We saw in Section 1.5 that if S is a generating set for a subspace W and no proper subset of S is a generating set for W, then S must be linearly independent. A linearly independent generating set for W possesses a very useful property—every vector in W can be expressed in one and only one way as a linear combination of the vectors in the set. (This property is proved below in Theorem 1.8.) It is this property that makes linearly independent generating sets the building blocks of vector spaces.
A basis β
Recalling that span (∅)={0}
In Fn
In Mm×n(F)
In Pn(F)
In P(F), the set {1, x, x2, …}
Observe that Example 5 shows that a basis need not be finite. In fact, later in this section it is shown that no basis for P(F) can be finite. Hence not every vector space has a finite basis.
The next theorem, which is used frequently in Chapter 2, establishes the most significant property of a basis.
Let V be a vector space and u1, u2, …, un
for unique scalars a1, a2, …, an
Proof. Let β
are two such representations of v. Subtracting the second equation from the first gives
Since β
The proof of the converse is an exercise.
Theorem 1.8 shows that if the vectors u1, u2, …, un
for appropriately chosen scalars a1, a2, …, an
In this book, we are primarily interested in vector spaces having finite bases. Theorem 1.9 identifies a large class of vector spaces of this type.
If a vector space V is generated by a finite set S, then some subset of S is a basis for V. Hence V has a finite basis.
Proof. If S=∅
(i) The set β=S
(ii) The set β
Because of the method by which the basis β
Let
It can be shown that S generates R3
Now we consider the set {(2,−3, 5), (1, 0,−2), (0, 2,−1)}
is linearly independent or linearly dependent. Because
we exclude (7, 2, 0) from our basis. We conclude that
is a subset of S that is a basis for R3
The corollaries of the following theorem are perhaps the most significant results in Chapter 1.
Let V be a vector space that is generated by a set G containing exactly n vectors, and let L be a linearly independent subset of V containing exactly m vectors. Then m≤n
Proof. The proof is by mathematical induction on m. The induction begins with m = 0
Now suppose that the theorem is true for some integer m≥0
Note that n−m>0
Let H={u2, …, un−m}
Because {v1, v2, …, vm, u1, u2, …, un−m}
Let V be a vector space having a finite basis. Then all bases for V are finite, and every basis for V contains the same number of vectors.
Proof. Suppose that β
If a vector space has a finite basis, Corollary 1 asserts that the number of vectors in any basis for V is an intrinsic property of V. This fact makes possible the following important definitions.
A vector space is called finite-dimensional if it has a basis consisting of a finite number of vectors. The unique integer n such that every basis for V contains exactly n elements is called the dimension of V and is denoted by dim(V). A vector space that is not finite-dimensional is called infinite-dimensional.
The following results are consequences of Examples 1 through 4.
The vector space {0} has dimension zero.
The vector space Fn
The vector space Mm×n(F)
The vector space Pn(F)
The following examples show that the dimension of a vector space depends on its field of scalars.
Over the field of complex numbers, the vector space of complex numbers has dimension 1. (A basis is {1} .)
Over the field of real numbers, the vector space of complex numbers has dimension 2. (A basis is {1, i}.)
In the terminology of dimension, the first conclusion in the replacement theorem states that if V is a finite-dimensional vector space, then no linearly independent subset of V can contain more than dim(V) vectors.
The vector space P(F) is infinite-dimensional because, by Example 5, it has an infinite linearly independent set, namely {1, x, x2, …}
In Example 13, the infinite linearly independent set {1, x, x2, …}
Just as no linearly independent subset of a finite-dimensional vector space V can contain more than dim(V) vectors, a corresponding statement can be made about the size of a generating set.
Let V be a vector space with dimension n.
(a) Any finite generating set for V contains at least n vectors, and a generating set for V that contains exactly n vectors is a basis for V.
(b) Any linearly independent subset of V that contains exactly n vectors is a basis for V.
