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1.6 Bases and Dimension

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.

Definition.

A basis β $β$ for a vector space V is a linearly independent subset of V that generates V. If β $β$ is a basis for V, we also say that the vectors of β $β$ form a basis for V.

Example 1

Recalling that span (∅)={0} $span (\emptyset) = {0}$ and ∅ $\emptyset$ is linearly independent, we see that ∅ $\emptyset$ is a basis for the zero vector space.

Example 2

In Fn $F^{n}$ , let e1=(1, 0, 0, …, 0), e2=(0, 1, 0, …, 0), …,en=(0, 0, …, 0, 1); {e1, e2, …, en} $e_{1} = (1, 0, 0, \dots, 0), e_{2} = (0, 1, 0, \dots, 0), \dots, e_{n} = (0, 0, \dots, 0, 1); {e_{1}, e_{2}, \dots, e_{n}}$ is readily seen to be a basis for Fn $F^{n}$ and is called the standard basis for Fn $F^{n}$ .

Example 3

In Mm×n(F) $M_{m \times n} (F)$ , let Eij $E^{i j}$ denote the matrix whose only nonzero entry is a 1 in the ith row and jth column. Then {Eij: 1≤i≤m, 1≤j≤n} ${E^{i j} : 1 \leq i \leq m, 1 \leq j \leq n}$ is a basis for Mm×n(F) $M_{m \times n} (F)$ .

Example 4

In Pn(F) $P_{n} (F)$ , the set {1, x, x2, …, xn} ${1, x, x^{2}, \dots, x^{n}}$ is a basis. We call this basis the standard basis for Pn(F) $P_{n} (F)$ .

Example 5

In P(F), the set {1, x, x2, …} ${1, x, x^{2}, \dots}$ is a basis.

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.

Theorem 1.8.

Let V be a vector space and u1, u2, …, un $u_{1}, u_{2}, \dots, u_{n}$ be distinct vectors in V. Then β={u1, u2, …, un} $β = {u_{1}, u_{2}, \dots, u_{n}}$ is a basis for V if and only if each v∈V $v \in V$ can be uniquely expressed as a linear combination of vectors of β $β$ , that is, can be expressed in the form

v = a 1 u 1 + a 2 u 2 + \dots + a n u n

$v = a_{1} u_{1} + a_{2} u_{2} + \dots + a_{n} u_{n}$

for unique scalars a1, a2, …, an $a_{1}, a_{2}, \dots, a_{n}$ .

Proof. Let β $β$ be a basis for V. If v∈V $v \in V$ , then v∈span(β) $v \in span (β)$ because span(β)=V $span (β) = V$ . Thus v is a linear combination of the vectors of β $β$ . Suppose that

v = a 1 u 1 + a 2 u 2 + \dots + a n u n and v = b 1 u 1 + b 2 u 2 + \dots + b n u n

$v = a_{1} u_{1} + a_{2} u_{2} + \dots + a_{n} u_{n} and v = b_{1} u_{1} + b_{2} u_{2} + \dots + b_{n} u_{n}$

are two such representations of v. Subtracting the second equation from the first gives

0 = (a 1 - b 1) u 1 + (a 2 - b 2) u 2 + \dots + (a n - b n) u n .

$0 = (a_{1} - b_{1}) u_{1} + (a_{2} - b_{2}) u_{2} + \dots + (a_{n} - b_{n}) u_{n} .$

Since β $β$ is linearly independent, it follows that a1−b1=a2−b2=⋯=an−bn=0 $a_{1} - b_{1} = a_{2} - b_{2} = \dots = a_{n} - b_{n} = 0$ . Hence a1=b1, a2=b2, …, an=bn $a_{1} = b_{1}, a_{2} = b_{2}, \dots, a_{n} = b_{n}$ , and so v is uniquely expressible as a linear combination of the vectors of β $β$ .

The proof of the converse is an exercise.

Theorem 1.8 shows that if the vectors u1, u2, …, un $u_{1}, u_{2}, \dots, u_{n}$ form a basis for a vector space V, then every vector in V can be uniquely expressed in the form

v = a 1 u 1 + a 2 u 2 + \dots + a n u n

$v = a_{1} u_{1} + a_{2} u_{2} + \dots + a_{n} u_{n}$

for appropriately chosen scalars a1, a2, …, an $a_{1}, a_{2}, \dots, a_{n}$ . Thus v determines a unique n-tuple of scalars (a1, a2, …, an) $(a_{1}, a_{2}, \dots, a_{n})$ and, conversely, each n-tuple of scalars determines a unique vector v∈V $v \in V$ by using the entries of the n-tuple as the coefficients of a linear combination of u1, u2, …, un $u_{1}, u_{2}, \dots, u_{n}$ . This fact suggests that V is like the vector space Fn $F^{n}$ , where n is the number of vectors in the basis for V. We see in Section 2.4 that this is indeed the case.

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.

Theorem 1.9.

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=∅ $S = \emptyset$ or S={0} $S = {0}$ , then V={0} $V = {0}$ and ∅ $\emptyset$ is a subset of S that is a basis for V. Otherwise S contains a nonzero vector u1 $u_{1}$ . By item 2 on page 38, {u1} ${u_{1}}$ is a linearly independent set. Continue, if possible, choosing vectors u2, …,uk $u_{2}, \dots, u_{k}$ in S such that {u1, u2, …, uk} ${u_{1}, u_{2}, \dots, u_{k}}$ is a linearly independent set of k vectors. Since S is a finite set, this process must end with a linearly independent set β={u1, u2, …,un} $β = {u_{1}, u_{2}, \dots, u_{n}}$ . There are two ways this could happen.

