Section VR  Vector Representations

From A First Course in Linear Algebra
Version 2.00
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/

We begin by establishing an invertible linear transformation between any vector space V of dimension m and m. This will allow us to “go back and forth” between the two vector spaces, no matter how abstract the definition of V might be.

Definition VR
Vector Representation
Suppose that V is a vector space with a basis B = v1,v2,v3,,vn. Define a function ρB: V n as follows. For w V define the column vector ρB w n by

w = ρB w1v1 + ρB w2v2 + ρB w3v3 + + ρB wnvn

(This definition contains Notation VR.)

This definition looks more complicated that it really is, though the form above will be useful in proofs. Simply stated, given w V , we write w as a linear combination of the basis elements of B. It is key to realize that Theorem VRRB guarantees that we can do this for every w, and furthermore this expression as a linear combination is unique. The resulting scalars are just the entries of the vector ρB w. This discussion should convince you that ρB is “well-defined” as a function. We can determine a precise output for any input. Now we want to establish that ρB is a function with additional properties - it is a linear transformation.

Theorem VRLT
Vector Representation is a Linear Transformation
The function ρB (Definition VR) is a linear transformation.

Proof   We will take a novel approach in this proof. We will construct another function, which we will easily determine is a linear transformation, and then show that this second function is really ρB in disguise. Here we go.

Since B is a basis, we can define T : V n to be the unique linear transformation such that T vi = ei, 1 i n, as guaranteed by Theorem LTDB, and where the ei are the standard unit vectors (Definition SUV). Then suppose for an arbitrary w V we have,

T wi = T j=1n ρ B wjvj i  Definition VR = j=1n ρ B wjT vj i  Theorem LTLC = j=1n ρ B wjej i = j=1n ρ B wjej i  Definition CVA = j=1n ρ B wj ej i  Definition CVSM = ρB wi ei i + j=1 ji n ρ B wj ej i  Property CC = ρB wi 1 + j=1 ji n ρ B wj 0  Definition SUV = ρB wi

As column vectors, Definition CVE implies that T w = ρB w. Since w was an arbitrary element of V , as functions T = ρB. Now, since T is known to be a linear transformation, it must follow that ρB is also a linear transformation.

The proof of Theorem VRLT provides an alternate definition of vector representation relative to a basis B that we could state as a corollary (Technique LC): ρB is the unique linear transformation that takes B to the standard unit basis.

Example VRC4
Vector representation in 4
Consider the vector y 4

y = 6 14 6 7

We will find several vector representations of y in this example. Notice that y never changes, but the representations of y do change.

One basis for 4 is

B = u1,u2,u3,u4 = 2 1 2 3 , 3 6 2 4 , 1 2 0 5 , 4 3 1 6

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 ρB y, we need to find scalars, a1,a2,a3,a4 such that

y = a1u1 + a2u2 + a3u3 + a4u4

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 B and with a vector of constants y. 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

a1 = 2 a2 = 1 a3 = 3 a4 = 4

Then by Definition VR, we have

ρB y = 2 1 3 4

Suppose now that we construct a representation of y relative to another basis of 4,

C = 15 9 4 2 , 16 14 5 2 , 26 14 6 3 , 14 13 4 6

As with B, it is easy to check that C is a basis. Writing y as a linear combination of the vectors in C 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

y = 6 14 6 7 = (28) 15 9 4 2 +(8) 16 14 5 2 +11 26 14 6 3 +0 14 13 4 6

so that Definition VR gives

ρC y = 28 8 11 0

We often perform representations relative to standard bases, but for vectors in m its a little silly. Let’s find the vector representation of y relative to the standard basis (Theorem SUVB),

D = e1,e2,e3,e4

Then, without any computation, we can check that

y = 6 14 6 7 = 6e1+14e2+6e3+7e4

so by Definition VR,

ρD y = 6 14 6 7

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

E = e3,e4,e2,e1

Then,

y = 6 14 6 7 = 6e3+7e4+14e2+6e1

so by Definition VR,

ρE y = 6 7 14 6

So for every possible basis of 4 we could construct a different representation of y.

Vector representations are most interesting for vector spaces that are not m.

