is small.
Theorem RPNDD
Rank Plus Nullity is Domain Dimension
Suppose that
is a linear transformation. Then
Proof Let and . Suppose that is a basis of the range of , , and is a basis of the kernel of , . Note that and are possibly empty, which means that some of the sums in this proof are “empty” and are equal to the zero vector.
Because the elements of are all in the range of , each must have a non-empty pre-image by Theorem RPI. Choose vectors , such that . So , . Consider the set
We claim that is a basis for .
To establish linear independence for , begin with a relation of linear dependence on . So suppose there are scalars and
Then
This is a relation of linear dependence on (Definition RLD), and since is a linearly independent set (Definition LI), we see that . Then the original relation of linear dependence on becomes
But this is again a relation of linear independence (Definition RLD), now on the set . Since is linearly independent (Definition LI), we have . Since we now know that all the scalars in the relation of linear dependence on must be zero, we have established the linear independence of through Definition LI.
To now establish that spans , choose an arbitrary vector . Then , so there are scalars such that
Use the scalars to define a vector ,
Then
So the vector is sent to the zero vector by and hence is an element of the kernel of . As such it can be written as a linear combination of the basis vectors for , the elements of the set . So there are scalars such that
Then
This says that for any vector, , from , there exist scalars () that form as a linear combination of the vectors in the set . In other words, spans (Definition SS).
So is a basis (Definition B) of with vectors, and thus
as desired.
Theorem RPNC said that the rank and nullity of a matrix sum to the number of columns of the matrix. This result is now an easy consequence of Theorem RPNDD when we consider the linear transformation defined with the matrix by . The range and kernel of are identical to the column space and null space of the matrix (Exercise ILT.T20, Exercise SLT.T20), so the rank and nullity of the matrix are identical to the rank and nullity of the linear transformation . The dimension of the domain of is the dimension of , exactly the number of columns for the matrix .
This theorem can be especially useful in determining basic properties of linear transformations. For example, suppose that is a linear transformation and you are able to quickly establish that the kernel is trivial. Then . First this means that is injective by Theorem NOILT. Also, Theorem RPNDD becomes
So the rank of is equal to the rank of the codomain, and by Theorem ROSLT we know is surjective. Finally, we know is invertible by Theorem ILTIS. So from the determination that the kernel is trivial, and consideration of various dimensions, the theorems of this section allow us to conclude the existence of an inverse linear transformation for .
Similarly, Theorem RPNDD can be used to provide alternative proofs for Theorem ILTD, Theorem SLTD and Theorem IVSED. It would be an interesting exercise to construct these proofs.
It would be instructive to study the archetypes that are linear transformations and see how many of their properties can be deduced just from considering only the dimensions of the domain and codomain. Then add in just knowledge of either the nullity or rank, and so how much more you can learn about the linear transformation. The table preceding all of the archetypes (Appendix A) could be a good place to start this analysis.
This subsection does not really belong in this section, or any other section, for that matter. It is just the right time to have a discussion about the connections between the central topic of linear algebra, linear transformations, and our motivating topic from Chapter SLE, systems of linear equations. We will discuss several theorems we have seen already, but we will also make some forward-looking statements that will be justified in Chapter R.
Archetype D and Archetype E are ideal examples to illustrate connections with linear transformations. Both have the same coefficient matrix,
To apply the theory of linear transformations to these two archetypes, employ matrix multiplication (Definition MM) and define the linear transformation,
Theorem MBLT tells us that is indeed a linear transformation. Archetype D asks for solutions to , where . In the language of linear transformations this is equivalent to asking for . In the language of vectors and matrices it asks for a linear combination of the four columns of that will equal . One solution listed is . With a non-empty preimage, Theorem KPI tells us that the complete solution set of the linear system is the preimage of ,
The kernel of the linear transformation is exactly the null space of the matrix (see Exercise ILT.T20), so this approach to the solution set should be reminiscent of Theorem PSPHS. The kernel of the linear transformation is the preimage of the zero vector, exactly equal to the solution set of the homogeneous system . Since has a null space of dimension two, every preimage (and in particular the preimage of ) is as “big” as a subspace of dimension two (but is not a subspace).
Archetype E is identical to Archetype D but with a different vector of constants, . We can use the same linear transformation to discuss this system of equations since the coefficient matrix is identical. Now the set of solutions to is the pre-image of , . However, the vector is not in the range of the linear transformation (nor is it in the column space of the matrix, since these two sets are equal by Exercise SLT.T20). So the empty pre-image is equivalent to the inconsistency of the linear system.
These two archetypes each have three equations in four variables, so either the resulting linear systems are inconsistent, or they are consistent and application of Theorem CMVEI tells us that the system has infinitely many solutions. Considering these same parameters for the linear transformation, the dimension of the domain, , is four, while the codomain, , has dimension three. Then
So the kernel of is nontrivial simply by considering the dimensions of the domain (number of variables) and the codomain (number of equations). Pre-images of elements of the codomain that are not in the range of are empty (inconsistent systems). For elements of the codomain that are in the range of (consistent systems), Theorem KPI tells us that the pre-images are built from the kernel, and with a non-trivial kernel, these pre-images are infinite (infinitely many solutions).