(i) The set β=S $β = S$ . In this case, S is both a linearly independent set and a generating set for V. That is, S is itself a basis for V.
(ii) The set β $β$ is a proper linearly independent subset of S such that adjoining to β $β$ any vector in S not in β $β$ produces a linearly dependent set. In this case, we claim that β $β$ is the desired subset of S that is a basis for V. Because β $β$ is linearly independent by construction, it suffices to show that β $β$ spans V. By Theorem 1.5 (p. 31), we need to show that S⊆span(β) $S \subseteq span (β)$ . Let v∈S $v \in S$ . If v∈β $v \in β$ , then clearly v∈span(β) $v \in span (β)$ . Otherwise, if v ∈ β $v \in β$ , then the preceding construction shows that β∪{v} $β \cup {v}$ is linearly dependent. So v∈span(β) $v \in span (β)$ by Theorem 1.7 (p. 40). Thus S⊆span(β) $S \subseteq span (β)$ , completing the proof.

Because of the method by which the basis β $β$ was obtained in the proof of Theorem 1.9, this theorem is often remembered as saying that a finite spanning set for V can be reduced to a basis for V. This method is illustrated in the next example.

Example 6

Let

S = {(2, - 3, 5), (8, - 12, 20), (1, 0, - 2), (0, 2, - 1), (7, 2, 0)} .

$S = {(2, - 3, 5), (8, - 12, 20), (1, 0, - 2), (0, 2, - 1), (7, 2, 0)} .$

It can be shown that S generates R3 $R^{3}$ . We can select a basis for R3 $R^{3}$ that is a subset of S by the technique used in proving Theorem 1.9. To start, select any nonzero vector in S, say (2,−3, 5) $(2, - 3, 5)$ , to be a vector in the basis. Since 4(2,−3, 5) = (8,−12, 20) $4 (2, - 3, 5) = (8, - 12, 20)$ , the set {(2,−3, 5), (8,−12, 20)} ${(2, - 3, 5), (8, - 12, 20)}$ is linearly dependent by Exercise 9 of Section 1.5. Hence we do not include (8,−12, 20) $(8, - 12, 20)$ in our basis. On the other hand, (1, 0,−2) $(1, 0, - 2)$ is not a multiple of (2,−3, 5) $(2, - 3, 5)$ and vice versa, so that the set {(2,−3, 5), (1, 0,−2)} ${(2, - 3, 5), (1, 0, - 2)}$ is linearly independent. Thus we include (1, 0,−2) $(1, 0, - 2)$ as part of our basis.

Now we consider the set {(2,−3, 5), (1, 0,−2), (0, 2,−1)} ${(2, - 3, 5), (1, 0, - 2), (0, 2, - 1)}$ obtained by adjoining another vector in S to the two vectors that we have already included in our basis. As before, we include (0, 2,−1) $(0, 2, - 1)$ in our basis or exclude it from the basis according to whether {(2,−3, 5), (1, 0,−2), (0, 2,−1)} ${(2, - 3, 5), (1, 0, - 2), (0, 2, - 1)}$ is linearly independent or linearly dependent. An easy calculation shows that this set is linearly independent, and so we include (0, 2,−1) $(0, 2, - 1)$ in our basis. In a similar fashion the final vector in S is included or excluded from our basis according to whether the set

{(2, - 3, 5), (1, 0, - 2), (0, 2, - 1), (7, 2, 0)}

${(2, - 3, 5), (1, 0, - 2), (0, 2, - 1), (7, 2, 0)}$

is linearly independent or linearly dependent. Because

2 (2, - 3, 5) + 3 (1, 0, - 2) + 4 (0, 2, - 1) - (7, 2, 0) = (0, 0, 0),

$2 (2, - 3, 5) + 3 (1, 0, - 2) + 4 (0, 2, - 1) - (7, 2, 0) = (0, 0, 0),$

we exclude (7, 2, 0) from our basis. We conclude that

{(2, - 3, 5), (1, 0, - 2), (0, 2, - 1)}

${(2, - 3, 5), (1, 0, - 2), (0, 2, - 1)}$

is a subset of S that is a basis for R3 $R^{3}$ .

The corollaries of the following theorem are perhaps the most significant results in Chapter 1.

Theorem 1.10. (Replacement Theorem)

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 $m \leq n$ and there exists a subset H of G containing exactly n−m $n - m$ vectors such that L∪H $L \cup H$ generates V.

Proof. The proof is by mathematical induction on m. The induction begins with m = 0 $m = 0$ ; for in this case L = ∅ $L = \emptyset$ , and so taking H=G $H = G$ gives the desired result.

Now suppose that the theorem is true for some integer m≥0 $m \geq 0$ . We prove that the theorem is true for m+1 $m + 1$ . Let L={v1, v2, …,vm+1} $L = {v_{1}, v_{2}, \dots, v_{m + 1}}$ be a linearly independent subset of V consisting of m+1 $m + 1$ vectors. By the corollary to Theorem 1.6 (p. 39), {v1, v2, …vm} ${v_{1}, v_{2}, \dots v_{m}}$ is linearly independent, and so we may apply the induction hypothesis to conclude that m≤n $m \leq n$ and that there is a subset {u1, u2, …, un−m} ${u_{1}, u_{2}, \dots, u_{n - m}}$ of G such that {v1, v2, …, vm}∪{u1, u2, …, un−m} ${v_{1}, v_{2}, \dots, v_{m}} \cup {u_{1}, u_{2}, \dots, u_{n - m}}$ generates V. Thus there exist scalars a1, a2, …, am, b1, b2, …, bn−m $a_{1}, a_{2}, \dots, a_{m}, b_{1}, b_{2}, \dots, b_{n - m}$ such that

a 1 v 1 + a 2 v 2 + \dots + a m v m + b 1 u 1 + b 2 u 2 + \dots + b n - m u n - m = v m + 1 .

$a_{1} v_{1} + a_{2} v_{2} + \dots + a_{m} v_{m} + b_{1} u_{1} + b_{2} u_{2} + \dots + b_{n - m} u_{n - m} = v_{m + 1} .$ (9)

Note that n−m>0 $n - m > 0$ , lest vm+1 $v_{m + 1}$ be a linear combination of v1, v2, …, vm $v_{1}, v_{2}, \dots, v_{m}$ , which by Theorem 1.7 (p. 40) contradicts the assumption that L is linearly independent. Hence n>m $n > m$ ; that is, n≥m+1 $n \geq m + 1$ . Moreover, some bi $b_{i}$ , say b1 $b_{1}$ , is nonzero, for otherwise we obtain the same contradiction. Solving (9) for u1 $u_{1}$ gives

u 1 = (- b - 1 1 a 1) v 1 + (- b - 1 1 a 2) v 2 + \dots + (- b - 1 1 a m) v m + (b - 1 1) v m + 1 + (- b - 1 1 b 2) u 2 + \dots + (- b - 1 1 b n - m) u n - m .

