From A First Course in Linear Algebra
Version 1.30
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/
We begin by establishing an invertible linear transformation between any vector space
of dimension
and
. This will
allow us to “go back and forth” between the two vector spaces, no matter how abstract the
definition of
might be.
Definition VR
Vector Representation
Suppose that is a vector
space with a basis .
Define a function
as follows. For ,
find scalars
so that
then define by setting
We need to show that is really a function (since “find scalars” sounds like it could be accomplished in many ways, or perhaps not at all) and right now we want to establish that is a linear transformation. We will wrap up both objectives in one theorem, even though the first part is working backwards to make sure that is well-defined.
Theorem VRLT
Vector Representation is a Linear Transformation
The function
(Definition VR) is a linear transformation.
Proof The definition of (Definition VR) appears to allow considerable latitude in selecting the scalars . However, since is a basis for , Theorem VRRB says this can be done, and done uniquely. So despite appearances, is indeed a function.
We will take a novel approach to establishing that is a linear transformation. We will construct another function, which we will easily determine is a linear transformation, and then show that this second function is really in disguise. Here we go.
Since is a basis, we can define to be the unique linear transformation such that , , as guaranteed by Theorem LTDB, and where the are the standard unit vectors (Definition SUV). Then suppose for an arbitrary
We have,
So, by Definition CVE, as elements of , . Since was arbitrary, . Now, since is known to be a linear transformation, it must follow that is also a linear transformation.
The proof of Theorem VRLT provides an alternate definition of vector representation relative to a basis : it is the unique linear transformation that takes to the standard unit basis.
Example VRC4
Vector representation in
Consider the vector
We will find several vector representations of in this example. Notice that never changes, but the representations of do change.
One basis for is
as can be seen by making these vectors the columns of a matrix, checking that the matrix is nonsingular and applying Theorem CNMB. To find , we need to find scalars, such that
By Theorem SLSLC the desired scalars are a solution to the linear system of equations with a coefficient matrix whose columns are the vectors in and with a vector of constants . With a nonsingular coefficient matrix, the solution is unique, but this is no surprise as this is the content of Theorem VRRB. This unique solution is
Then by Definition VR, we have
Suppose now that we construct a representation of relative to another basis of ,
As with , it is easy to check that is a basis. Writing as a linear combination of the vectors in leads to solving a system of four equations in the four unknown scalars with a nonsingular coefficient matrix. The unique solution can be expressed as
so that Definition VR gives
We often perform representations relative to standard bases, but for vectors in its a little silly. Let’s find the vector representation of relative to the standard basis (Theorem SUVB),
Then, without any computation, we can check that
so by Definition VR,
which is not very exciting. Notice however that the order in which we place the vectors in the basis is critical to the representation. Let’s keep the standard unit vectors as our basis, but rearrange the order we place them in the basis. So a fourth basis is
Then,
so by Definition VR,
So for every possible basis of we could construct a different representation of .
Vector representations are most interesting for vector spaces that are not .
Example VRP2
Vector representations in
Consider the vector
from the vector space of polynomials with degree at most 2 (Example VSP). A nice
basis for
is
so that
so by Definition VR
Another nice basis for is
so that now it takes a bit of computation to determine the scalars for the representation. We want so that
Performing the operations in on the right-hand side, and equating coefficients, gives the three equations in the three unknown scalars,
The coefficient matrix of this sytem is nonsingular, leading to a unique solution (no surprise there, see Theorem VRRB),
so by Definition VR
While we often form vector representations relative to “nice” bases, nothing prevents us from forming representations relative to “nasty” bases. For example, the set
can be verified as a basis of by checking linear independence with Definition LI and then arguing that 3 vectors from , a vector space of dimension 3 (Theorem DP), must also be a spanning set (Theorem G). Now we desire scalars so that
Performing the operations in on the right-hand side, and equating coefficients, gives the three equations in the three unknown scalars,
The coefficient matrix of this sytem is nonsingular, leading to a unique solution (no surprise there, see Theorem VRRB),
so by Definition VR
Theorem VRI
Vector Representation is Injective
The function
(Definition VR) is an injective linear transformation.
