Introduction

In Example 5 of Section 2.4, we encountered avector space that was contained in a larger vector space and sharedwith the larger space its definitions of vector addition and scalar multiplication. This is a common phenomenon, deserving our special study.

A subspace is nonempty by the definition of the term ‘‘subspace.’’ If an element x is in the subspace, then 0 is in the subspace because 0 = 0 · x. Thus, a subspace must contain the zero element of the larger space. We can state these requirements for a subspace U as follows:

One sees immediately that under those circumstances, U is a vector space, too. All the axioms listed in Section 2.4 will be true for U. Most of these axioms are true automatically because U is a subset of V, and V obeys the axioms. The closure axioms are different in nature, however. Furthermore, we want to be sure that U is nonempty. The element that must be there is, of course, 0.

We notice that two extreme cases of subspaces occur in any vector space V: namely, the subspace 0 and V itself. Any other subspace, U, must lie between these extremes: 0 ⊆ U ⊆ V. The subspace of V that contains only the zero vector is denoted by 0 or {0}. We summarize all this in a formal statement:

PROOF Let the vector space be V and let S be a nonempty subset of V. If x and y are two elements of Span(S), then each is a linear combination of elements of S. Simple algebra shows that the vector αx + βy will also be a linear combination of elements in S. This verifies the two closure axioms for Span(S).

There are many examples of linear subspaces, such as subspaces of polynomials, subspaces of trigonometric polynomials, subspaces of matrices ^m×ⁿ, subspaces of upper or lower triangular matrices, and subspaces of the image or the kernel of some linear operators such as differentiation and integration.

SOLUTION This is the vector space . It is a subspace of . (The zero-polynomial has no degree.)

SOLUTION The set of polynomials of the form t at + bt² + ct⁵ + dt⁷ is a subspace of . ( is defined in Example 1.)

EXAMPLE 3

Let s and t be variables that range independently over . Consider the vectors of the form

Is the set of all vectors of that form a subspace of ?

SOLUTION We have no theorem that directly addresses such a problem. However, we may be able to see that the set in question is the linear span of asetof vectors. If so, Theorem 1 will apply. Observe that

Hence, the set of vectors considered at the beginning is identical to the span of a pair of vectors:

This pair isobviously linearly independent, and,therefore, generates a two-dimensional subspace of . (The concept of dimension is taken up in Section 5.2.)

From two subspaces X and Y in a vector space, we can construct two more subspaces, X + Y and X ∩ Y. (These arise in General Exercises 13–14.) The vector sum, X + Y, and the intersection, X ∩ Y, are defined as follows:

The union of two subspaces is usually not a subspace! (See General Exercises 15–16.)

PROOF Let X and Y be two subspaces of a vector space V. Then X and Y contain the zero element of V. Therefore, X + Y contains the zero element of V. Is X + Y closed under addition? Let v and be two points in X + Y. Then v = x + y and for suitable points x, y, , and . Thus, we obtain . A similar equation shows that X + Y is closed under scalar multiplication.

SOLUTION The vector sum U + V consists of all vectors expressible as αu + βv. In other words, this is the span of the pair {u, v}.

The proofis left as General Exercise 14.

SOLUTION U ∩ V is the set of all x such that x₁ +3x₂ − 4x₃ = 0 and 2x₁ +3x₂ = 0. Geometrically, we are considering two planes in and their intersection, which in this case is a line in through the origin.

Images and inverse images of subspaces by linear transformations are also subspaces.

DEFINITION

If f is a mapping of a set X to a set Y, then for any set U in X, f[U] is defined to be the set of images of points in U. In symbols, we have

PROOF Since U is a subspace, it contains 0.Since f is linear, f(0) = 0. Thus, 0 is in f[U]. If x and y are in U, then so are αx + βy. Hence, we have f(αx + βy) f[U]. Since f is linear, αf(x) + βf(y) f[U]. Thus, f[U] is closed under addition and scalar multiplication.

EXAMPLE 6

This example illustrates Theorem 5 with specific vectors and subspaces. Define a linear transformation f from to by the equation

Let W be the span of a set of two vectors in :

Thus, f maps into . Find a simple description of f[W].

SOLUTION Using the formula for f we obtain

It follows that

DEFINITION

If f is a mapping from one set X to another set Y, then for any subset S in Y, f⁻¹[S] is defined to be the set of all points in X that are mapped by f into S. This set is the inverse image of S in X. In symbols, we have

See Figure 5.1 for an illustration of f[U] and f⁻¹[S]. The notation f⁻¹[S] involves a set S. Its use here does not imply that f is an invertible mapping.

PROOF Since V is a subspace in Y, the zero vectoris in V. Since f is a linear transformation and X is a vector space, 0 X and f(0) = 0. Hence, we have 0 f⁻¹[V]. If x and y are in f⁻¹[V], then x = f(z) and y f(w), for some z and w in V. Thus, we obtain

This proves that αx + βy f⁻¹[V] and that f⁻¹[V] is closed under vector addition and scalar multiplication. Hence, f⁻¹[V] is a subspace of X.

FIGURE 5.1 Sketch of (a) f[U] and (b) f⁻¹[V].

SOLUTION Letting (3x₁, 2x₂ − x₁, x₄) = (1, 1, 0), we obtain , , and x₄ = 0. Hence, we have . Similarly, letting (3x₁, 2x₂ − x₁, x₄) = (0, 1, 1), we have x₁ = 0, , and x₄ = 1 and . Consequently, we obtain

for any s

Linear Transformations

Recall the notion of null space (or kernel) of a linear transformation. It consists of all vectors that are mapped into 0 by the transformation.

PROOF Let T be a linear transformation from one vector space V to another vector space W. The kernel or null space of T is the set

It is obviously a subset of V. It is a subspace of V because first, 0 ∈ Ker(T). Second, if x and y belong to the kernel of T, then αx + βy also belong to the kernel because

EXAMPLE 8

Let (c₁, c₂, …, c_n) be any fixed vector in . Define a linear transformation T: → by the following equation, in which x = (x₁, x₂, …, x_n):

Describe the kernel of T.

SOLUTION It is the set of all vectors x that satisfy the equation

If n = 3, we can visualize this set as a plane through the origin in . This is true if the vector c is not 0. If c is zero, the subspace here is all of . If n = 2, the set in question is a line through 0 if c ≠ 0. If c = 0, the set is all of .

SOLUTION The kernel of T consists of all vectors in for which 3x₁ − 2x₂ = 0. Thus, we have . The graph of this equation is the straight line through the origin with slope . The two points (0, 0) and (2, 3) determine this line. See Figure 5.2.