$\begin{array}{l} u_{1} = (- b_{1}^{- 1} a_{1}) v_{1} + (- b_{1}^{- 1} a_{2}) v_{2} + \dots + (- b_{1}^{- 1} a_{m}) v_{m} + (b_{1}^{- 1}) v_{m + 1} \\ + (- b_{1}^{- 1} b_{2}) u_{2} + \dots + (- b_{1}^{- 1} b_{n - m}) u_{n - m} . \end{array}$

Let H={u2, …, un−m} $H = {u_{2}, \dots, u_{n - m}}$ . Then u1∈span(L∪H) $u_{1} \in span (L \cup H)$ , and because v1, v2, …, vm, u2, …, un−m $v_{1}, v_{2}, \dots, v_{m}, u_{2}, \dots, u_{n - m}$ are clearly in span(L∪H) $span (L \cup H)$ , it follows that

{v 1, v 2, \dots, v m, u 1, u 2, \dots, u n - m} \subseteq span (L \cup H) .

${v_{1}, v_{2}, \dots, v_{m}, u_{1}, u_{2}, \dots, u_{n - m}} \subseteq span (L \cup H) .$

Because {v1, v2, …, vm, u1, u2, …, un−m} ${v_{1}, v_{2}, \dots, v_{m}, u_{1}, u_{2}, \dots, u_{n - m}}$ generates V, Theorem 1.5 (p. 31) implies that span(L∪H)=V $span (L \cup H) = V$ . Since H is a subset of G that contains (n−m)−1=n−(m+1) $(n - m) - 1 = n - (m + 1)$ vectors, the theorem is true for m+1 $m + 1$ . This completes the induction.

Corollary 1.

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 $β$ is a finite basis for V that contains exactly n vectors, and let γ $γ$ be any other basis for V. If γ $γ$ contains more than n vectors, then we can select a subset S of γ $γ$ containing exactly n+1 $n + 1$ vectors. Since S is linearly independent and β $β$ generates V, the replacement theorem implies that n+1≤n $n + 1 \leq n$ , a contradiction. Therefore γ $γ$ is finite, and the number m of vectors in γ $γ$ satisfies m≤n $m \leq n$ . Reversing the roles of β $β$ and γ $γ$ and arguing as above, we obtain n≤m $n \leq m$ . Hence m=n $m = n$ .

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.

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.

Example 7

The vector space {0} has dimension zero.

Example 8

The vector space Fn $F^{n}$ has dimension n.

Example 9

The vector space Mm×n(F) $M_{m \times n} (F)$ has dimension mn.

Example 10

The vector space Pn(F) $P_{n} (F)$ has dimension n+1 $n + 1$ .

The following examples show that the dimension of a vector space depends on its field of scalars.

Example 11

Over the field of complex numbers, the vector space of complex numbers has dimension 1. (A basis is {1} .)

Example 12

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.

Example 13

The vector space P(F) is infinite-dimensional because, by Example 5, it has an infinite linearly independent set, namely {1, x, x2, …} ${1, x, x^{2}, \dots}$ .

In Example 13, the infinite linearly independent set {1, x, x2, …} ${1, x, x^{2}, \dots}$ is, in fact, a basis for P(F). Yet nothing that we have proved in this section guarantees an infinite-dimensional vector space must have a basis. In Section 1.7 it is shown, however, that every vector space has a basis.

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.

Corollary 2.

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 $β$ of V such that L⊆β $L \subseteq β$ .

Proof. Let β $β$ be a basis for V.

(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 $H = G$ , so that G is a basis for V.
(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 $β$ containing n−n=0 $n - n = 0$ vectors such that L∪H $L \cup H$ generates V. Thus H=∅ $H = \emptyset$ , and L generates V. Since L is also linearly independent, L is a basis for V.
(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 $β$ containing exactly n−m $n - m$ vectors such that L∪H $L \cup H$ generates V. Now L∪H $L \cup H$ contains at most n vectors; therefore (a) implies that L∪H $L \cup H$ contains exactly n vectors and that L∪H $L \cup H$ is a basis for V.

Example 14

It follows from Example 4 of Section 1.4 and (a) of Corollary 2 that

{x 2 + 3 x - 2, 2 x 2 + 5 x - 3, - x 2 - 4 x + 4}

${x^{2} + 3 x - 2, 2 x^{2} + 5 x - 3, - x^{2} - 4 x + 4}$

is a basis for P2(R) $P_{2} (R)$ .

Example 15

It follows from Example 5 of Section 1.4 and (a) of Corollary 2 that

{(1110), (1011), (1101), (0111)}

${(\begin{array}{r} 1 & 1 \\ 1 & 0 \end{array}), (\begin{array}{r} 1 & 1 \\ 0 & 1 \end{array}), (\begin{array}{r} 1 & 0 \\ 1 & 1 \end{array}), (\begin{array}{r} 0 & 1 \\ 1 & 1 \end{array})}$

is a basis for M2×2(R) $M_{2 \times 2} (R)$ .

Example 16

It follows from Example 3 of Section 1.5 and (b) of Corollary 2 that

{(1, 0, 0, - 1), (0, 1, 0, - 1), (0, 0, 1, - 1), (0, 0, 0, 1)}

${(1, 0, 0, - 1), (0, 1, 0, - 1), (0, 0, 1, - 1), (0, 0, 0, 1)}$

is a basis for R4 $R^{4}$ .

Example 17

For k=0, 1, …, n $k = 0, 1, \dots, n$ , let pk(x)=xk+xk+1+⋯+xn $p_{k} (x) = x^{k} + x^{k + 1} + \dots + x^{n}$ . It follows from Example 4 of Section 1.5 and (b) of Corollary 2 that

{p 0 (x), p 1 (x), \dots, p n (x)}

${p_{0} (x), p_{1} (x), \dots, p_{n} (x)}$

is a basis for Pn(F) $P_{n} (F)$ .