(c) Every linearly independent subset of V can be extended to a basis for V, that is, if L is a linearly independent subset of V, then there is a basis β
Proof. Let β
(a) Let G be a finite generating set for V. By Theorem 1.9 some subset H of G is a basis for V. Corollary 1 implies that H contains exactly n vectors. Since a subset of G contains n vectors, G must contain at least n vectors. Moreover, if G contains exactly n vectors, then we must have H=G
(b) Let L be a linearly independent subset of V containing exactly n vectors. It follows from the replacement theorem that there is a subset H of β
(c) If L is a linearly independent subset of V containing m vectors, then the replacement theorem asserts that there is a subset H of β
It follows from Example 4 of Section 1.4 and (a) of Corollary 2 that
is a basis for P2(R)
It follows from Example 5 of Section 1.4 and (a) of Corollary 2 that
is a basis for M2×2(R)
It follows from Example 3 of Section 1.5 and (b) of Corollary 2 that
is a basis for R4
For k=0, 1, …, n
is a basis for Pn(F)
A procedure for reducing a generating set to a basis was illustrated in Example 6. In Section 3.4, when we have learned more about solving systems of linear equations, we will discover a simpler method for reducing a generating set to a basis. This procedure also can be used to extend a linearly independent set to a basis, as (c) of Corollary 2 asserts is possible.
Theorem 1.9 as well as the replacement theorem and its corollaries contain a wealth of information about the relationships among linearly independent sets, bases, and generating sets. For this reason, we summarize here the main results of this section in order to put them into better perspective.
A basis for a vector space V is a linearly independent subset of V that generates V. If V has a finite basis, then every basis for V contains the same number of vectors. This number is called the dimension of V, and V is said to be finite-dimensional. Thus if the dimension of V is n, every basis for V contains exactly n vectors. Moreover, every linearly independent subset of V contains no more than n vectors and can be extended to a basis for V by including appropriately chosen vectors. Also, each generating set for V contains at least n vectors and can be reduced to a basis for V by excluding appropriately chosen vectors. The Venn diagram in Figure 1.6 depicts these relationships.
Our next result relates the dimension of a subspace to the dimension of the vector space that contains it.
Let W be a subspace of a finite-dimensional vector space V. Then W is finite-dimensional and dim(W)≤dim(V)
Proof. Let dim(V)=n
If dim(W)=n
Let
It is easily shown that W is a subspace of F5
as a basis. Thus dim(W)=3
The set of diagonal n×n
where Eij
We saw in Section 1.3 that the set of symmetric n×n
where Aij
If W is a subspace of a finite-dimensional vector space V, then any basis for W can be extended to a basis for V.
Proof. Let S be a basis for W. Because S is a linearly independent subset of V, Corollary 2 of the replacement theorem guarantees that S can be extended to a basis for V.
The set of all polynomials of the form
where a18, a16, …, a2, a0∈F
We can apply Theorem 1.11 to determine the subspaces of R2
If a point of R2
Similarly, the subspaces of R3
In many applications, we have a collection of data obtained from experiments or samples. For example, we may know the locations of an airplane flying from New York to London at certain times and would like to be able to approximate the locations of the plane at one or more intermediate times. The process of estimating intermediate values of a variable from known values is called interpolation.
Corollary 2 of the replacement theorem (Theorem 1.10) can be applied to obtain a useful formula that enables us to approximate the values of an unknown function by a polynomial function. Let c0, c1, …, cn
are called the Lagrange polynomials (associated with c0, c1, …, cn
This property of Lagrange polynomials can be used to show that β={f0, f1, …, fn}
where 0 denotes the zero function. Then
But also
by (10). Hence aj=0
Because β
It follows that
so
is the unique representation of g as a linear combination of elements of β
is the unique polynomial in Pn(F)
and
Hence the desired polynomial is
An important consequence of the Lagrange interpolation formula is the following result: If f∈Pn(F)
Label the following statements as true or false.
(a) The zero vector space has no basis.
(b) Every vector space that is generated by a finite set has a basis.
(c) Every vector space has a finite basis.
(d) A vector space cannot have more than one basis.
(e) If a vector space has a finite basis, then the number of vectors in every basis is the same.
(f) The dimension of Pn(F)
(g) The dimension of Mm×n(F)
(h) Suppose that V is a finite-dimensional vector space, that S1
(i) If S generates the vector space V, then every vector in V can be written as a linear combination of vectors in S in only one way.
(j) Every subspace of a finite-dimensional space is finite-dimensional.
(k) If V is a vector space having dimension n, then V has exactly one subspace with dimension 0 and exactly one subspace with dimension n.
(l) If V is a vector space having dimension n, and if S is a subset of V with n vectors, then S is linearly independent if and only if S spans V.