EXAMPLE 10

We can obtain a subspace by this description:

Take all the vectors x = (x₁, x₂, x₃, x₄) in that are linear combinations of (3, 7, 4, 2) and (2, 0, 1, 3), and in addition obey the equation

Here we are taking the intersection of two subspaces to get a third subspace.

SOLUTION We have

FIGURE 5.2 Kernel of T(x) = 3x₁ − 2x₂.

We also set

Thus, we obtain . In the first preceding equation, replace b by this multiple of a, and arrive at

The solution is that the points described are all the scalar multiples of the vector (11, 35, 18, 4).

EXAMPLE 11

In , let U be the span of the single vector u = (3, 2, 4, 0). Let V be the kernel of the matrix

Find a simple description of the subspace U + V in .

SOLUTION To understand the kernel of A, we use the standard row-reduction process to get

The general solution of the associated homogeneous system is given by

As usual, we write this as

where s and t are free parameters. Thus, the subspace in question is the span of a set of three vectors; namely,

PROOF If T is a linear transformation, say from U to V, then its range is

The vector 0 of V is in Range(T) because 0 = T(0). Thus, V is nonempty. If x and y are in Range(T), then x = T(u) and y = T(w), for suitable u and w in U. It follows that

This shows that Range(T) is closed under vector addition and scalar multiplication. Notice that we used the linearity of T and the hypotheses that U and V are vector spaces.

SOLUTION Every point in the range of T is of the form Ax for some x. Thus, the range of T is the span of the columns of A. If the columns of A form a linearly dependent set, then some columns are not needed in describing the range of T. A row reduction shows that

Because the systems Ax = 0 and Bx = 0 have the same solutions, we can use columns 1 and 2 in A to span the range of T:

Notice that we cannot use the columns of B to describe the range of A in this problem. Each of those columns has 0 as its last entry, whereas the range of A obviously contains some vectors that do not have that property.

Revisiting Kernels and Null Spaces

Recall from your study of elementary algebra the important topic of solving equations. For example, you learned how to find the ‘‘roots’’ of an equation such as 7x² + 3x − 15 = 0. In linear algebra, the corresponding topic is the finding of solutions of an equation of the form L(x) = v, where L is a linear transformation. Even the simple case, L(x) = 0, requires some study. Several of the preceding chapters have dealt with the matrix version of that topic.

Recall that the terms null space and kernel are interchangable. They both refer to

The first case is more general as it does not require a matrix, but only a linear mapping. We define the null space of a matrix A as the set

which we can find by solving Ax = 0.

EXAMPLE 13

Let

Are the two vectors, u = [3, 7, −8, 9]^T and v = [7, 1, 4, −5]^T, in the null space of C? Are these vectors w = [5, −3] and z = [0, 0] in the left null space of C^T?

SOLUTION The question is answered by computing Cu =[10, 38]^T and Cv =[0, 0]^T. Thus, we obtain v Null(C), but u ∉ Null(C). Also, we find that w is not in the null space of C^T, but z is because C^T w ≠ 0 and C^Tz = 0.

SOLUTION The null space of C consists of all solutions to the equation Cx = 0. One can use row reduction on the augmented matrix:

For the associated homogeneous system, the general solution is

This can be written as

where x₃ and x₄ are free variables. Thus, the null space is the span of a pair of vectors; namely, w = [−2, −1, 1, 0]^T and z = [−3, −1, 0, 1]^T.

With regard to the null space of C^T, we obtain

Thus the null space of C^T is {0}.

The Row Space and Column Space of a Matrix

Two important vector subspaces associated with a given matrix are its row space andits column space.

Translating these definitions into alternative terminology, we can say that

Here we have assumed that A is an m × n matrix whose rows are r_i and whose columns are a_j.

PROOF Let A be any matrix. We will prove that any single row operation performed on A to produce a new matrix à has no effect on the row space of A. Once this has been established, it will follow that a sequence of row operations performed on A will also have no effect on the row space. In carrying out this proof, it suffices to consider just the two row operations labeled as Types 1 and 2 in Section 1.1. Consider first the operation of adding to one row a multiple of another (replacement). There is no loss of generality in considering the operation r₁ ← r₁ + αr₂. This notation means replace the first row by row 1 plus α times row 2. Any element in the row space of à has the form

This i sclearly a linear combination of rows r₁, r₂, … r_n. Hence, we have Row(Ã) ⊆ Row(A). The reverse inclusion is also true because the row operation in question is reversible by another row operation of the same type (explicitly, use r₁ ← r₁ − αr₂). A similar analysis handles any row operation of Type 2 (scale, multiplying a row by a nonzero scalar).

EXAMPLE 15

Find simple descriptions of the row space and column space of this matrix:

SOLUTION Because row operations generally simplify a matrix and the row space is not affected, we carry out a row reduction, arriving at

The row space of A is the row space of B, and it is spanned by the first three rowsof B. We can write

the latter being a more elegant description. The rows of an augmented matrix correspond to equations in a linear system. In the row-reduction process, each system has the same solution. Consequently, once the pivot positions are determined, we can use the corresponding rows in any of these coefficient matrices to span the row space! Conversely, the column space of A is given only by the span of the columns from A corresponding to the pivot positions. In this example, we obtain:

Think about why this is so!

EXAMPLE 16

Do these matrices have the same row space and column space?

SOLUTION We examine the reduced row echelon form of each matrix.

Calling upon mathematical software and changing some numbers into simple fractions, we arrive at

The answer is yes! The row spaces are the same, and the three rows of this last matrix form a simple set of three vectors that spans the common row space. The first three columns of G form a linearly independent set that spans Col(G) = . Similarly, the first three columns of H span Col(H) = .

PROOF Recall that the product Ax is defined to be a linear combination of the columns of A, in which the coefficients are the components of the vector x. Thus, with the notation of the preceding paragraphs, we have

(We suppose that A has n columns, a_j.) Because x runs over all possible vectors in , we get all the linear combinations of columns of A. In other words, we get the span of the set of columns in A.

EXAMPLE 17

Is the vector b = [1, 20, 16]^T in the column space of the following matrix?

SOLUTION We are asking whether some vector x exists for which Ax = b. This question is answered by attempting to solve that equation. The appropriate augmented matrix is

Yes, the system is consistent and has the solution x = [3, 2]^T. Checking with an independent verification, we have 3[3, 6, 2]^T + 2[−4, 1, 5]^T = [1, 20, 16]^T.