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.

An Overview of Dimension and Its Consequences

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.

A Venn diagram with 2 circles that intersect. The inside left circle is labeled Linearly independent sets. Inside the intersection is the label Bases. The inside right circle is labeled Generating sets. — Figure 1.6

The Dimension of Subspaces

Our next result relates the dimension of a subspace to the dimension of the vector space that contains it.

Theorem 1.11.

Let W be a subspace of a finite-dimensional vector space V. Then W is finite-dimensional and dim(W)≤dim(V) $\dim (W) \leq \dim (V)$ . Moreover, if dim(W)=dim(V) $\dim (W) = \dim (V)$ , then V=W $V = W$ .

Proof. Let dim(V)=n $\dim (V) = n$ . If W={0} $W = {0}$ , then W is finite-dimensional and dim(W)=0≤n $\dim (W) = 0 \leq n$ . Otherwise, W contains a nonzero vector x1 $x_{1}$ ; so {x1} ${x_{1}}$ is a linearly independent set. Continue choosing vectors, x1, x2, …, xk $x_{1}, x_{2}, \dots, x_{k}$ in W such that {x1, x2, …, xk} ${x_{1}, x_{2}, \dots, x_{k}}$ is linearly independent. Since no linearly independent subset of V can contain more than n vectors, this process must stop at a stage where k≤n $k \leq n$ and {x1, x2, …, xk} ${x_{1}, x_{2}, \dots, x_{k}}$ is linearly independent but adjoining any other vector from W produces a linearly dependent set. Theorem 1.7 (p. 40) implies that {x1, x2, …, xk} ${x_{1}, x_{2}, \dots, x_{k}}$ generates W, and hence it is a basis for W. Therefore dim(W)=k≤n $\dim (W) = k \leq n$ .

If dim(W)=n $\dim (W) = n$ , then a basis for W is a linearly independent subset of V containing n vectors. But Corollary 2 of the replacement theorem implies that this basis for W is also a basis for V; so W=V $W = V$ .

Example 18

Let

W = {(a 1, a 2, a 3, a 4, a 5) \in F 5 : a 1 + a 3 + a 5 = 0, a 2 = a 4} .

$W = {(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}) \in F^{5} : a_{1} + a_{3} + a_{5} = 0, a_{2} = a_{4}} .$

It is easily shown that W is a subspace of F5 $F^{5}$ having

{(- 1, 0, 1, 0, 0), (- 1, 0, 0, 0, 1), (0, 1, 0, 1, 0)}

${(- 1, 0, 1, 0, 0), (- 1, 0, 0, 0, 1), (0, 1, 0, 1, 0)}$

as a basis. Thus dim(W)=3 $\dim (W) = 3$ .

Example 19

The set of diagonal n×n $n \times n$ matrices is a subspace W of Mn×n(F) $M_{n \times n} (F)$ (see Example 3 of Section 1.3). A basis for W is

{E 11, E 22, \dots, E n n},

${E^{11}, E^{22}, \dots, E^{n n}},$

where Eij $E^{i j}$ is the matrix in which the only nonzero entry is a 1 in the ith row and jth column. Thus dim(W)=n $\dim (W) = n$ .

Example 20

We saw in Section 1.3 that the set of symmetric n×n $n \times n$ matrices is a subspace W of Mn×n(F) $M_{n \times n} (F)$ . A basis for W is

{A i j : 1 \leq i \leq j \leq n},

${A^{i j} : 1 \leq i \leq j \leq n},$

where Aij $A^{i j}$ is the n×n $n \times n$ matrix having 1 in the ith row and jth column, 1 in the jth row and ith column, and 0 elsewhere. It follows that

dim (W) = n + (n - 1) + \dots + 1 = 1 2 n (n + 1) .

$\dim (W) = n + (n - 1) + \dots + 1 = \frac{1}{2} n (n + 1) .$

Corollary.

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.

Example 21

The set of all polynomials of the form

a 18 x 18 + a 16 x 16 + \dots + a 2 x 2 + a 0,

$a_{18} x^{18} + a_{16} x^{16} + \dots + a_{2} x^{2} + a_{0},$

where a18, a16, …, a2, a0∈F $a_{18}, a_{16}, \dots, a_{2}, a_{0} \in F$ , is a subspace W of P18(F) $P_{18} (F)$ . A basis for W is {1, x2, …, x16, x18} ${1, x^{2}, \dots, x^{16}, x^{18}}$ , which is a subset of the standard basis for P18(F) $P_{18} (F)$ .

We can apply Theorem 1.11 to determine the subspaces of R2 $R^{2}$ and R3 $R^{3}$ . Since R2 $R^{2}$ has dimension 2, subspaces of R2 $R^{2}$ can be of dimensions 0, 1, or 2 only. The only subspaces of dimension 0 or 2 are {0} and R2 $R^{2}$ , respectively. Any subspace of R2 $R^{2}$ having dimension 1 consists of all scalar multiples of some nonzero vector in R2 $R^{2}$ (Exercise 11 of Section 1.4).

If a point of R2 $R^{2}$ is identified in the natural way with a point in the Euclidean plane, then it is possible to describe the subspaces of R2 $R^{2}$ geometrically: A subspace of R2 $R^{2}$ having dimension 0 consists of the origin of the Euclidean plane, a subspace of R2 $R^{2}$ with dimension 1 consists of a line through the origin, and a subspace of R2 $R^{2}$ having dimension 2 is the entire Euclidean plane.

Similarly, the subspaces of R3 $R^{3}$ must have dimensions 0, 1, 2, or 3. Interpreting these possibilities geometrically, we see that a subspace of dimension zero must be the origin of Euclidean 3-space, a subspace of dimension 1 is a line through the origin, a subspace of dimension 2 is a plane through the origin, and a subspace of dimension 3 is Euclidean 3-space itself.