Proof We will appeal to Theorem KILT. Suppose is a vector space of dimension , so vector representation is of the form . Let be the basis of used in the definition of . Suppose . Finally, since is a basis for , by Theorem VRRB there are (unique) scalars, such that
Then for
So
Thus an arbitrary vector, , from the kernel ,, must equal the zero vector of . So and by Theorem KILT, is injective.
Theorem VRS
Vector Representation is Surjective
The function
(Definition VR) is a surjective linear transformation.
Proof We will appeal to Theorem RSLT. Suppose is a vector space of dimension , so vector representation is of the form . Let be the basis of used in the definition of . Suppose . Define the vector by
Then for
so the entries of vectors and are equal and Definition CVE yields the vector equality . This demonstrates that , so . Since by Definition RLT, we have and Theorem RSLT says is surjective.
We will have many occasions later to employ the inverse of vector representation, so we will record the fact that vector representation is an invertible linear transformation.
Theorem VRILT
Vector Representation is an Invertible Linear Transformation
The function
(Definition VR) is an invertible linear transformation.
Proof The function (Definition VR) is a linear transformation (Theorem VRLT) that is injective (Theorem VRI) and surjective (Theorem VRS) with domain and codomain . By Theorem ILTIS we then know that is an invertible linear transformation.
Informally, we will refer to the application of as coordinatizing a vector, while the application of will be referred to as un-coordinatizing a vector.
Limiting our attention to vector spaces with finite dimension, we now describe every possible vector space. All of them. Really.
Theorem CFDVS
Characterization of Finite Dimensional Vector Spaces
Suppose that is a vector
space with dimension .
Then is
isomorphic to .
Proof Since has dimension we can find a basis of of size (Definition D) which we will call . The linear transformation is an invertible linear transformation from to , so by Definition IVS, we have that and are isomorphic.
Theorem CFDVS is the first of several surprises in this chapter, though it might be a bit demoralizing too. It says that there really are not all that many different (finite dimensional) vector spaces, and none are really any more complicated than . Hmmm. The following examples should make this point.
Example TIVS
Two isomorphic vector spaces
The vector space of polynomials with degree 8 or less,
,
has dimension 9 (Theorem DP). By Theorem CFDVS,
is isomorphic
to .
Example CVSR
Crazy vector space revealed
The crazy vector space,
of Example CVS, has dimension 2 by Example DC. By Theorem CFDVS,
is isomorphic to
. Hmmmm. Not really
so crazy after all?
Example ASC
A subspace characterized
In Example DSP4 we determined that a certain subspace
of
has dimension
. By
Theorem CFDVS,
is isomorphic to .
Theorem IFDVS
Isomorphism of Finite Dimensional Vector Spaces
Suppose
and
are both finite-dimensional vector spaces. Then
and
are isomorphic
if and only if .
Proof () This is just the statement proved in Theorem IVSED.
() This is the advertised converse of Theorem IVSED. We will assume and have equal dimension and discover that they are isomorphic vector spaces. Let be the common dimension of and . Then by Theorem CFDVS there are isomorphisms and .
is therefore an invertible linear transformation by Definition IVS. Similarly, is an invertible linear transformation, and so is an invertible linear transformation (Theorem IILT). The composition of invertible linear transformations is again invertible (Theorem CIVLT) so the composition of with is invertible. Then is an invertible linear transformation from to and Definition IVS says and are isomorphic.
Example MIVS
Multiple isomorphic vector spaces
,
,
and
are
all vector spaces and each has dimension 10. By Theorem IFDVS each is
isomorphic to any other.
The subspace of that contains all the symmetric matrices (Definition SYM) has dimension , so this subspace is also isomorphic to each of the four vector spaces above.
With available as an invertible linear transformation, we can translate between vectors in a vector space of dimension and . Furthermore, as a linear transformation, respects the addition and scalar multiplication in , while respects the addition and scalar multiplication in . Since our definitions of linear independence, spans, bases and dimension are all built up from linear combinations, we will finally be able to translate fundamental properties between abstract vector spaces () and concrete vector spaces ().
Theorem CLI
Coordinatization and Linear Independence
Suppose that is a vector
space with a basis
of size . Then
is a linearly independent
subset of if and
only if is a linearly
independent subset of .