To describethe row space of a matrix inasimilar manner, weobserve that the row space of an m × n matrix A isthecolumn space of the n × m matrix A^T . One may object that vectors inthe row space should be row vectors, whereas vectors in the column space should be column vectors. However, these distinctions come into play only when we insist on a vector being treated as a matrix. Then a column vector and a row vector must be interpreted differently. But we can take transposes to get rows from columns, and vice versa. Thus, we have

EXAMPLE 18

Find simple descriptions of the row and column spaces of the matrix

SOLUTION Because elementary row operations do not change the row space (Theorem 7), weare free to carry out the row-reduction process, obtaining

The row space is the span of the set consisting of the row vectors in matrix B:

The column space is spanned by columns 1, 3, and 5 of matrix A:

EXAMPLE 19

Let

Also, set v =[3, 12]^T and w = [−4, 15, 22, 11]. Is v in the column space of F? Is w in the row space of F?

SOLUTION We see that thevector v is inthecolumn space of F because

So we have −3[1, −2]^T + 2[3, 3]^T = [3, 12]^T.

To determine whether thevector w is in the row space of F, we take the transpose of the matrix and vector.

The row reduction shows that column three is a linear combination of columns one and two. Checking, we find the relationship 2[1, 3, 2, −5] + 3[−2, 3, 6, 7] = [−4, 15, 22, 11].

Here is another way to answer the question in Example 19 about a row space. The two rows of the matrix F form a linearly independent pair because neither row is a scalar multiple of the other. If we insert the vector w as a third row in this matrix, we can call the resulting matrix G. In going from F to G, the row space will either grow or remain unchanged. In the first case, the vector w is not inthe row space of F, but in the second case it is. We leave the details as General Exercise 38.

Caution

If A ~ B and the matrix B is in reducedrow echelon form, then Col(A) is thespan of the set of pivot columns in A, whereas Row(A) is the span of the set of pivot rows in B.

SUMMARY 5.1

KEY CONCEPTS 5.1

Subspaces, kernel, vector sum, vector intersection, images, inverse image, range, row space, column space, null space, creating subspaces, row equivalent

GENERAL EXERCISES 5.1

1. Determine whether the set of matrices having the form is a subspace of ^2×2.

2. All invertible n × n matrices have the same row space. What is it?

3. Verify that the set {p ∈ : p′(3) = 2p(5)} is a subspace of the space of all polynomials.

4. In , consider the set of all vectors that have the form (3α −2β, α + β + 2, −2β + α, 5α − πβ, α + β. Is this set a subspace of ?

5. Is the set of all vectorsof the form (s³ − t, 5s³ + 2t, 0) a subspace of ? Here s and t run over .

6. Determine whether the set of all 2 × 2 matrices of the form is a subspace ^2×2.

7. In Example 1, we noted that is a subspace of . Explain why we do not say that is a subspace of .

8. Do these two matrices have the same row space?

9. Define T : → by T(x₁, x₂, x₃) = (3x₁ − 2x₂ +5x₃, 2x₁ + x₂ − x₃). Define U = {x ∈ : x₁ + x₂ − 2x₃ = 0}. Find a simple description of T[U].

10. Describe in simple terms the row and column spaces of

11. What is the simplest example of two matrices that are row equivalent to each other yet have different column spaces?

12. The trace of a square matrix is the sum of its diagonal elements. Thus, if A is n × n, then . In the vector space , consisting of all n × n matrices, is the set of matrices with zero trace a subspace?

13. Give an example and a geometric interpretation of two subspaces U and W of the vector space . Note that their sum is also a subspace of . The sum is defined by U + W = {u + w : u ∈ U and w ∈ W}. This illustrates Theorem 3.

14. Establish that if U and W are subspaces of a vector space V, then their intersection is also a subspace of V. The intersection is defined by U ∩ W = {x : x ∈ U and x ∈ W}. This is Theorem 4.

15. Is the union of two subspaces in a vector space always a subspace? Explain why or find a counterexample. The union is defined by U ∪ W = {x : x ∈ U or x ∈ W}. (In mathematics, we always use the inclusive or. Thus, p or q is true when either p is true or q is true orboth are true.)

16. (Continuation.) Find the exact conditions on subspaces U and W in a vector space V in order that their union be a subspace.

17. Let be the vector space of all polynomials and let be the vector space of all polynomials of degree at most n. Explain why may be viewed as a subspace of . Let be the collection of polynomials with only even powers of t. Is a subspace of ? (Assume that a constant polynomial c ct⁰ is an even power of t.)

18. Fix a set of vectors {u₁, u₂, …, u_n} in some vector space. Explain why the set of n-tuples (c₁, c₂, …, c_n) such that is a subspace of .

19. Argue that if A ~ B, then each row of B is a linear combination of rows in A.

20. Give a proof of Theorem 1.

21. Show why the set of all non invertible 2 × 2 matrices (with the usual operations) is not a vector space.

22. Explain why if T : U → V is a linear transformation, the inverse image of a subspace W in V via the transformation T is a subspace of U. The inverse image is defined by T⁻¹[W] = {x ∈ U : T(x) ∈ W}.

Caution: The notation T⁻¹[W] does not insinuate that T has an inverse. Every map (invertible or not) has inverse images as defined in the preceding equation.

23. Establish that if T is a linear transformation, then the image of a subspace in the domain of T is a subspace in the co-domain of T.

24. Establish that the span of a set in a vector space is the smallest subspace containing that set.

25. In a vector space, suppose that w = u + v and z = u − v. Explain why the spans of {u, v} and {w, z} are the same.

26. What conclusion can be drawn about a set S in a vector space if it satisfies the equation Span(S) = S?

27. Use the standard polynomials p₀, p₁, …, where p_i(t) = tⁱ. Then we write = Span{p₀, p₁, … p_n}. Let D denote the differentiation operator. Establish that . Also, show that Span{p₁, p₂, … p_n} is the same as the set {p ∈ : p(0) = 0}.

28. Explain why two matrices cannot have the same column space if one matrix has a row of zeros and the other does not.

29. If A and B are row equivalent to each other, does it follow that A and B have the same column space? (A proof or counterexample is needed.)

30. Explain why the ith row of AB is x^TB, where x^T is the ith row of A.

31. Establish that if u_i is the ith column of A and if is the ith row of B, then AB = . (Notice that if A is m × n and if B is n × p, then each u_i is m × 1 and each is 1× p. Hence each is m × p.)

32. If f : X → Y and if U ⊆ X, then we can create a new function g by the equation g(x) = f(x) for all x in U. If U is a proper subset of X, then g ≠ f because they have different domains. The function g is called the restriction of f to U. The notation g = f|U is often used. Suppose now that X and Y are linear spaces and L is a linear map from X to Y. Let U be a subspace of X. Is L|U linear? What is its domain? Do L and L|U have the same range? (Arguments and examples are needed.)

33. Can this be used as the definition of a subspace? A subspace in a vector space is a subset that is closed under the operations of vector addition and multiplication of a vector by an arbitrary scalar.