The Lagrange Interpolation Formula

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 $c_{0}, c_{1}, \dots, c_{n}$ be distinct scalars in an infinite field F. The polynomials f0(x), f1(x), …, fn(x) $f_{0} (x), f_{1} (x), \dots, f_{n} (x)$ defined by

f i (x) = ( x - c 0 ) \dots ( x - c i - 1 ) ( x - c i + 1 ) \dots ( x - c n ) ( c i - c 0 ) \dots ( c i - c i - 1 ) ( c i - c i + 1 ) \dots ( c i - c n ) = \prod k = 0 k \neq i n x - c k c i - c k

$f_{i} (x) = \frac{(x - c_{0}) \dots (x - c_{i - 1}) (x - c_{i + 1}) \dots (x - c_{n})}{(c_{i} - c_{0}) \dots (c_{i} - c_{i - 1}) (c_{i} - c_{i + 1}) \dots (c_{i} - c_{n})} = \prod_{\begin{array}{l} k = 0 \\ k \neq i \end{array}}^{n} \frac{x - c_{k}}{c_{i} - c_{k}}$

are called the Lagrange polynomials (associated with c0, c1, …, cn $c_{0}, c_{1}, \dots, c_{n}$ ). Note that each fi(x) $f_{i} (x)$ is a polynomial of degree n and hence is in Pn(F) $P_{n} (F)$ . If we regard fi(x) $f_{i} (x)$ as a polynomial function fi: F→F $f_{i} : F \to F$ , we see that

f i (c j) = {01 if i \neq j if i = j .

$f_{i} (c_{j}) = {\begin{array}{l} 0 & if i \neq j \\ 1 & if i = j . \end{array}$ (10)

This property of Lagrange polynomials can be used to show that β={f0, f1, …, fn} $β = {f_{0}, f_{1}, \dots, f_{n}}$ is a linearly independent subset of Pn(F) $P_{n} (F)$ . Suppose that

\sum i = 0 n a i f i = 0 for some scalars a 0, a 1, \dots, a n,

$\sum_{i = 0}^{n} a_{i} f_{i} = 0 for some scalars a_{0}, a_{1}, \dots, a_{n},$

where 0 denotes the zero function. Then

\sum i = 0 n a i f i (c j) = 0 for j = 0, 1, \dots, n .

$\sum_{i = 0}^{n} a_{i} f_{i} (c_{j}) = 0 for j = 0, 1, \dots, n .$

But also

\sum i = 0 n a i f i (c j) = a j

$\sum_{i = 0}^{n} a_{i} f_{i} (c_{j}) = a_{j}$

by (10). Hence aj=0 $a_{j} = 0$ for j=0, 1, ...., n; $j = 0, 1, ...., n;$ so β $β$ is linearly independent. Since the dimension of Pn(F) $P_{n} (F)$ is n+1 $n + 1$ , it follows from Corollary 2 of the replacement theorem that β $β$ is a basis for Pn(F) $P_{n} (F)$ .

Because β $β$ is a basis for Pn(F) $P_{n} (F)$ , every polynomial function g in Pn(F) $P_{n} (F)$ is a linear combination of polynomial functions of β $β$ , say,

g = \sum i = 0 n b i f i .

$g = \sum_{i = 0}^{n} b_{i} f_{i} .$

It follows that

g (c j) = \sum i = 0 n b i f i (c j) = b j;

$g (c_{j}) = \sum_{i = 0}^{n} b_{i} f_{i} (c_{j}) = b_{j};$

g = \sum i = 0 n g (c i) f i

$g = \sum_{i = 0}^{n} g (c_{i}) f_{i}$

is the unique representation of g as a linear combination of elements of β $β$ . This representation is called the Lagrange interpolation formula. Notice that the preceding argument shows that if b0, b1, ..., bn $b_{0}, b_{1}, ..., b_{n}$ are any n+1 $n + 1$ scalars in F (not necessarily distinct), then the polynomial function

g = \sum i = 0 n b i f i

$g = \sum_{i = 0}^{n} b_{i} f_{i}$

is the unique polynomial in Pn(F) $P_{n} (F)$ such that g(cj)=bj $g (c_{j}) = b_{j}$ . Thus we have found the unique polynomial of degree not exceeding n that has specified values bj $b_{j}$ at given points cj $c_{j}$ in its domain (j=0, 1, ..., n) $(j = 0, 1, ..., n)$ . For example, let us construct the real polynomial g of degree at most 2 whose graph contains the points (1, 8), (2, 5), and (3,−4) $(3, - 4)$ . (Thus, in the notation above, c0=1, c1=2, c2=3, b0=8, b1=5 $c_{0} = 1, c_{1} = 2, c_{2} = 3, b_{0} = 8, b_{1} = 5$ , and b2=−4 $b_{2} = - 4$ .) The Lagrange polynomials associated with c0, c1 $c_{0}, c_{1}$ , and c2 $c_{2}$ are

f 0 (x) = ( x - 2 ) ( x - 3 ) ( 1 - 2 ) ( 1 - 3 ) = 1 2 (x 2 - 5 x + 6), f 1 (x) = ( x - 1 ) ( x - 3 ) ( 2 - 1 ) ( 2 - 3 ) = - 1 (x 2 - 4 x + 3),

$\begin{array}{l} f_{0} (x) = \frac{(x - 2) (x - 3)}{(1 - 2) (1 - 3)} = \frac{1}{2} (x^{2} - 5 x + 6), \\ f_{1} (x) = \frac{(x - 1) (x - 3)}{(2 - 1) (2 - 3)} = - 1 (x^{2} - 4 x + 3), \end{array}$

and

f 2 (x) = ( x - 1 ) ( x - 2 ) ( 3 - 1 ) ( 3 - 2 ) = 1 2 (x 2 - 3 x + 2) .

$f_{2} (x) = \frac{(x - 1) (x - 2)}{(3 - 1) (3 - 2)} = \frac{1}{2} (x^{2} - 3 x + 2) .$

Hence the desired polynomial is

g (x) = = = \sum i = 0 2 b i f i (x) = 8 f 0 (x) + 5 f 1 (x) - 4 f 2 (x) 4 (x 2 - 5 x + 6) - 5 (x 2 - 4 x + 3) - 2 (x 2 - 3 x + 2) - 3 x 2 + 6 x + 5.