Proof The linear transformation is an isomorphism between and (Theorem VRILT). As an invertible linear transformation, is an injective linear transformation (Theorem ILTIS), and is also an injective linear transformation (Theorem IILT, Theorem ILTIS).
() Since is an injective linear transformation and is linearly independent, Theorem ILTLI says that is linearly independent.
() If we apply to each element of , we will create the set . Since we are assuming is linearly independent and is injective, Theorem ILTLI says that is linearly independent.
Theorem CSS
Coordinatization and Spanning Sets
Suppose that is a vector
space with a basis
of size .
Then if and
only if .
Proof () Suppose . Then there are scalars, , such that
Then,
which says that .
() Suppose that . Then there are scalars such that
Recall that is invertible (Theorem VRILT), so
which says that .
Here’s a fairly simple example that illustrates a very, very important idea.
Example CP2
Coordinatizing in
In Example VRP2 we needed to know that
is a basis for . With Theorem CLI and Theorem CSS this task is much easier. First, choose a known basis for , a basis that forms vector representations easily. We will choose
Now, form the subset of that is the result of applying to each element of ,
and ask if is a linearly independent spanning set for . This is easily seen to be the case by forming a matrix whose columns are the vectors of , row-reducing to the identity matrix , and then using the nonsingularity of to assert that is a basis for (Theorem CNMB). Now, since is a basis for , Theorem CLI and Theorem CSS tell us that is also a basis for .
Example CP2 illustrates the broad notion that computations in abstract vector spaces can be reduced to computations in . You may have noticed this phenomenon as you worked through examples in Chapter VS or Chapter LT employing vector spaces of matrices or polynomials. These computations seemed to invariably result in systems of equations or the like from Chapter SLE, Chapter V and Chapter M. It is vector representation, , that allows us to make this connection formal and precise.
Knowing that vector representation allows us to translate questions about linear combinations, linear indepencence and spans from general vector spaces to allows us to prove a great many theorems about how to translate other properties. Rather than prove these theorems, each of the same style as the other, we will offer some general guidance about how to best employ Theorem VRLT, Theorem CLI and Theorem CSS. This comes in the form of a “principle”: a basic truth, but most definitely not a theorem (hence, no proof).
The Coordinatization Principle Suppose that
is a vector space
with a basis of
size . Then any
question about ,
or its elements, which ultimately depends on the vector addition or scalar multiplication
in , or
depends on linear independence or spanning, may be translated into the same question in
by application of the
linear transformation
to the relevant vectors. Once the question is answered in
, the answer may be
translated back to
(if necessary) through application of the inverse linear transformation
.
Example CM32
Coordinatization in
This is a simple example of the Coordinatization Principle, depending
only on the fact that coordinatizing is an invertible linear transformation
(Theorem VRILT). Suppose we have a linear combination to perform in
, the vector
space of
matrices, but we are adverse to doing the operations of
(Definition MA, Definition MSM). More specifically, suppose we are faced with
the computation
We choose a nice basis for (or a nasty basis if we are so inclined),
and apply to each vector in the linear combination. This gives us a new computation, now in the vector space ,
which we can compute with the operations of (Definition CVA, Definition CVSM), to arrive at
We are after the result of a computation in , so we now can apply to obtain a matrix,
which is exactly the matrix we would have computed had we just performed the matrix operations in the first place. So this was not meant to be an easier way to compute a linear combination of two matrices, just a different way.
Compute .
C10 In the vector space , compute the vector representation for the basis and vector below.
Contributed by Robert Beezer Solution [1538]
C20 Rework Example CM32 replacing the basis by the basis
Contributed by Robert Beezer Solution [1539]
M10 Prove that the set below is a basis for the vector space of matrices, . Do this choosing a natural basis for and coordinatizing the elements of with respect to this basis. Examine the resulting set of column vectors from and apply the Coordinatization Principle.
Contributed by Andy Zimmer
C10 Contributed by Robert Beezer Statement [1536]
We need to express the vector
as a linear combination of the vectors in
.
Theorem VRRB tells us we will be able to do this, and do it uniquely. The vector
equation
becomes (via Theorem SLSLC) a system of linear equations with augmented matrix,
This system has the unique solution , , . So by Definition VR,
C20 Contributed by Robert Beezer Statement [1536]
The following computations replicate the computations given in Example CM32, only using
the basis .