34. Let A and B be arbitrary matrices subject only to the condition that the product AB exist. Use the notation Col to indicate the column space of a matrix. Consider these two inclusion relations: Col(AB) ⊆ Col(A) and Col(AB) ⊆ Col(B). Select the one of these that is always correct and prove it.

35. Give an example in which Col(AB) = Col(A). Give another example in which Col(AB) ≠ Col(A). Finally, give an example in which Col(AB) = Col(B).

36. Establish that each row of AB is a linear combination of the rows of B, and thus establish that Row(AB) ⊆ Row(B).

37. Let ^∞ be the space of all infinite sequences of the form [x₁, x₂, …], where the x_i are arbitrary real numbers. Let X be the subset of ^∞ consisting of all sequences that have only finitely many nonzero terms. Is X a subspace of ^∞?

38. In the last part of Example 19, use row reduction on the matrix G to show that w is inthe row space of F.

39. Use in this exercise.

40. Is the set of all vectors x =[x₁, x₂, x₃, x₄]^T that are linear combinations of vectors[4, 2, 0, 1]^T and [6, 3, −1, 2]^T, and in addition satisfy the equation x₁ = 2x₂ a subspace of ? Explain why or why not.

41. Write out the details in the solution to Example 2.

42. Write out the details in the alternative solution to Example 19.

43. Explain why f[Span(W)] = Span(f[W]) when f is linear. Hint: This is needed in Example 6.

44. Let p₀, p₁, p₂, …, p_n be polynomials such that the degree of p_k is k, for 0 ≤ k ≤ n. Explain why {p₀, p₁, p₂…., p_n} is a basis for .

45. Let Find the null space of each of these matrices as well as for A^T and B^T.

46. Let U consist of all vectors in whose entries are equal. Explain why U is a subspace of and describe U geometrically.

47. Let U consist of the span of all 2 × 2 matrices whose second row is zero. Let V be the span of those whose second column is zero. Explain why U + W and U ∩ W are vector subspaces of _2×2, which is the vector space of all 2 × 2 matrices.

48. Let W be the subspace of spanned by these vectors: u₁ = (1, 2, 1, 3, 2), u₂ = (1, 3, 3, 5, 3), u₃ = (3, 8, 7, 13, 8), u₄ = (1, 4, 6, 9, 7), and u₅ = (5, 13, 13, 25, 19). Find a basis for W.

49. Consider

Find bases for the row space and the column space of A.

50. Explain and discuss the following.

COMPUTER EXERCISES 5.1

1. Consider

Find simple descriptions of the

2. Consider the linear transformation T(x) = Bx where

Describe in simple terms the

3. Consider

Find simple descriptions of the

4. Consider

Find simple descriptions of the

Basis for a Vector Space

In this section, we explore the different ways in which the elements of a vector space can be represented by using various coordinate systems. The unifying concept is that of a basis for a vector space.

SOLUTION Yes! This set spans and is linearly independent. Thus, it is a basis. It is often referred to as the standard basis for .

SOLUTION We ask: “Can we solve any equation Ax = b, when b is an arbitrary vector in ?” Yes; we start with the fourth equation, which gives x₁ = b₄ (and there is no other choice!). Then go to the third equation. It can be solved for x₂ because x₁ is known. Again, there is no other possible value for x₂. Continue upward through the set offour equations, noting that there are no choices to be made. The solution is uniquely determined by the vector b. Consequently, if b = 0, then x = 0 necessarily follows. We have proved that the columns formal in early independent set of vectors that spans .

SOLUTION Put the vectors into a matrix A as columns. We ask first whether the equation Ax = 0 has any nontrivial solutions. Row reduction leads to

(Here we omit the righthand-side zero-vector.) This shows that the equation Ax = 0 can be true only if x = 0. Hence, the set of columns is linearly independent. The row reduction also shows that the equation Ax = b can be solved for any b in . Therefore, the setof columns spans and provides a basis for .

SOLUTION We happen to notice that column three in this matrix is equal to column two minus column one. The set of columns is, therefore, a linearly dependent set. Hence, it is not a basis for anything. Of course, this set of columns spans a two-dimensional subspace in , but it is not a basis for that subspace because of its linear dependence. Columns one and two in this matrix, taken together, provide a basis for the column space of A. That is a two-dimensional subspace in . In otherwords, a plane in three-space passing through 0. (In this discussion, the term dimension is being used in an informal way. Precise definitions are forthcoming.)

The preceding four examples suggest that a general theorem can be proved.

PROOF Let A be such a matrix. Do its columns span ? Can we solve the system of equations Ax = b for any b in ? Yes, because A⁻¹b solves the system. Thus, thecolumns of A span . What about the linear independence of the set of columns? Is there a non zero solution to the equation Ax = 0? No, because if Ax = 0, then x = 0, as is seen by multiplying by A⁻¹. So the columns constitute a linearly independent set that spans .

EXAMPLE 5

Do the first four Legendre polynomials, defined by

constitute a basis for ?

SOLUTION Let us investigate the spanning property first. Given any polynomial q in , say q(t) = a₀ + a₁t + a₂t² + a₃t³, can we express it in terms of Legendre polynomials? In otherwords, can we solve this equation for c₀, c₁, c₂, c₃?

Putting inmore detail, we have

Collecting similar terms on the left brings us to

From elementary algebra, we recall that two polynomials are the same if and only if the coefficients of each power of t are the same. Thus, we must have 2c₀ − c₂ = 2a₀, 2c₁ − 3c₃ = 2a₁, 3c₂ = 2a₂, and 5c₃ = 2a₃. We can indeed solve for the coefficients c_i using these equations in reverse order. We arrive at , , , and . This proves that the span of the first four Legendre polynomials is . For the linear independence question, just substitute p = 0 in the preceding calculation to see that all c_j must be zero(because a₀ = a₁ = a₂ = a₃ = 0).

PROOF This is established by Theorem 6 in Section 2.4. The reader can probably supply a proof without referring to Section 2.4. The proof relies on the concepts of basis, spanning, and linear independence.

Let’s look at a more provocative example, where the vectors involved are again polynomials.

EXAMPLE 6

Find a basis for the vector space spanned by the four polynomials p₁, p₂, p₃, p₄, where

SOLUTION The correspondence between these polynomials and their terms can be displayed as follows:

The four column vectors headed by p₁, p₂, p₃, and p₄, in this table need to be analyzed to uncover any linear dependencies. A row reduction is called for:

(There is no need to keep the zero row.) The row-reduced form shows that column four is a linear combination of the first three columns. This is true of the beginning matrixaswell as the final matrix. (The row-reduction process does not disturb linear dependencies among the columns of a matrix.) Hence, a basis is {p₁, p₂, p₃}. (We do not need p₄.)