$\begin{array}{rcl} g (x) & = & \sum_{i = 0}^{2} b_{i} f_{i} (x) = 8 f_{0} (x) + 5 f_{1} (x) - 4 f_{2} (x) \\ = & 4 (x^{2} - 5 x + 6) - 5 (x^{2} - 4 x + 3) - 2 (x^{2} - 3 x + 2) \\ = & - 3 x^{2} + 6 x + 5. \end{array}$

An important consequence of the Lagrange interpolation formula is the following result: If f∈Pn(F) $f \in P_{n} (F)$ and f(ci)=0 $f (c_{i}) = 0$ for n+1 $n + 1$ distinct scalars c0, c1, ..., cn $c_{0}, c_{1}, ..., c_{n}$ in F, then f is the zero function.

Exercises

Label the following statements as true or false.
1. (a) The zero vector space has no basis.
2. (b) Every vector space that is generated by a finite set has a basis.
3. (c) Every vector space has a finite basis.
4. (d) A vector space cannot have more than one basis.
5. (e) If a vector space has a finite basis, then the number of vectors in every basis is the same.
6. (f) The dimension of Pn(F) $P_{n} (F)$ is n.
7. (g) The dimension of Mm×n(F) $M_{m \times n} (F)$ is m+n $m + n$ .
8. (h) Suppose that V is a finite-dimensional vector space, that S1 $S_{1}$ is a linearly independent subset of V, and that S2 $S_{2}$ is a subset of V that generates V. Then S1 $S_{1}$ cannot contain more vectors than S2 $S_{2}$ .
9. (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.
10. (j) Every subspace of a finite-dimensional space is finite-dimensional.
11. (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.
12. (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 $R^{3}$ .
1. (a) {(1, 0,−1), (2, 5, 1), (0,−4, 3)} ${(1, 0, - 1), (2, 5, 1), (0, - 4, 3)}$
2. (b) {(2,−4, 1), (0, 3,−1), (6, 0,−1)} ${(2, - 4, 1), (0, 3, - 1), (6, 0, - 1)}$
3. (c) {(1, 2,−1), (1, 0, 2), (2, 1, 1)} ${(1, 2, - 1), (1, 0, 2), (2, 1, 1)}$
4. (d) {(−1, 3, 1), (2,−4,−3), (−3, 8, 2)} ${(- 1, 3, 1), (2, - 4, - 3), (- 3, 8, 2)}$
5. (e) {(1,−3,−2), (−3, 1, 3), (−2,−10,−2)} ${(1, - 3, - 2), (- 3, 1, 3), (- 2, - 10, - 2)}$
Determine which of the following sets are bases for P2(R) $P_{2} (R)$ .
1. (a) {−1−x+2x2, 2+x−2x2, 1−2x+4x2} ${- 1 - x + 2 x^{2}, 2 + x - 2 x^{2}, 1 - 2 x + 4 x^{2}}$
2. (b) {1+2x+x2, 3+x2, x+x2} ${1 + 2 x + x^{2}, 3 + x^{2}, x + x^{2}}$
3. (c) {1−2x−2x2,−2+3x−x2, 1−x+6x2} ${1 - 2 x - 2 x^{2}, - 2 + 3 x - x^{2}, 1 - x + 6 x^{2}}$
4. (d) {−1+2x+4x2, 3−4x−10x2,−2−5x−6x2} ${- 1 + 2 x + 4 x^{2}, 3 - 4 x - 10 x^{2}, - 2 - 5 x - 6 x^{2}}$
5. (e) {1+2x−x2, 4−2x+x2,−1+18x−9x2} ${1 + 2 x - x^{2}, 4 - 2 x + x^{2}, - 1 + 18 x - 9 x^{2}}$
Do the polynomials x3−2x2+1, 4x2−x+3 $x^{3} - 2 x^{2} + 1, 4 x^{2} - x + 3$ , and 3x−2 $3 x - 2$ generate P3(R) $P_{3} (R)$ ? Justify your answer.
Is {(1, 4,−6), (1, 5, 8), (2, 1, 1), (0, 1, 0)} ${(1, 4, - 6), (1, 5, 8), (2, 1, 1), (0, 1, 0)}$ a linearly independent subset of R3 $R^{3}$ ? Justify your answer.
Give three different bases for F2 $F^{2}$ and for M2×2(F) $M_{2 \times 2} (F)$ .
The vectors u1=(2,−3, 1), u2=(1, 4,−2), u3=(−8, 12,−4), u4=(1, 37,−17) $u_{1} = (2, - 3, 1), u_{2} = (1, 4, - 2), u_{3} = (- 8, 12, - 4), u_{4} = (1, 37, - 17)$ , and u5=(−3,−5, 8) $u_{5} = (- 3, - 5, 8)$ generate R3 $R^{3}$ . Find a subset of the set {u1, u2, u3, u4, u5} ${u_{1}, u_{2}, u_{3}, u_{4}, u_{5}}$ that is a basis for R3 $R^{3}$ .
Let W denote the subspace of R5 $R^{5}$ consisting of all the vectors having coordinates that sum to zero. The vectors