Determine which of the following sets are bases for R3
(a) {(1, 0,−1), (2, 5, 1), (0,−4, 3)}
(b) {(2,−4, 1), (0, 3,−1), (6, 0,−1)}
(c) {(1, 2,−1), (1, 0, 2), (2, 1, 1)}
(d) {(−1, 3, 1), (2,−4,−3), (−3, 8, 2)}
(e) {(1,−3,−2), (−3, 1, 3), (−2,−10,−2)}
Determine which of the following sets are bases for P2(R)
(a) {−1−x+2x2, 2+x−2x2, 1−2x+4x2}
(b) {1+2x+x2, 3+x2, x+x2}
(c) {1−2x−2x2,−2+3x−x2, 1−x+6x2}
(d) {−1+2x+4x2, 3−4x−10x2,−2−5x−6x2}
(e) {1+2x−x2, 4−2x+x2,−1+18x−9x2}
Do the polynomials x3−2x2+1, 4x2−x+3
Is {(1, 4,−6), (1, 5, 8), (2, 1, 1), (0, 1, 0)}
Give three different bases for F2
The vectors u1=(2,−3, 1), u2=(1, 4,−2), u3=(−8, 12,−4), u4=(1, 37,−17)
Let W denote the subspace of R5
generate W. Find a subset of the set {u1, u2, ..., u8}
The vectors u1=(1, 1, 1, 1), u2=(0, 1, 1, 1), u3=(0, 0, 1, 1)
In each part, use the Lagrange interpolation formula to construct the polynomial of smallest degree whose graph contains the following points.
(a) (−2,−6), (−1, 5), (1, 3)
(b) (−4, 24), (1, 9), (3, 3)
(c) (−2, 3), (−1,−6), (1, 0), (3,−2)
(d) (−3,−30), (−2, 7), (0, 15), (1, 10)
Let u and v be distinct vectors of a vector space V. Show that if {u, v} is a basis for V and a and b are nonzero scalars, then both {u+v, au}
Let u, v, and w be distinct vectors of a vector space V. Show that if {u, v, w} is a basis for V, then {u+v+w, v+w, w}
The set of solutions to the system of linear equations
is a subspace of R3
Find bases for the following subspaces of F5
and
What are the dimensions of W1
The set of all n×n
The set of all upper triangular n×n
The set of all skew-symmetric n×n
Let V denote the vector space of all sequences in F, as defined in Example 5 of Section 1.2. Find a basis for the subspace W of V consisting of the sequences (an)
Complete the proof of Theorem 1.8.
† Let V be a vector space having dimension n, and let S be a subset of V that generates V.
(a) Prove that there is a subset of S that is a basis for V. (Be careful not to assume that S is finite.)
(b) Prove that S contains at least n vectors.
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Prove that a vector space is infinite-dimensional if and only if it contains an infinite linearly independent subset.
Let W1
Let v1, v2, …, vk, v
(a) Find necessary and sufficient conditions on v such that dim(W1)=dim(W2)
(b) State and prove a relationship involving dim(W1)
Let f(x) be a polynomial of degree n in Pn(R)
where f(n)(x)
Let V, W, and Z be as in Exercise 21 of Section 1.2. If V and W are vector spaces over F of dimensions m and n, determine the dimension of Z.
For a fixed a∈R
Let W1
Let V be a finite-dimensional vector space over C with dimension n. Prove that if V is now regarded as a vector space over R, then dim V=2n
Exercises 29–34 require knowledge of the sum and direct sum of subspaces, as defined in the exercises of Section 1.3.
(a) Prove that if W1
(b) Let W1
Let
and
Prove that W1
Let W1
(a) Prove that dim(W1∩W2)≤n
(b) Prove that dim(W1+W2)≤m+n
Find examples of subspaces W1
(a) dim(W1∩W2)=dim(W2);
(b) dim(W1+W2)=dim(W1)+dim(W2);
(c) dim(W1+W2)<dim(W1)+dim(W2).
(a) Let W1
(b) Conversely, let β1
(a) Prove that if W1
(b) Let V=R2
The following exercise requires familiarity with Exercise 31 of Section 1.3.
Let W be a subspace of a finite-dimensional vector space V, and consider the basis {u1, u2, ..., uk}
(a) Prove that {uk+1+W, uk+2+W, …, un+W}
(b) Derive a formula relating dim(V), dim(W), and dim(V/W).