Another question for the same example: What non trivial linear relationship exists among p₁, p₂, p₃, p₄? We continue the reduction process on the preceding matrix, ending at the reduced echelon form:

We obtain , , and x₃ = −3.2x₄. Leting x₄ = 5, we have x = (8, 6, −16, 5). Therefore, we have 8p₁ + 6p₂ − 16p₃ + 5p₄ = 0. This verifies that p₄ can be expressed in terms of the basis.

Coordinate Vector

We almost always restrict our attention to ordered bases, so that we can refer to their elements by subscript notation. For example, we might write

If this is a basis for a vector space V, then each vector x in V can be written in one and only one way as . The n-tuple (c₁, c₂, …, c_n) that arises in this way is called the coordinate vector of x associated with the (ordered) basis . It is denoted by .

SOLUTION This problem can be turned into a system of linear equations that must be solved:

Row reduction of the augmented matrix leads to c = (3, 4, −5), as shown next:

An independent verification consists in computing

PROOF There are two aspects of linearity, which we can combine and address as follows:

where α and β are scalars. To see that this equation is true, let and . Then these three equations follow:

Hence, we obtain

PROOF Is this map surjective (onto)? Is every vector in an image of some x in V? Let c be any vector in , and write c = (c₁, c₂, …, c_n). Is c equal to for some x in V? Of course: it is the image of the vector . (Refer to Section 2.3, Theorem 4.)

Is our mapping injective (one-to-one)? Because we know it to be linear, it suffices to verify that only the element 0 in V maps into the vector 0 of . But this is easy, because if , then . (Refer to Section 2.3, Theorem 5.)

Isomorphism and Equivalence Relations

In Theorem 4, we have observed a pair of vector spaces, V and , that are related by a linear, injective, and surjective map. Such a map is called an isomorphism, and the spaces are said to be isomorphic to each other. The relationship of two spaces being isomorphic to each other is an example of an equivalence relation.

PROOF There are three properties to verify.

We discuss equivalence relations briefly in Section 1.1 and will return to them again in Section 5.3.

SOLUTION If we think that these spaces are isomorphic to each other, we must invent an isomorphism, and prove that it is one. Given a polynomial in , write it in detail—for example, in the standard form

With the polynomial p, we can now associate the vectorin

This is another example of selecting a basis for a space and then identifying each element in the space with its coordinate vector relative to the basis. In this example, we associate with p, and the isomorphism is the map . Theorems 3 and 4 indicate that we have an isomorphism. What basis for are we using? It is the (ordered) set of functions t 1, t, t², …, tⁿ.

SOLUTION Consider the map .

We get all of , and L is surjective. If , then . Hence, L is injective. Thus, L is one-to-one and maps onto . The mapping L is linear. Hence, it is an isomorphism.

SOLUTION Let p(t) = a₀ + a₁t + a₂t² + a₃t³ + a₄t⁴ + a₅t⁵, so that p is a generic element of . The vector of coefficients [a_0, a₁, a₂, a₃, a₄, a₅]^T is in , and the matrix is in . The mapping from p to the 2 × 3 matrix is linear, injective, and surjective.

PROOF I. Let be a finite basis for the vector space being considered. Let be another basis for the same space. Since is linearly independent and spans the space, the number of elements in is not greater than the number of elements in , by Theorem 10 in Section 2.4. Conversely, since is linearly independent and spans the space, the number of elements in is no greater than the number of elements in .

We give another, slightly different, proof. This is by the method of contradiction, and it uses coordinate vectors.

PROOF II. Let n denote the number of elements in , and suppose that has more than n elements. Select v₁, v₂, …, v_m in , where m > n. We intend to show that is linearly dependent, hence not a basis. Consider the equation

Each is a column vector having n components, because n is the number of elements in . Thus, the preceding vector equation has n equations and m unknowns. By Corollary 2 in Section 1.3, this system has a nontrivial solution. By the linearity of the map , we have

nontrivially. Thus, is linearly dependent.

Finite-Dimensional and Infinite-Dimensional Vector Spaces

A vector space having a finite basis is said to be finite dimensional. The number of elements in any basis for that space is called the dimension of the space. Thus, we can now say that is n-dimensional, and has dimension n + 1. In both cases, it suffices to exhibit a basis of the space and count thenumber of elements in it. In the same way, we find that the dimension of is mn. If V is a finite-dimensional vector space, we use Dim(V) for its dimension. Thus, for example, we have

The space , defined in Section 2.4, Example 3, is not finite dimensional. This fact is most easily established by noting that the vectors e_n defined by (e_n)_i = δ_ni are infinite in number yet form a linearly independent set. By Theorem 6 in Section 2.4, this vector space cannot have a finite spanning set. Naturally, we call such vector spaces infinite dimensional. Many such spaces are needed in applied mathematics and they have been assiduously studied for more than a century. Hilbert spaces and Banach spaces are examples. So are Sobolev spaces. (These topics are beyond the scope of this book.)

The linear space of all continuous real-valued functions on the interval [0, 1] is infinite dimensional. To be convinced of this, it suffices to exhibit an infinite linearly independent family of continuous functions on [0, 1]. The set of monomials, p_n(t) = tⁿ, where n = 0, 1, 2, …, serves this purpose.

PROOF First, suppose that the twospaces U and V are finite dimensional and have the same dimension. Select bases {u₁, u₂, …, u_n} for U and {v₁, v₂, …, v_n} for V. Define a linear transformation T by writing T(u_i) = v_i, for i = 1, 2, …, n. Thus, the effect of T on any point

in U is like this:

Is T injective? Suppose T(u) = 0. Thus, we have

Yes, T is one-to-one by Theorem 5 in Section 2.3 because it follows that . Is T surjective? For any v in V, write

Yes, T is onto by Theorem 4 in Section 2.3.

For the other half of the proof, let T be an isomorphism from U onto V. Let {u₁, u₂, …, u_n} be a basis for U. Then {T(u_i) : 1 ≤ i ≤ n} is a basis for V. This assertion has two parts, each requiring proof. First, is that set linearly independent? If

then

But, since T is an isomorphism, it is injective, and so and by the linear independence of {u_i : 1 ≤ i ≤ n}. Second, does thesetin question span V ? If v ∈ V, then v = T(u) for some u ∈ U, because T is surjective. Then it follows that

SOLUTION The two matrix spaces have dimension mn, and those two spaces are therefore isomorphic to each other by Theorem 7. The other pair of spaces is isomorphic for the same reason: they have the same dimension, namely 21.