$u 1 = (2, - 3, 4, - 5, 2), u 2 = (- 6, 9, - 12, 15, - 6), u 3 = (3, - 2, 7, - 9, 1), u 4 = (2, - 8, 2, - 2, 6), u 5 = (- 1, 1, 2, 1, - 3), u 6 = (0, - 3, - 18, 9, 12), u 7 = (1, 0, - 2, 3, - 2), u 8 = (2, - 1, 1, - 9, 7)$ $\begin{array}{l} u_{1} = (2, - 3, 4, - 5, 2), u_{2} = (- 6, 9, - 12, 15, - 6), \\ u_{3} = (3, - 2, 7, - 9, 1), u_{4} = (2, - 8, 2, - 2, 6), \\ u_{5} = (- 1, 1, 2, 1, - 3), u_{6} = (0, - 3, - 18, 9, 12), \\ u_{7} = (1, 0, - 2, 3, - 2), u_{8} = (2, - 1, 1, - 9, 7) \end{array}$

generate W. Find a subset of the set {u1, u2, ..., u8} ${u_{1}, u_{2}, ..., u_{8}}$ that is a basis for W.
The vectors u1=(1, 1, 1, 1), u2=(0, 1, 1, 1), u3=(0, 0, 1, 1) $u_{1} = (1, 1, 1, 1), u_{2} = (0, 1, 1, 1), u_{3} = (0, 0, 1, 1)$ , and u4=(0, 0, 0, 1) $u_{4} = (0, 0, 0, 1)$ form a basis for F4 $F^{4}$ . Find the unique representation of an arbitrary vector (a1, a2, a3, a4) $(a_{1}, a_{2}, a_{3}, a_{4})$ in F4 $F^{4}$ as a linear combination of u1, u2, u3 $u_{1}, u_{2}, u_{3}$ , and u4 $u_{4}$ .
In each part, use the Lagrange interpolation formula to construct the polynomial of smallest degree whose graph contains the following points.
1. (a) (−2,−6), (−1, 5), (1, 3) $(- 2, - 6), (- 1, 5), (1, 3)$
2. (b) (−4, 24), (1, 9), (3, 3) $(- 4, 24), (1, 9), (3, 3)$
3. (c) (−2, 3), (−1,−6), (1, 0), (3,−2) $(- 2, 3), (- 1, - 6), (1, 0), (3, - 2)$
4. (d) (−3,−30), (−2, 7), (0, 15), (1, 10) $(- 3, - 30), (- 2, 7), (0, 15), (1, 10)$
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} ${u + v, a u}$ and {au, bv} ${a u, b v}$ are also bases for V.
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} ${u + v + w, v + w, w}$ is also a basis for V.
The set of solutions to the system of linear equations

$x 1 2 x 1 - - 2 x 2 3 x 2 + + x 3 x 3 = = 00$ $\begin{array}{rcrcrcl} x_{1} & - & 2 x_{2} & + & x_{3} & = & 0 \\ 2 x_{1} & - & 3 x_{2} & + & x_{3} & = & 0 \end{array}$

is a subspace of R3 $R^{3}$ . Find a basis for this subspace.
Find bases for the following subspaces of F5 $F^{5}$ :

$W 1 = {(a 1, a 2, a 3, a 4, a 5) \in F 5 : a 1 - a 3 - a 4 = 0}$ $W_{1} = {(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}) \in F^{5} : a_{1} - a_{3} - a_{4} = 0}$

and

$W 2 = {(a 1, a 2, a 3, a 4, a 5) \in F 5 : a 2 = a 3 = a 4 and a 1 + a 5 = 0} .$ $W_{2} = {(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}) \in F^{5} : a_{2} = a_{3} = a_{4} and a_{1} + a_{5} = 0} .$

What are the dimensions of W1 $W_{1}$ and W2 $W_{2}$ ?
The set of all n×n $n \times n$ matrices having trace equal to zero is a subspace W of Mn×n(F) $M_{n \times n} (F)$ (see Example 4 of Section 1.3). Find a basis for W. What is the dimension of W?
The set of all upper triangular n×n $n \times n$ matrices is a subspace W of Mn×n(F) $M_{n \times n} (F)$ (see Exercise 12 of Section 1.3). Find a basis for W. What is the dimension of W?
The set of all skew-symmetric n×n $n \times n$ matrices is a subspace W of Mn×n(F) $M_{n \times n} (F)$ (see Exercise 28 of Section 1.3). Find a basis for W. What is the dimension of W?
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) $(a_{n})$ that have only a finite number of nonzero terms an $a_{n}$ . Justify your answer.
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.
1. (a) Prove that there is a subset of S that is a basis for V. (Be careful not to assume that S is finite.)
2. (b) Prove that S contains at least n vectors.
  
  Visit goo.gl/wE2wwA for a solution.
Prove that a vector space is infinite-dimensional if and only if it contains an infinite linearly independent subset.
Let W1 $W_{1}$ and W2 $W_{2}$ be subspaces of a finite-dimensional vector space V. Determine necessary and sufficient conditions on W1 $W_{1}$ and W2 $W_{2}$ so that dim(W1∩W2)=dim(W1) $\dim (W_{1} \cap W_{2}) = \dim (W_{1})$ .
Let v1, v2, …, vk, v $v_{1}, v_{2}, \dots, v_{k}, v$ be vectors in a vector space V, and define W1=span({v1, v2, …, vk}) $W_{1} = span ({v_{1}, v_{2}, \dots, v_{k}})$ , and W2=span({v1, v2, …, vk, v}) $W_{2} = span ({v_{1}, v_{2}, \dots, v_{k}, v})$ .
1. (a) Find necessary and sufficient conditions on v such that dim(W1)=dim(W2) ${dim(W}_{1}) = \dim (W_{2})$ .
2. (b) State and prove a relationship involving dim(W1) ${dim(W}_{1})$ and dim(W2) $\dim (W_{2})$ in the case that dim(W1)≠dim(W2) $\dim (W_{1}) \neq \dim (W_{2})$ .
Let f(x) be a polynomial of degree n in Pn(R) $P_{n} (R)$ . Prove that for any g(x)∈Pn(R) $g (x) \in P_{n} (R)$ there exist scalars c0, c1, ...., cn $c_{0}, c_{1}, ...., c_{n}$ such that

$g (x) = c 0 f (x) + c 1 f' (x) + c 2 f'' (x) + \dots + c n f (n) (x),$ $g (x) = c_{0} f (x) + c_{1} f^{'} (x) + c_{2} f^{″} (x) + \dots + c_{n} f^{(n)} (x),$

where f(n)(x) $f^{(n)} (x)$ denotes the nth derivative of f(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 $a \in R$ , determine the dimension of the subspace of Pn(R) $P_{n} (R)$ defined by {f∈Pn(R): f(a)=0} ${f \in P_{n} (R) : f (a) = 0}$ .
Let W1 $W_{1}$ and W2 $W_{2}$ be the subspaces of P(F) defined in Exercise 25 in Section 1.3. Determine the dimensions of the subspaces W1∩Pn(F) $W_{1} \cap P_{n} (F)$ and W2∩Pn(F) $W_{2} \cap P_{n} (F)$ .
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 $\dim V = 2 n$ . (See Examples 11 and 12.)