When do systems of equations have unique solutions? Consider the system of linear equations and the linear transformation . If has a trivial kernel, then pre-images will either be empty or be finite sets with single elements. Correspondingly, the coefficient matrix will have a trivial null space and solution sets will either be empty (inconsistent) or contain a single solution (unique solution). Should the matrix be square and have a trivial null space then we recognize the matrix as being nonsingular. A square matrix means that the corresponding linear transformation, , has equal-sized domain and codomain. With a nullity of zero, is injective, and also Theorem RPNDD tells us that rank of is equal to the dimension of the domain, which in turn is equal to the dimension of the codomain. In other words, is surjective. Injective and surjective, and Theorem ILTIS tells us that is invertible. Just as we can use the inverse of the coefficient matrix to find the unique solution of any linear system with a nonsingular coefficient matrix (Theorem SNCM), we can use the inverse of the linear transformation to construct the unique element of any pre-image (proof of Theorem ILTIS).
The executive summary of this discussion is that to every coefficient matrix of a system of linear equations we can associate a natural linear transformation. Solution sets for systems with this coefficient matrix are preimages of elements of the codomain of the linear transformation. For every theorem about systems of linear equations there is an analogue about linear transformations. The theory of linear transformations provides all the tools to recreate the theory of solutions to linear systems of equations.
We will continue this adventure in Chapter R.
C10 The archetypes below are linear transformations of the form that are invertible. For each, the inverse linear transformation is given explicitly as part of the archetype’s description. Verify for each linear transformation that
Archetype R,
Archetype V,
Archetype W
Contributed by Robert Beezer
C20 Determine if the linear transformation is (a) injective, (b) surjective, (c) invertible.
Contributed by Robert Beezer Solution [1666]
C21 Determine if the linear transformation is (a) injective, (b) surjective, (c) invertible.
Contributed by Robert Beezer Solution [1667]
C25 For each linear transformation below: (a) Find the matrix representation of , (b) Calculate , (c) Calculate , (d) Graph the image in either or as appropriate, (e) How many dimensions are lost?, and (f) How many dimensions are preserved?
Contributed by Chris Black
C50 Consider the linear transformation from the set of matrices to the set of polynomials of degree at most 1, defined by
Prove that is invertible. Then show that the linear transformation
is the inverse of ,
that is .
Contributed by Robert Beezer Solution [1670]
M30 The linear transformation below is invertible. Find a formula for the inverse linear transformation, .
Contributed by Robert Beezer Solution [1672]
M31 The linear transformation is invertible. Determine a formula for the inverse linear transformation .
Contributed by Robert Beezer Solution [1674]
M50 Rework Example CIVLT, only in place of the basis
for
, choose instead
to use the basis .
This will complicate writing a generic element of the domain of
as a linear combination of the basis elements, and the algebra will be
a bit messier, but in the end you should obtain the same formula for
. The
inverse linear transformation is what it is, and the choice of a particular basis
should not influence the outcome.
Contributed by Robert Beezer
T05 Prove that the identity linear transformation (Definition IDLT) is both
injective and surjective, and hence invertible.
Contributed by Robert Beezer
T15 Suppose that
is a surjective linear transformation and
. Prove
that
is injective.
Contributed by Robert Beezer Solution [1676]
T16 Suppose that
is an injective linear transformation and
. Prove
that
is surjective.
Contributed by Robert Beezer
T30 Suppose that
and
are isomorphic vector spaces. Prove that there are infinitely many isomorphisms
between
and .
Contributed by Robert Beezer Solution [1677]
C20 Contributed by Robert Beezer Statement [1659]
(a) We will compute the kernel of .
Suppose that .
Then
and matrix equality (Theorem ME) yields the homogeneous system of four equations in three variables,
The coefficient matrix of this system row-reduces as
From the existence of non-trivial solutions to this system, we can infer non-zero polynomials in . By Theorem KILT we then know that is not injective.
(b) Since , by Theorem SLTD is not surjective.
(c) Since is not surjective, it is not invertible by Theorem ILTIS.
C21 Contributed by Robert Beezer Statement [1660]
(a) To check injectivity, we compute the kernel of
. To this end,
suppose that ,
so
this creates the homogeneous system of four equations in four variables,
The coefficient matrix of this system row-reduces as,
We recognize the coefficient matrix as being nonsingular, so the only solution to the system is , and the kernel of is trivial, . By Theorem KILT, we see that is injective.
(b) We can establish that is surjective by considering the rank and nullity of .
So, is a subspace of (Theorem RLTS) whose dimension equals that of . By Theorem EDYES, we gain the set equality . Theorem RSLT then implies that is surjective.
(c) Since is both injective and surjective, Theorem ILTIS says is invertible.
C50 Contributed by Robert Beezer Statement [1662]
Determine the kernel of
first. The condition that
becomes .