EXAMPLE 12

Consider the three vectors

What can be said of Span{u, v, w}?

SOLUTION Notice that v = 2u −3w; namely, (13, 3, 4) = 2(2, 3, −1) − 3(−3, 1, −2). Thus, if we have a generic element of the span, we can do some simplifying, like this:

This equation shows that

The vector v, which was a linear combination of u and w, is not needed in describing the span. This calculation can be done in greater generality, as shown in the next result.

PROOF Let S be the set, and suppose that

where the vectors v₀, v₁, …, v_n are distinct members of S. Let w be any vector in the span of S. Write

We may require that m ≥ n, and we permit some coefficients a_i to be zero. Then rewrite

where, if necessary, we have inserted some zero coefficients. Now we have

This shows that w is representable as a linear combination of elements in S, but without using v₀.

PROOF If S is the set and its span is V, we have only to prove that S is linearly independent. Suppose S is linearly dependent. Then some element of S is a linear combination of the other elements of S. Removing that element does not affect the span of the set, by Theorem 8. By Theorem 6 in Section 2.4, every linearly independent set in V can have at most n − 1 elements. Hence, we have Dim(V) ≤ n − 1, a contradiction.

PROOF If S is the given set and fails to be a basis, it must fail to span the vector space. Therefore, some point x in the vector space is not in the span of S. Now the set S ∪ {x} is linearly independent, and the dimension of the space must be greater than n.

EXAMPLE 13

Find asimple basis for the row space of

SOLUTION Again, we use the principle that row-equivalent matrices have the same row space. The row-reduction process yields this reduced echelon matrix:

The dimension of the row space is three, and the first three rows of the matrix B provide a basis for the row space of A. It may or may not be true that the first three rows in A give a basis for the row space. Do not assume that this is true. (The next example illustrates this.)

EXAMPLE 14

Find a simple basis for the row space of

SOLUTION A row reduction leads to

The first two rows of B give a simple basis for the row space of A, but the first two rowsof A do not span the row space. (Why?)

PROOF According to Theorem 6 in Section 2.4, a linearly independent set in a vector space can have no more elements than a spanning set. Hence, a basis can have no more elements than a spanning set.

PROOF Let {v₁, v₂, …, v_n} be a linearly independent set in a vector space V. By Theorem 6 in Section 2.4, any spanning set in V must have at least n elements. Hence, any basis for V must have at least n elements, and Dim(V) ≥ n.

PROOF Let A be any matrix, and denote by B its reduced echelon form. The pivot columns in B form a linearly independent set because no pivot column in B is a linear combination of columns that precede it. (Look at Example 15, which shows a typical case.) Because A is row equivalent to B, the pivot columns of A form a linearly independent set. In this situation, remember that the vectors x that satisfy the equation Ax = 0 are the same as the vectors that satisfy Bx = 0. Each nonpivot column in A is a linear combination of the pivot columns of A. The set of pivot columns in A spans Col(A) and is linearly independent. Hence, it is a basis for the column space of A.

EXAMPLE 15

Find a basis for the column space of

SOLUTION The row-reduction process on A yields a matrix B:

Notice that in B, columns three and four are linear combinations of columns one and two. Hence, the same is true for the columns of A. We conclude that columns one, two, and five in A constitute a basis for Col(A). Observe that columns one, two, and five in B certainly do not constitute a basis for Col(A). (Why?)

SOLUTION From f(t) = 5t³ + 7t² − 8t + 9, we find

Suppose ; that is,

or

Using the coefficients in these equations, we obtain this linear system

From forward substitution, we find that c₀ = 0, c₁ = 0, c₂ = 0, and c₃ = 0. The set is linearly independent. Therefore, the set is a basis for .

EXAMPLE 17

Is it possible to extend this pair of vectors to get a basis for ?

SOLUTION We create this matrix and transform it to reduced echelon form

Use columns one, two, four, and five in A as a basis. Notice that a linear combination of the first two columns gives the third column in A. The columns of B do not solve the stated problem because B does not contain the two given vectors.

Linear Transformation of a Set

The next theorem asserts that the application of a linear transformation to a set cannot increase its dimension. Stated in other terms, the dimension of the range of a linear transformation cannot be greater than the dimension of its domain.

THEOREM 14

For any finite-dimensional vector space U, for any vector space V, and for any linear transformation L from U to V, we have

PROOF Select a basis {u₁, u₂, …, u_n} for U. If x ∈ U, then for suitable c_i. Consequently, we have

Since x was arbitrary in U, we have

Hence, we obtain Dim(L[U]) ≤ n = Dim(U).

PROOF If x is any point in U, its coordinates (a₁, a₂, …, a_n) with respect to the basis {u₁, u₂, …, u_n} are uniquely determined. Then

Since T is linear,

A common pitfall is to assume that a linear transformation can be defined by specifying T(w_i) = y_i for i = 1, 2, …, n for any vectors w_i and y_i. Theorem 15 does not say that, does it?

EXAMPLE 18

Find the linear transformation T such that T(u_i) = v_i, where

SOLUTION We are tempted to use Theorem 15. But the set {u₁, u₂, u₃} is linearly dependent, in as much as

If T is a linear transformation, then

Putting in the details, we eventually get a contradiction:

Thus there is no such linear transformation.

PROOF Let {u₁, u₂, …, u_k} be linearly independent. If {u₁, u₂, …, u_k} spans the given vector space V, then it is a basis (because it is linearly independent and spans V). If{u₁, u₂, …, u_k} does not span V, find u_k+1 that is in V but not in Span{u₁, u₂, …, u_k}. Then {u₁, u₂, …, u_k, u_k+1} is linearly independent. Repeat this process. See Theorem 11 in Section 2.4.

Dimensions of Various Subspaces

If T is a linear transformation defined on one vector space U and taking values in another vector space V, we already have two spaces related to T: its domain (U) and its co-domain (V). Then we can go on to define the kernel of T by writing

This is also called the null space of T and is denoted by Null(T). The nullity of a linear transformation T is the dimension of its null space. The kernel of T consists of all the vectors in U that are mapped by T into 0. Next, we can define the range of T,

This is the set of all images created by T when it acts on all the vectors in U. The kernel of T is a subspace of U, and the range of T is a subspace of V. These assertions have been proved in Section 5.1. (See Theorems 7 and 8 in Section 5.1.) The preceding definitions are valid without restricting U or V to be finite dimensional. However, if U is finite dimensional, we can establish some relations among the dimensions of these spaces. (All of these dimension numbers are non negative integers.) Begin by observing that if U is of dimension n, then the range of T cannot be of dimension greater than n. (Recall Theorem 14, which asserts that one cannot increase the dimension of a set by applying a linear transformation to it.)