Example VRP2
Vector representations in P2
Consider the vector u = 15 + 10x 6x2 P 2 from the vector space of polynomials with degree at most 2 (Example VSP). A nice basis for P2 is

B = 1,x,x2

so that

u = 15 + 10x 6x2 = 15(1) + 10(x) + (6)(x2)

so by Definition VR

ρB u = 15 10 6

Another nice basis for P2 is

B = 1,1 + x,1 + x + x2

so that now it takes a bit of computation to determine the scalars for the representation. We want a1,a2,a3 so that

15 + 10x 6x2 = a 1(1) + a2(1 + x) + a3(1 + x + x2)

Performing the operations in P2 on the right-hand side, and equating coefficients, gives the three equations in the three unknown scalars,

15 = a1 + a2 + a3 10 = a2 + a3 6 = a3

The coefficient matrix of this sytem is nonsingular, leading to a unique solution (no surprise there, see Theorem VRRB),

a1 = 5 a2 = 16 a3 = 6

so by Definition VR

ρC u = 5 16 6

While we often form vector representations relative to “nice” bases, nothing prevents us from forming representations relative to “nasty” bases. For example, the set

D = 2 x + 3x2,1 2x2,5 + 4x + x2

can be verified as a basis of P2 by checking linear independence with Definition LI and then arguing that 3 vectors from P2, a vector space of dimension 3 (Theorem DP), must also be a spanning set (Theorem G). Now we desire scalars a1,a2,a3 so that

15 + 10x 6x2 = a 1(2 x + 3x2) + a 2(1 2x2) + a 3(5 + 4x + x2)

Performing the operations in P2 on the right-hand side, and equating coefficients, gives the three equations in the three unknown scalars,

15 = 2a1 + a2 + 5a3 10 = a1 + 4a3 6 = 3a1 2a2 + a3

The coefficient matrix of this sytem is nonsingular, leading to a unique solution (no surprise there, see Theorem VRRB),

a1 = 2 a2 = 1 a3 = 2

so by Definition VR

ρD u = 2 1 2

Theorem VRI
Vector Representation is Injective
The function ρB (Definition VR) is an injective linear transformation.

Proof   We will appeal to Theorem KILT. Suppose U is a vector space of dimension n, so vector representation is of the form ρB: Un. Let B = u1,u2,u3,,un be the basis of U used in the definition of ρB. Suppose u KρB. We write u as a linear combination of the vectors in the basis B where the scalars are the components of the vector representation, ρB u.

u = ρB u1u1 + ρB u2u2 + ρB u3u3 + + ρB unun Definition VR = 0 1u1 + 0 2u2 + 0 3u3 + + 0 nun  Definition KLT = 0u1 + 0u2 + 0u3 + + 0un  Definition ZCV = 0 + 0 + 0 + + 0  Theorem ZSSM = 0  Property Z

Thus an arbitrary vector, u, from the kernel ,KρB, must equal the zero vector of U. So KρB = 0 and by Theorem KILT, ρB is injective.

Theorem VRS
Vector Representation is Surjective
The function ρB (Definition VR) is a surjective linear transformation.

Proof   We will appeal to Theorem RSLT. Suppose U is a vector space of dimension n, so vector representation is of the form ρB: Un. Let B = u1,u2,u3,,un be the basis of U used in the definition of ρB. Suppose v n. Define the vector u by

u = v1u1 + v2u2 + v3u3 + + vnun

Then for 1 i n

ρB ui = ρB v1u1 + v2u2 + v3u3 + + vnun i = vi  Definition VR

so the entries of vectors ρB u and v are equal and Definition CVE yields the vector equality ρB u = v. This demonstrates that v ρB, so n ρ B. Since ρB n by Definition RLT, we have ρB = n and Theorem RSLT says ρB 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 ρB (Definition VR) is an invertible linear transformation.

Proof   The function ρB (Definition VR) is a linear transformation (Theorem VRLT) that is injective (Theorem VRI) and surjective (Theorem VRS) with domain V and codomain n. By Theorem ILTIS we then know that ρB is an invertible linear transformation.

Informally, we will refer to the application of ρB as coordinatizing a vector, while the application of ρB1 will be referred to as un-coordinatizing a vector.

Subsection CVS: Characterization of Vector Spaces

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 V is a vector space with dimension n. Then V is isomorphic to n.

Proof   Since V has dimension n we can find a basis of V of size n (Definition D) which we will call B. The linear transformation ρB is an invertible linear transformation from V to n, so by Definition IVS, we have that V and n 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 n. Hmmm. The following examples should make this point.

Example TIVS
Two isomorphic vector spaces
The vector space of polynomials with degree 8 or less, P8, has dimension 9 (Theorem DP). By Theorem CFDVS, P8 is isomorphic to 9.

Example CVSR
Crazy vector space revealed
The crazy vector space, C of Example CVS, has dimension 2 by Example DC. By Theorem CFDVS, C is isomorphic to 2. Hmmmm. Not really so crazy after all?