Equating coefficients of these polynomials yields the system
This homogeneous system has a nonsingular coefficient matrix, so the only solution is , and thus
By Theorem KILT, we know is injective. With we employ Theorem RPNDD to find
Since and , we can apply Theorem EDYES to obtain the set equality and therefore is surjective.
One of the two defining conditions of an invertible linear transformation is (Definition IVLT)
That is similar.
M30 Contributed by Robert Beezer Statement [1663]
(Another approach to this solution would follow Example CIVLT.)
Suppose that has a form given by
where are unknown scalars. Then
Equating coefficients of these two polynomials, and then equating coefficients on and , gives rise to 4 equations in 4 variables,
This system has a unique solution: , , , . So the desired inverse linear transformation is
Notice that the system of 4 equations in 4 variables could be split into two systems, each with two equations in two variables (and identical coefficient matrices). After making this split, the solution might feel like computing the inverse of a matrix (Theorem CINM). Hmmmm.
M31 Contributed by Robert Beezer Statement [1663]
(Another approach to this solution would follow Example CIVLT.)
We are given that is invertible. The inverse linear transformation can be formulated by considering the pre-image of a generic element of the codomain. With injectivity and surjectivity, we know that the pre-image of any element will be a set of size one — it is this lone element that will be the output of the inverse linear transformation.
Suppose that we set as a generic element of the codomain, . Then if ,
So we obtain the system of two equations in the two variables and ,
With a nonsingular coefficient matrix, we can solve the system using the inverse of the coefficient matrix,
So we define,
T15 Contributed by Robert Beezer Statement [1664]
If is surjective, then
Theorem RSLT says ,
so . In turn, the
hypothesis gives .
Then, using Theorem RPNDD,
With a null space of zero dimension, , and by Theorem KILT we see that is injective. is both injective and surjective so by Theorem ILTIS, is invertible.
T30 Contributed by Robert Beezer Statement [1665]
Since
and are
isomorphic, there is at least one isomorphism between them (Definition IVS), say
. As
such,
is an invertible linear transformation.
For define the linear transformation by . Convince yourself that when , is an invertible linear transformation (Definition IVLT). Then the composition, , is an invertible linear transformation by Theorem CIVLT. Once convinced that each non-zero value of gives rise to a different functions for , then we have constructed infinitely many isomorphisms from to .
Theorem MBLT
You give me an
matrix and I’ll give you a linear transformation
. This
is our first hint that there is some relationship between linear transformations and
matrices.
Theorem MLTCV
You give me a linear transformation
and I’ll give you an
matrix. This is our second hint that there is some relationship between linear
transformations and matrices. Generalizing this relationship to arbitrary vector spaces
(i.e. not just
and )
will be the most important idea of Chapter R.
Theorem LTLC
A simple idea, and as described in Exercise LT.T20, equivalent to the
Definition LT. The statement is really just for convenience, as we’ll quote this one
often.
Theorem LTDB
Another simple idea, but a powerful one. “It is enough to know what
a linear transformation does to a basis.” At the outset of Chapter R,
Theorem VRRB will help us define a very important function, and then
Theorem LTDB will allow us to understand that this function is also a linear
transformation.
Theorem KPI
The pre-image will be an important construction in this chapter, and this is one of
the most important descriptions of the pre-image. It should remind you of
Theorem PSPHS, which is described in Acronyms V. See Theorem RPI, which is
also described below.
Theorem KILT
Kernels and injective linear transformations are intimately related. This result is
the connection. Compare with Theorem RSLT below.
Theorem ILTB
Injective linear transformations and linear independence are intimately
related. This result is the connection. Compare with Theorem SLTB
below.
Theorem RSLT
Ranges and surjective linear transformations are intimately related. This result is
the connection. Compare with Theorem KILT above.
Theorem SSRLT
This theorem provides the most direct way of forming the range of a linear
transformation. The resulting spanning set might well be linearly dependent, and
beg for some clean-up, but that doesn’t stop us from having very quickly formed a
reasonable description of the range. If you find the determination of spanning
sets or ranges difficult, this is one worth remembering. You can view
this as the analogue of forming a column space by a direct application of
Definition CSM.
Theorem SLTB
Surjective linear transformations and spanning sets are intimately related. This
result is the connection. Compare with Theorem ILTB above.
Theorem RPI
This is the analogue of Theorem KPI. Membership in the range is equivalent to
nonempty pre-images.
Theorem ILTIS
Injectivity and surjectivity are independent concepts. You can have one without
the other. But when you have both, you get invertibility, a linear transformation
that can be run “backwards.” This result might explain the entire structure of the
four sections in this chapter.
Theorem RPNDD
This is the promised generalization of Theorem RPNC about matrices. So the
number of columns of a matrix is the analogue of the dimension of the domain.
This will become even more precise in Chapter R. For now, this can be a powerful
result for determining dimensions of kernels and ranges, and consequently, the
injectivity or surjectivity of linear transformations. Never underestimate a
theorem that counts something.