The theorem we will prove can be illustrated in a simple case where the dimensions are easily discerned.

EXAMPLE 19

Consider a linear transformation T : → defined by the following matrix A. Recall that this terminology means T(x) = Ax. We show also the reduced row echelon form of A as the matrix Ã.

What are the dimensions of the various subspaces associated with A?

SOLUTION The dimension of the kernel of A will be two because there are two free variables in Ã. Columns one, three, and five in à form a linearly independent set because they are pivot columns. Therefore, columns one, three, and five in A form a linearly independent set. These columns in A, (not in Ã) give a basis for the range of T. Hence, the dimension of the range of T is 3. The sum of the dimensions of the range and kernel is 5, which is also the dimension of the domain of T.

The preceding calculation should generalize to any matrix. Here is the general case, for an arbitrary matrix.

SOLUTION There will be k columns having pivots and n − k columns without pivots. Each column without a pivot adds to the dimension of the null space because those columns correspond to free variables. Hence, we have

The columns of A that have pivot positions form a basis for the range of A. Hence, the equation (n − k) + k = n allows us to conclude that

This theoretical result can be formulated without referring to matrices as follows.

THEOREM 17

If T is a linear transformation whose domain is an n-dimensional vector space, then

PROOF Select a basis {y₁, y₂, …, y_m} for the range of T. Select a basis {x₁, x₂, …, x_k} for the kernel of T. Select vectors u_i so that T(u_i) = y_i for i = 1, 2, …, m. Define

It is to be proved that is a basis for the domain of T. Let the domain of T be the space U. First, we will prove that spans U. Let v ∈ U. Then T(v) is in the range of T. Therefore, we obtain

This shows that

and that is in the kernel of T. From this it follows that , whence . Next, we prove that is linearly independent. Suppose that . Then

Hence, all b_j are zero. Then and all a_i are zero. Conclusion:

The purely matrix form of Theorem 17 can be stated as follows.

Think of this theorem as stating that each column of a matrix contributes either to the dimension of the null space or the dimension of the column space. Ultimately, this depends on the very simple observation that each column either has a pivot position orit does not. The columns with pivots add to the dimension of the column space, whereas thecolumns without pivots add to the dimension of the null space (because they correspond to free variables in solving the homogeneous problem Ax = 0). In otherwords, for an m × n matrix, we have

The next theorem is an important one in linear algebra. It may even be surprising, because it draws a connection between the row space and the column space of a matrix. In general, these are subspaces in completely different vector spaces, and , if the matrix in question is m × n.

THEOREM 19

The row space and column space of a matrix have the same dimension:

PROOF Let A be an m × n matrix of rank r. Let à denote its reduced row echelon form. Then r is the number of nonzero rows in Ã, by the definition of the term rank. Each nonzero row in à must have a pivot element, and these rows form a linearly independent set. Hence, they constitute a basis for the row space of A. Thus, r is the dimension of the row space of A. Because r rows have pivots, r columns have pivots. The remaining n − r columns do not have pivots. Each of these columns gives rise to a free variable, and the dimension of the null space of A equals this number n − r. The column space of A has dimension r, because the pivotal columns (of A) constitute a basis for the column space.

Another way to see that the column space has dimension r is to use Theorem 17. The dimension of the domain of A (interpreted as a linear transformation in the usual fashion x Ax) is n. The kernel of A has dimension n − r. Therefore, the range of A is of dimension r. The range of A is its column space.

In some books and technical papers, you will encounter the terminology row rank and column rank of a matrix. These are the dimension of the row space and the dimension of the column space, respectively. Thus, Theorem 18 can be stated in the form ‘‘The row rank and column rank of a matrix are equal.’’

SOLUTION Before attempting to construct such an example, we test the equation

The information given would lead to 9 + 7 = 17, which is incorrect. No such linear transformation exists.

SOLUTION In this inequality, let U be the domain of T. The notation T[U] represents the image of the vector space U under the linear transformation T. Hence, it is the range of T. By Theorem 17, the inequality follows at once, because the kernel of T cannot have a negative dimension.

SOLUTION We find this row equivalence:

The reduced echelon form revealsthat insolving theequation Ax = 0 we obtain x₁ + 2x₂ + 3x₃ = 0 and , or

A basis for the null space is {u, v, w}. Therefore, we obtain Dim(Null(A)) = 3. A basis for the column space of A is given by columns one and four of A or

Consequently, we obtain Dim(Col(A)) = 2. A basis for the row space of A is given by rows one and two of B or

Finally, as a check, we have Dim Col(A)) + Dim (Null(A)) = 2 + 3 = 5 = n, where A is m × n = 4 × 5.

Caution

If A ~ B, where B is the reduced echelon form of A, then select pivot columns from A for the basis of Col(A), but select pivot rows from B for Row(A). Remember that row operations preserve the linear dependence among the columns but not among the rows!

SUMMARY 5.2

KEY CONCEPTS 5.2

Basis for avector space, standard basis, coordinate vector of an element in a vector space , isomorphism, isomorphic, equivalence relations, dimension of a vector space, finite dimensional and infinite dimensional, kernel, domain, range, co-domain, dimensions of various subspaces, column space, row space, null space, rank, nullity

GENERAL EXERCISES 5.2

COMPUTER EXERCISES 5.2

1. Let

Numerically compute the following:

2. Consider this n × n matrix

Determine numerically for what values of m this matrix is noninvertible.

3. Consider this matrix

Numerically compute the following:

4. Consider this augmented matrix:

5. Consider the matrices

Determine numerically which of these matrices have the same row space.

Coordinate Vectors

In Section 5.2, we considered an arbitrary n-dimensional vector space V and a prescribed ordered basis for V: = {u₁, u₂, …, u_n}. Every vector x in V has a uniquely determined set of coefficients (a₁, a₂, …, a_n) such that . This set of coefficients is denoted by and is called the coordinate vector of x relative to the basis , or the coordinatization of x associated with . It is a vector in . Thus, the equation = a is true if and only if

Here, of course, we have written a = (a₁, a₂, …, a_n). One must think of as an ordered set {u₁, u₂, …, u_n} so that the coefficients a_i will match with the vectors u_i.

SOLUTION We place the vectors in an augmented matrix and find its reduced echelon form:

The result is , and one can verify it by computing 2v₁ + 5v₂ − 3v₃ = w.

Changing Coordinates

Now there is the question of how to change coordinates from one basis to another. Thus, two bases will be active in a single vector space, and we shall seek a method to go back and forth between the two types of coordinatization. Suppose, then, that the vector space V is finite-dimensional, and that two bases, and , have been prescribed. How can be computed from ? Here is the appropriate analysis.