Exercises 29–34 require knowledge of the sum and direct sum of subspaces, as defined in the exercises of Section 1.3.

1. (a) Prove that if W1 $W_{1}$ and W2 $W_{2}$ are finite-dimensional subspaces of a vector space V, then the subspace W1+W2 $W_{1} + W_{2}$ is finite-dimensional, and dim(W1+W2)=dim(W1)+dim(W2)−dim(W1∩W2) $\dim (W_{1} + W_{2}) = \dim (W_{1}) + \dim (W_{2}) - \dim (W_{1} \cap W_{2})$ . Hint: Start with a basis {u1, u2, ..., uk} ${u_{1}, u_{2}, ..., u_{k}}$ for W1∩W2 $W_{1} \cap W_{2}$ and extend this set to a basis {u1, u2, ..., uk, v1, v2, ..., vm} ${u_{1}, u_{2}, ..., u_{k}, v_{1}, v_{2}, ..., v_{m}}$ for W1 $W_{1}$ and to a basis {u1, u2, ..., uk, w1, w2, ..., wp} ${u_{1}, u_{2}, ..., u_{k}, w_{1}, w_{2}, ..., w_{p}}$ for W2 $W_{2}$ .
2. (b) Let W1 $W_{1}$ and W2 $W_{2}$ be finite-dimensional subspaces of a vector space V, and let V=W1+W2 $V = W_{1} + W_{2}$ . Deduce that V is the direct sum of W1 $W_{1}$ and W2 $W_{2}$ if and only if dim(V)=dim(W1)+dim(W2) $\dim (V) = \dim (W_{1}) + \dim (W_{2})$ .
Let

$V = M 2 \times 2 (F), W 1 = {(a c b a) \in V : a, b, c \in F},$ $V = M_{2 \times 2} (F), W_{1} = {(\begin{array}{r} a & b \\ c & a \end{array}) \in V : a, b, c \in F},$

and

$W 2 = {(0 - a a b) \in V : a, b \in F} .$ $W_{2} = {(\begin{array}{r} 0 & a \\ - a & b \end{array}) \in V : a, b \in F} .$

Prove that W1 $W_{1}$ and W2 $W_{2}$ are subspaces of V, and find the dimensions of W1, W2, W1+W2 $W_{1}, W_{2}, W_{1} + W_{2}$ , and W1∩W2 $W_{1} \cap W_{2}$ .
Let W1 $W_{1}$ and W2 $W_{2}$ be subspaces of a vector space V having dimensions m and n, respectively, where m≥n $m \geq n$ .
1. (a) Prove that dim(W1∩W2)≤n $\dim (W_{1} \cap W_{2}) \leq n$ .
2. (b) Prove that dim(W1+W2)≤m+n $\dim (W_{1} + W_{2}) \leq m + n$ .
Find examples of subspaces W1 $W_{1}$ and W2 $W_{2}$ of R3 $R^{3}$ such that dim(W1)>dim(W2)>0 $\dim (W_{1}) > \dim (W_{2}) > 0$ and
1. (a) dim(W1∩W2)=dim(W2); $\dim (W_{1} \cap W_{2}) = \dim (W_{2});$
2. (b) dim(W1+W2)=dim(W1)+dim(W2); $\dim (W_{1} + W_{2}) = \dim (W_{1}) + \dim (W_{2});$
3. (c) dim(W1+W2)<dim(W1)+dim(W2). $\dim (W_{1} + W_{2}) < \dim (W_{1}) + \dim (W_{2}) .$
1. (a) Let W1 $W_{1}$ and W2 $W_{2}$ be subspaces of a vector space V such that V=W1⊕W2 $V = W_{1} \oplus W_{2}$ . If β1 $β_{1}$ and β2 $β_{2}$ are bases for W1 $W_{1}$ and W2 $W_{2}$ , respectively, show that β1∩β2=∅ $β_{1} \cap β_{2} = \emptyset$ and β1∪β2 $β_{1} \cup β_{2}$ is a basis for V.
2. (b) Conversely, let β1 $β_{1}$ and β2 $β_{2}$ be disjoint bases for subspaces W1 $W_{1}$ and W2 $W_{2}$ , respectively, of a vector space V. Prove that if β1∪β2 $β_{1} \cup β_{2}$ is a basis for V, then V=W1⊕W2 $V = W_{1} \oplus W_{2}$ .
1. (a) Prove that if W1 $W_{1}$ is any subspace of a finite-dimensional vector space V, then there exists a subspace W2 $W_{2}$ of V such that V=W1⊕W2 $V = W_{1} \oplus W_{2}$ .
2. (b) Let V=R2 $V = R^{2}$ and W1={(a1, 0): a1∈R} $W_{1} = {(a_{1}, 0) : a_{1} \in R}$ . Give examples of two different subspaces W2 $W_{2}$ and W′2 ${W^{'}}_{2}$ such that V=W1⊕W2 $V = W_{1} \oplus W_{2}$ and V=W1⊕W′2 $V = W_{1} \oplus {W^{'}}_{2}$ .

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} ${u_{1}, u_{2}, ..., u_{k}}$ for W. Let {u1, u2, ..., uk, uk+1, ..., un} ${u_{1}, u_{2}, ..., u_{k}, u_{k + 1}, ..., u_{n}}$ be an extension of this basis to a basis for V.
1. (a) Prove that {uk+1+W, uk+2+W, …, un+W} ${u_{k + 1} + W, u_{k + 2} + W, \dots, u_{n} + W}$ is a basis for V/W.
2. (b) Derive a formula relating dim(V), dim(W), and dim(V/W).

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Table of Contents for 1.6 Bases and Dimension

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Table of Contents for
1.6 Bases and Dimension