Example ASC
A subspace characterized
In Example DSP4 we determined that a certain subspace W of P4 has dimension 4. By Theorem CFDVS, W is isomorphic to 4.

Theorem IFDVS
Isomorphism of Finite Dimensional Vector Spaces
Suppose U and V are both finite-dimensional vector spaces. Then U and V are isomorphic if and only if dim U = dim V .

Proof   ( ) This is just the statement proved in Theorem IVSED.

( ) This is the advertised converse of Theorem IVSED. We will assume U and V have equal dimension and discover that they are isomorphic vector spaces. Let n be the common dimension of U and V . Then by Theorem CFDVS there are isomorphisms T : Un and S : V n.

T is therefore an invertible linear transformation by Definition IVS. Similarly, S is an invertible linear transformation, and so S1 is an invertible linear transformation (Theorem IILT). The composition of invertible linear transformations is again invertible (Theorem CIVLT) so the composition of S1 with T is invertible. Then S1 T : UV is an invertible linear transformation from U to V and Definition IVS says U and V are isomorphic.

Example MIVS
Multiple isomorphic vector spaces
10, P9, M2,5 and M5,2 are all vector spaces and each has dimension 10. By Theorem IFDVS each is isomorphic to any other.

The subspace of M4,4 that contains all the symmetric matrices (Definition SYM) has dimension 10, so this subspace is also isomorphic to each of the four vector spaces above.

Subsection CP: Coordinatization Principle

With ρB available as an invertible linear transformation, we can translate between vectors in a vector space U of dimension m and m. Furthermore, as a linear transformation, ρB respects the addition and scalar multiplication in U, while ρB1 respects the addition and scalar multiplication in m. 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 (U) and concrete vector spaces (m).

Theorem CLI
Coordinatization and Linear Independence
Suppose that U is a vector space with a basis B of size n. Then S = u1,u2,u3,,uk is a linearly independent subset of U if and only if R = ρB u1 ,ρB u2 ,ρB u3 ,,ρB uk is a linearly independent subset of n.

Proof   The linear transformation ρB is an isomorphism between U and n (Theorem VRILT). As an invertible linear transformation, ρB is an injective linear transformation (Theorem ILTIS), and ρB1 is also an injective linear transformation (Theorem IILT, Theorem ILTIS).

( ) Since ρB is an injective linear transformation and S is linearly independent, Theorem ILTLI says that R is linearly independent.

( ) If we apply ρB1 to each element of R, we will create the set S. Since we are assuming R is linearly independent and ρB1 is injective, Theorem ILTLI says that S is linearly independent.

Theorem CSS
Coordinatization and Spanning Sets
Suppose that U is a vector space with a basis B of size n. Then u u1,u2,u3,,uk if and only if ρB u ρB u1 ,ρB u2 ,ρB u3 ,,ρB uk.

Proof   ( ) Suppose u u1,u2,u3,,uk. Then there are scalars, a1,a2,a3,,ak, such that

u = a1u1 + a2u2 + a3u3 + + akuk

Then,

ρB u = ρB a1u1 + a2u2 + a3u3 + + akuk = a1ρB u1 + a2ρB u2 + a3ρB u3 + + akρB uk Theorem LTLC

which says that ρB u ρB u1 ,ρB u2 ,ρB u3 ,,ρB uk.

( ) Suppose that ρB u ρB u1 ,ρB u2 ,ρB u3 ,,ρB uk. Then there are scalars b1,b2,b3,,bk such that

ρB u = b1ρB u1 + b2ρB u2 + b3ρB u3 + + bkρB uk

Recall that ρB is invertible (Theorem VRILT), so

u = IU u  Definition IDLT = ρB1 ρ B u  Definition IVLT = ρB1 ρ B u  Definition LTC = ρB1 b 1ρB u1 + b2ρB u2 + b3ρB u3 + + bkρB uk = b1ρB1 ρ B u1 + b2ρB1 ρ B u2 + b3ρB1 ρ B u3 + + bkρB1 ρ B uk  Theorem LTLC = b1IU u1 + b2IU u2 + b3IU u3 + + bkIU uk  Definition IVLT = b1u1 + b2u2 + b3u3 + + bkuk  Definition IDLT

which says that u u1,u2,u3,,uk.

Here’s a fairly simple example that illustrates a very, very important idea.

Example CP2
Coordinatizing in P2
In Example VRP2 we needed to know that

D = 2 x + 3x2,1 2x2,5 + 4x + x2

is a basis for P2. With Theorem CLI and Theorem CSS this task is much easier. First, choose a known basis for P2, a basis that forms vector representations easily. We will choose

B = 1,x,x2

Now, form the subset of 3 that is the result of applying ρB to each element of D,

F = ρB 2 x + 3x2 ,ρ B 1 2x2 ,ρ B 5 + 4x + x2 = 2 1 3 , 1 0 2 , 5 4 1

and ask if F is a linearly independent spanning set for 3. This is easily seen to be the case by forming a matrix A whose columns are the vectors of F, row-reducing A to the identity matrix I3, and then using the nonsingularity of A to assert that F is a basis for 3 (Theorem CNMB). Now, since F is a basis for 3, Theorem CLI and Theorem CSS tell us that D is also a basis for P2.

Example CP2 illustrates the broad notion that computations in abstract vector spaces can be reduced to computations in m. 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, ρB, 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 m 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 U is a vector space with a basis B of size n. Then any question about U, or its elements, which ultimately depends on the vector addition or scalar multiplication in U, or depends on linear independence or spanning, may be translated into the same question in n by application of the linear transformation ρB to the relevant vectors. Once the question is answered in n, the answer may be translated back to U (if necessary) through application of the inverse linear transformation ρB1.

Example CM32
Coordinatization in M32
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 M32, the vector space of 3 × 2 matrices, but we are adverse to doing the operations of M32 (Definition MA, Definition MSM). More specifically, suppose we are faced with the computation

6 3 7 2 4 0 3 +2 13 4 8 2 5

We choose a nice basis for M32 (or a nasty basis if we are so inclined),

B = 10 0 0 0 0 , 00 1 0 0 0 , 00 0 0 1 0 , 01 0 0 0 0 , 00 0 1 0 0 , 00 0 0 0 1

and apply ρB to each vector in the linear combination. This gives us a new computation, now in the vector space 6,

6 3 2 0 7 4 3 +2 1 4 2 3 8 5

which we can compute with the operations of 6 (Definition CVA, Definition CVSM), to arrive at

16 4 4 48 40 8

We are after the result of a computation in M32, so we now can apply ρB1 to obtain a 3 × 2 matrix,

16 10 0 0 0 0 +(4) 00 1 0 0 0 +(4) 00 0 0 1 0 +48 01 0 0 0 0 +40 00 0 1 0 0 +(8) 00 0 0 0 1 = 1648 440 48

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.

Subsection READ: Reading Questions

  1. The vector space of 3 × 5 matrices, M3,5 is isomorphic to what fundamental vector space?
  2. A basis for 3 is
    B = 1 2 1 , 3 1 2 , 1 1 1

    Compute ρB 5 8 1 .

  3. What is the first “surprise,” and why is it surprising?

Subsection EXC: Exercises

C10 In the vector space 3, compute the vector representation ρB v for the basis B and vector v below.

B = 2 2 2 , 1 3 1 , 3 5 2 v = 11 5 8

 
Contributed by Robert Beezer Solution [1582]

C20 Rework Example CM32 replacing the basis B by the basis

C = 149 10 10 6 2 , 74 5 5 3 1 , 31 0 2 1 1 , 74 3 2 1 0 , 4 2 33 2 1 , 0 0 12 1 1

 
Contributed by Robert Beezer Solution [1583]

M10 Prove that the set S below is a basis for the vector space of 2 × 2 matrices, M22. Do this choosing a natural basis for M22 and coordinatizing the elements of S with respect to this basis. Examine the resulting set of column vectors from 4 and apply the Coordinatization Principle.

S = 3399 78 9 , 1647 36 2 , 1027 17 3 , 27 6 4

 
Contributed by Andy Zimmer

Subsection SOL: Solutions

C10 Contributed by Robert Beezer Statement [1580]
We need to express the vector v as a linear combination of the vectors in B. Theorem VRRB tells us we will be able to do this, and do it uniquely. The vector equation

a1 2 2 2 +a2 1 3 1 +a3 3 5 2 = 11 5 8

becomes (via Theorem SLSLC) a system of linear equations with augmented matrix,

2 1311 235 5 2 12 8

This system has the unique solution a1 = 2, a2 = 2, a3 = 3. So by Definition VR,

ρB v = ρB 11 5 8 = ρB 2 2 2 2 + (2) 1 3 1 + 3 3 5 2 = 2 2 3

C20 Contributed by Robert Beezer Statement [1580]
The following computations replicate the computations given in Example CM32, only using the basis C.

ρC 3 7 2 4 0 3 = 9 12 6 7 2 1 ρC 13 4 8 2 5 = 11 34 4 1 16 5 6 9 12 6 7 2 1 + 2 11 34 4 1 16 5 = 76 140 44 40 20 4 ρC1 76 140 44 40 20 4 = 1648 430 48