First, we must name the elements in the first basis. = {u₁, u₂, …, u_n}. Let x be any element of V, and let = (a₁, a₂, …, a_n). Then and

This equation used crucially the linear property of coordinatization (Theorem 3 in Section 5.2). Now recall that a linear combination of vectors can be written as a matrix times a column vector. The columns of the matrix are the vectors figuring in the linear combination, whereas the entries of the single additional column vector are the coefficients in this linear combination. Accordingly, we construct a matrix P whose columns are , and write

where

The matrix P is called the transition matrix that goes from basis to basis . This discussion has proved the following theorem.

This theorem can be illustrated with the following diagram

in which the transition matrix P goes from basis to basis .

SOLUTION By Theorem 1, the columns of P should be and . Concentrating on the first column of P, we ask for the coefficients x₁ and x₂ such that

The augmented matrix for this problem is . The solution is x₁ = 1 and x₂ = −1. Hence, column one in P is .

Going on to column two in P, we must find the coefficients y₁ and y₂ such that

The augmented matrix for this system of equations is .

The solution is y₁ = −2 and y₂ = 3. Hence, column two in P is . It follows that the P-matrix is .

A more efficient approach is to note that there are two systems of equations to solve and that they share a coefficient matrix, A. The two systems of equations are Ax = u₁ and Ay = u₂. One should think of this problem as

For the augmented matrix, the steps in row reduction are like this:

One obtains the same P as found previously.

EXAMPLE 3

For , let us use two bases, where the first consists of Chebyshev¹ polynomials:

For the second basis, the simple monomials are used. This basis is called the natural basis or the standard basis:

What is the matrix needed for transforming coordinates to coordinates?

SOLUTION Call this matrix P. The columns of P should be , where 0 ≤ i ≤ 3. We observe that

Hence, we obtain

SOLUTION Let’s not call it P because that will confuse the two examples. Use instead Q. The columns of Q should be , and . Consider first p₀. How can it be expressed as a linear combination of T₀, T₁, T₂, T₃? Call the unknown coefficients z₀, z₁, z₂, z₃. We want z₀T₀ + z₁T₁ + z₂T₂ + z₃T₃ = p₀. Representing the polynomials just by their coefficients, we must have

This is a system of equations that can be solved by row reduction. Each T_i must be treated in the same way, and the coefficient matrix remains the same for each. Hence, we should solve the system whose augmented matrix is

It is clear that we are finding the inverse of P by this calculation! But now it is obvious from Theorem 1 that the inverse of P is indeed what is wanted: From , we have immediately . Thus, we obtain Q = P⁻¹. For the record, the answer is

Examples 3 and 4 can be illustrated with the following diagram:

Here the transition matrix P goes from basis to basis , and its inverse P⁻¹ goes from basis back to basis .

Linear Transformations

Theorem 3 in Section 2.3 asserted that a linear transformation from to is completely determined by the images of the standard basis elements e_i in . A more general result along these lines is possible; it is the same as Theorem 15 in Section 5.2 except that there the basis is finite.

PROOF We employ the usual notation, L : X → Y, to name the map, the domain, and the co-domain. Let be a basis for the domain. (It need not be finite.) If x is any point in X, then for a suitable finite set of elements u_i in and accompanying scalars a_i, we have . By linearity, it follows that

SOLUTION We only have to be sure that the two vectors u₁ and u₂ form a linearly independent set. That they do so is obvious because neither is a multiple of the other. These two vectors forma basis for their span. Notice that AU = V where

SOLUTION If the linear transformation is given by a 3 × 3 matrix X, then we want

The equation XB = C is not the sort of problem we have become skilled at solving. Ta ketransposes to get a familiar problem, B^TX^T = C^T. The accompanying augmented matrix is

The system is obviously inconsistent. There is a linear relation among the columns of B, and any linear transformation must preserve that. In other words, the third vector is not free. The following theorem is pertinent.

PROOF We have at once

A linear transformation mapping a finite-dimensional vector space into another vector space (not necessarily finite-dimensional) can be described by a matrix, after bases have been chosen for the domain and range of the transformation. Suppose that the transformation is L : X → Y and that bases have been chosen: = {u₁, u₂, …, u_n} for X, and = {v₁, v₂, …, v_m} for the range of L. By Theorem 11 (in Section 5.2), we know that m ≤ n.

For any x in X, we can write . The scalars α_i are the coordinates of x with respect to the basis . Because L is linear, we can proceed to the equation . Now take the -coordinates of both sides, using the fact that this process is also linear (Theorem 3 of Section 5.2). The result is . Form a matrix A by putting as the columns of A. The preceding equation is then

because the α_i are the entries in the coordinate vector for x. We have established the following result.

SOLUTION To answer this, we must find L(p_i) and . We have L(p₀) = 0, L(p₁) = 1, L(p₂) = 4x, L(p₃) = 12x² − 3, L(p₄) = 32x³ − 16x. Therefore, the matrix sought is

If we desire to know the effect of applying L to a typical random polynomial x = 3T₀ − 2T₁ + 7T₂ − 4T₃ + 6T₄, the answer is , which is computed from the matrix A:

In other words, the resulting polynomial is x 10 − 68x − 48x² + 192x³. Checking with an independent verification:

The derivative is 10 − 68x − 48x² + 192x³.

EXAMPLE 8

Consider the linear transformation defined by

Suppose that the basis for is chosen to be

where

and the basis for is

where

Find the matrix A such that .

SOLUTION The transformation is

Associate with each matrix in the basis the vector ; namely,

We seek the matrix A such that . Matrix A has columns for i = 1, 2, 3, 4. We find

The matrix A is therefore

Mapping a Vector Space into Itself

The final question that we want to address pertains to the case of a linear transformation mapping a vector space into itself. We want to understand the various matrices that can represent our mapping when the bases are changed. Let L : X → X be a linear map, where X is an n-dimensional vector space. Let X be given two bases, and . We already know how to get the matrix for L when we restrict our attention to , using it twice, once for the domain and once for the co-domain. Specifically, we form the matrix A whose columns are , where u_i are the vectors in . Having done so, we have, by Theorem 4,

In the same way, there is a matrix B such that for all x,

Finally, another matrix P enters as the transition matrix in Theorem 1. Thus, P does the following task:

for all x. Putting this all together, we have

Because can be any vector in , we conclude that B = PAP⁻¹. This is the relationship we sought to uncover. The result is the following theorem.

Similar Matrices

The relationship between A and B in Theorem 5 is called similarity. Thus, as often happens, a common English word receives a highly technical meaning in mathematics.

Similarity is an equivalence relation, as discussed in Section 5.2. Thus, if we express ‘‘A is similar to B’’ by A B, we have these properties: