Chapter 2: Linear Independence and Bases

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1 MATH20300: Linear Algebra 2 (2016 Chapter 2: Linear Independence and Bases 1 Linear Combinations and Spans Example 11 Consider the vector v (1, 1 R 2 What is the smallest subspace of (the real vector space R 2 to which v belongs? If W is any subspace of R 2 which contains v then, since W is closed under scalar multiplication, W must contain all scalar multiples a v, any a R; ie W {av a R} We also require W to be closed under addition Note, however, that the sum of any two scalar multiples of a given vector v is again a scalar multiple of v by Axiom 7: Thus the set a 1 v + a 2 v (a 1 + a 2 v {av a R} {a (1, 1 a R} {(a, a a R} is nonempty (it contains v 1 v, closed under scalar multiplication and closed under addition It is a subspace of R 2 containing v By the arguments above it is the smallest subspace of R 2 to which v belongs, in the sense that it is contained in any other subspace of R 2 of which v is an element The subspace {av a R} of R 2 is denoted span{v} (or span R {v} and is called the (linear span of the vector v Note that, geometrically, span{v} L is just the line through 0 which contains v (in this case, the particular line is given by the equation x + y 0 More generally, let V be a vector space over the field K and let v be any vector in v Then span{v} span K {v} : {a v a K} This is the span (over K of the vector v When v 0, we also refer to it as the line (through 0 spanned by v As you can easily verify, span{v} is a subspace of V containing v and contained in any other subspace of V which contains v Example 12 Let V C and v ι 1 V is a complex vector space Note that span C {v} span C {ι} {z ι z C} C (since if w is any complex number then w ( wι ι Thus v spans the whole vector space C 1

2 Example 13 On the other hand, V C is also a real vector space Note that span R {ι} {x ι x R} is precisely the imaginary axis in C (and not all of C The imaginary axis is a real subspace of C, but not a complex subspace of C Example 14 Consider the vectors v (1, 0, 2 and w (2, 2, 0 in (the real vector space R 3 What is the smallest subspace containing both of these vectors? Let W be any subspace of V R 3 which contains v and w Then certainly, W contains the lines span{v} and span{w} But W also contains many other vectors: For example 2 v (2, 0, 4 span{v} W and 3 w (6, 6, 0 span{w} W, so that 2 v + 3 w (8, 6, 4 W since W is closed under addition (Note that (8, 0, 4 does not lie on either of the lines span{v} and span{w} More generally, W contains any vector of the form av + bw, for any scalars a, b R; ie W span R {v, w} span{v, w} : {av + bw a, b R} Note that, in this example, span{v, w} {av + bw a, b R} {(a + 2b, 2b, 2a R 3 a, b R 2 } is precisely the unique plane through 0 that contains v and w (An equation of this plane is 2x 2y z 0 - see below Thus span{v, w} is itself a subspace of R 3, and hence is the smallest subspace of R 3 containing v and w These examples generalize to any vector space V and any collection of vectors in V : Definition 15 (Linear Combination Let V be a vector space over a field K and let v 1,, v n be any elements of V Then any vector of the form a 1 v a n v n, where a 1,, a n K is said to be a linear combination of v 1,, v n Remark 16 Observe that any subspace W of V which contains all of the vectors v 1,, v n also contains all linear combinations of v 1,, v n (Why? Note 17 Note that the zero vector is trivially a linear combination of any collection v 1,, v n : 0 0 v v v n Definition 18 ((Linear span Let V be a vector space over a field K and let v 1,, v n be any elements of V Then the (linear span of v 1,, v n (over K is the set span K {v 1,, v n } span{v 1,, v 2 } which consists of all linear combinations of v 1,, v n ; ie span K {v 1,, v n } : {a 1 v a n v n a 1,, a n K} We also, for convenience, define span K { } to be the zero vector space {0} 2

3 Proposition 19 Let V be a vector space over a field K and let v 1,, v n be any elements of V Then span{v 1,, v n } is a subspace of V Furthermore, if W is any subspace of V such that v 1,, v n W then span{v 1,, v n } W Proof: Let S span{v 1,, v n } S is nonempty since 0 0 v v n S S is closed under scalar multiplication: Let v S and let b K v a 1 v a n v n for some a 1,, a n K Thus By definition, bv b(a 1 v a n v n (ba 1 v (ba n v n (using Axioms 6 and 8 Thus bv is a linear combination of v 1,, v n and hence lies in S S is closed under addition: Let v, w S Then v a 1 v a n v n, w b 1 v b n v n for some a 1,, a n, b 1,, b n K Thus v + w (a 1 v a n v n + (b 1 v b n v n (a 1 + b 1 v (a n + b n v n S as required (We used Axioms 2 and 7 here So we have shown that S is a subspace of V Now let W be any subspace of V containing v 1,, v n Let a 1,, a n be any elements of K Since W is closed under scalar multiplication W contains each of the elements a 1 v 1,, a n v n Since W is closed under addition, it follows (by a simple induction on n that W contains the element a 1 v a n v n Thus W contains all linear combinations of v 1,, v n ; ie W span{v 1,, v n } Remark 110 Observe that if v 1,, v m,, v n (1 m < n is a collection of vectors in a vector space V then span{v 1,, v m } span{v 1,, v n } since any linear combination of v 1,, v m is also a linear combination of v 1,, v n : a 1 v a m v m a 1 v a m v m + 0 v m v n In other words if S is any finite subset of V and if T S then span{t } span{s} Note that it follows that span{t } is a subspace of span{s} Problem 111 Let S and T be any two finite subsets of a vector space V Then span{s} and span{t } are subspaces of V Show that span{s T } span{s} + span{t } 3

4 Definition 112 (Spanning set Let V be a vector space We say that the vectors v 1,, v n is spanning set of V, or that these vectors span V, if span{v 1,, v n } V Equivalently, v 1,, v n span the vector space V if every vector in v can be expressed as a linear combination of v 1,, v n Example 113 The vectors e 1 (1, 0, 0, e 2 (0, 1, 0 e 3 (0, 0, 1 span the real vector space R 3 : Let (a, b, c be any vector in R 3 Then (a, b, c a (1, 0, 0 + b (0, 1, 0 + c (0, 0, 1 ae 1 + be 2 + ce 3 span{e 1, e 2, e 3 } Example 114 We will show that the vectors v (1, 2 and w (1, 1 span the vector space R 2 : Let (a, b R 2 We must show that there exist x, y R such that (a, b xv + yw (x + y, 2x y Equivalently, we must show that the system of equations x + y a 2x y b has a solution We easily find that is a solution x a + b 3, y 2a b 3 Example 115 The vectors v (1, 0, 2 and w (2, 2, 0 do not span R 3 since span{v, w} is contained in the plane 2x 2y z 0 (see Example 14 Example 116 Again, let v (1, 0, 2 and w (2, 2, 0 in R 3 Let H be the plane 2x 2y z 0 in R 3 So H is a subspace of R 3, hence a vector space in its own right Also v, w H In fact span{v, w} H, so that the vectors v, w span H: Let u (a, b, c H Then, from the defining equation of H a b c Thus u ( 1 2 c c + b, b, c ( 1 2, 0, 1 + b (1, 1, 0 c 2 (1, 0, 2 + b 2 (2, 2, 0 c 2 v + b 2 w which lies in span{v, w} So H span{v, w} as claimed 4

5 Example 117 Let D be the real vector space of all differentiable functions R R Then cos x, sin x D So span{cos x, sin x} {f D f(x a cos x + b sin x, a, b R} sin x and cos x do not span the vector space D: For example, the identity function x does not lie in span{cos x, sin x} There are many ways to see this For example, suppose that there were scalars a, b such that x a cos x + b sin x for all x R Differentiating both sides of this equation twice gives the equation 0 a cos x b sin x for all x R Letting x 0 and x π/2 in this equation forces a 0 b Returning to the previous equation, we deduce the false statement: x 0 for all x Thus, no such a, b exist [Exercise: Find a different way of showing that x span{cos x, sin x}] Example 118 Let W be the vector space of solutions of the differential equation y + y 0 Then cos x, sin x W You learn in your calculus classes that every solution of this differential equation is of the form a cos x + b sin x for some a, b R In other words, W span{cos x, sin x} cos x and sin x span the vector space W 2 Linear Independence Definition 21 Let V be a vector space over K Let v 1,, v n V We say that the vectors v 1,, v n are linearly dependent if there exist a 1,, a n K, not all 0, such that 0 a 1 v a n v n Such an equation is then called a dependence relation for v 1,, v n If v 1,, v n are not linearly dependent, they are said to be linearly independent Example 22 The vectors (1, 0, 0 and (0, 1, 0 in R 3 are linearly independent since the equation (0, 0, 0 a 1 (1, 0, 0 + a 2 (0, 1, 0 (a 1, a 2, 0 forces a 1 a 2 0 Example 23 The three vectors v 1 (1, 0, 1, v 2 (3, 1, 1 v 3 (5, 1, 1 in R 3 are linearly dependent since v 3 2v 1 + v 2 and hence we have the dependence relation 2v 1 + v 2 + ( 1 v 3 0 5

6 Example 24 Let V be a vector space The set {0}, consisting of the zero vector, is linearly dependent since it satisfies the dependence relation (Here a Example 25 Let V be the real vector space R Then the set {1, 2 is linearly dependent: it satisfies the dependence relation ( Example 26 Let V be vector space R considered as a vector space over Q Then the set {1, 2 is linearly independent To see this: Suppose, for the sake of contradiction, it satisfies a dependence relation a 1 + b 2 0 with a, b Q, (a, b not both 0 Then clearly, b 0 (otherwise a 0 and we deduce 2 a b Q, a contradiction Example 27 In the 1882 the mathematician Ferdinand von Lindemann proved that the real number π is not the root of any polynomial whose coefficients are rational; ie π is a transcendental number (Note, however, that π is a root of many polynomials with real coefficients Give an example We deduce: For any n 1, the set {1, π,, π n } R is linearly independent over Q For suppose that there were a nontrivial dependence relation: a a 1 π + + a n π n 0 where a 0,, a n Q Then π would be a root of the polynomial contradicting Lindemann s theorem p(x a 0 + a 1 x + + a n x n Q[x] Example 28 If V is a vector space and if v V is any nonzero vector, then the set {v} is linearly independent For if a 0 then av 0 since ( 1 (av 1 v v 0 a Example 29 Let V be a vector space Let u, v V Then the set {u, v} is linearly dependent if and only if either u av for some scalar a or v bu for some scalar b: For suppose that u, v are linearly dependent Then they satisfy a dependence relation a 1 u + a 2 v 0 where not both a 1, a 2 are zero Suppose that a 2 0 for example Then ( a 2 v a 1 u v a 1 u bu a 2 6

7 Example 210 cos x and sin x are linearly independent elements of the vector space C, or of W, since certainly sin x a cos x and cos x b sin x (Prove this Since linear independence is such a central notion in linear algebra, it is useful to have several charaterizations Proposition 211 Let V be a vector space over a field K Let v 1,, v n V The following statements are equivalent (each logically implies the others: 1 v 1,, v n are linearly independent 2 Whenever a 1,, a n K satisfy then a 1 a 2 a n 0 a 1 v a n v n 0 3 Every vector in span{v 1,, v n } has a unique representation as a linear combination of v 1,, v n : ie whenever a 1,, a n, b 1,, b n K satisfy then a 1 b 1, a 2 b 2,, a n b n a 1 v a n v n b 1 v b n v n Proof: (1 (2: This follows from the definition of linear dependence Statement (2 simply says that v 1,, v n satisfy no non-trivial dependence relation (3 (2 Suppose that (3 holds and that a 1 v a n v n 0 We must show that all the a i are 0 We have a 1 v a n v n 0 v v n Since (3 holds, it follows that a 1 b 1 0, a 2 0,, a n 0 (2 (3 Suppose (2 holds and that This equation can be re-written as and hence as a 1 v a n v n b 1 v b n v n a 1 v a n v n + ( 1 (b 1 v b n v n 0 (a 1 b 1 v 1 + (a n b n v n 0 By (2, it follows that 0 a 1 b 1 a 2 b 2 a n b n, and hence a 1 b 1, a 2 b 2,, a n b n, as required Proposition 212 Let V be a vector space over a field K Let S {v 1,, v m } be a linearly independent subset of V Let v V, v S Then v span{s} if and only if {v 1,, v m, v} is a linearly dependent set 7

8 Proof: Suppose first that v 1,, v m, v is linearly dependent By definition, there is a dependence relation a 1 v a m v m + av 0 (1 with not all of a 1,, a m, a equal to 0 Now if a 0 in (1 then a 1 v a m v m 0 with not all a i 0, contradicting linear independence of the v i Thus a 0 in (1 Then we have and thus as required av ( a 1 v ( a m v m v ( a 1 /av ( a m /av m span{s} Conversely, suppose that v span{v 1,, v m } Then there are scalars b 1,, b m such that v b 1 v b m v m Therefore we have the dependence relation 0 b 1 v b m v m + ( 1v m 3 Bases and Dimension Example 31 Recall that the real vector space R 3 is spanned by the vectors e 1 (1, 0, 0, e 2 (0, 1, 0, e 3 (0, 0, 1 (These are of course the unit vectors in the direction of each of the three axes Indeed, (a, b, c ae 1 + be 2 + ce 3 for any (a, b, c R 3 Clearly this expression of a given vector as a linear combination of e 1, e 2, e 3 is unique, since each vector in R 3 is determined by its components It follows from Proposition 211 (3 that the vectors e 1, e 2, e 3 are also linearly independent Example 32 This last example generalizes greatly: Let V be the vector space K n over the field K Let e i (0,, 0, 1, 0, 0 1 in the i-th position i 1,, n Then every vector v (a 1,, a n can be expressed in one and only one way as a linear combination of e 1,, e n : (a 1,, a n a 1 e a n e n n a i e i i1 It follows that the set {e 1,, e n } is linearly independent and spans K n 8

9 Definition 33 (Basis of a vector space Let V be a vector space over a field K A subset B {v 1,, v n } of V is said to be a basis (over K if 1 B spans V and 2 B is linearly independent (over K Remark 34 Note that to say B spans V is equivalent to saying that every vector in v can be expressed as a linear combination of the vectors in B To then say that B is linearly independent is to say that each such expression is unique; for a given v there are not two different linear combinations both equal to v To summarize: B is a basis for the vector space V if and only if each vector in V can be expressed in one and only one way as a linear combination of the vectors in B Remark 35 This definition also makes sense if B is an infinite set spaces have infinite bases, as we will see Many vector Example 36 The set {e 1,, e n }, as in Example 32, is a basis of K n As we will see, K n has many different bases The basis E {e 1,, e n } is called the standard basis of K n Example 37 The set {sin x, cos x} is a basis of the vector space W of solutions of the differential equation y + y 0 Example 38 Let v (1, 0, 2, w (2, 2, 0 R 3 Recall that v and w lie in the subspace H defined by the equation 2x 2y z 0 Then B {v, w} is a basis of the vector space H: We have already seen that B spans H (Example 116 Furthermore B is linearly independent since neither one of v, w is a scalar multiple of the other Example 39 is a basis of the zero vector space V {0} Example 310 We will show that the three vectors u (1, 1, 1, v (1, 0, 2, w (2, 2, 0 are a basis of R 3 : We begin by showing that they span R 3 : A typical linear combination of u, v, w has the form xu + yv + zw x (1, 1, 1 + y (1, 0, 2 + z (2, 2, 0 (x + y + 2z, x + 2z, x + 2y Thus, to show that (a, b, c R 3 is a linear combination of u, v, w, we must show that the system x + y + 2z a x + +2z b x + 2y c 9

10 is always solvable Now write this system in the form x a A y b z c where A is the 3 3 matrix A Note that A is the 3 3 matrix whose columns are the vectors u, v, w Now det A 2 0 So the matrix A has an inverse A 1 and the system is solvable: x a y A 1 b z c Finally, note that since A 1 is unique, the system has a unique solution (for any given a, b, c In other words, any vector (a, b, c R 3 can be expressed in one and only one way as a linear combination of u, v, w This says precisely that u, v, w are linearly independent Since u, v, w span R 3 and are linearly independent, by definition they form a basis of R 3 The methods of Example 310 generalise greatly: Proposition 311 Let V be the vector space K n over the field K Let v 1,, v n V (It is crucial that n is also the size of the vector space in this proposition We define the scalars a ij by v 1 (a 11,, a n1, v 2 (a 12,, a n2, v j (a 1j,, a nj, v n (a 1n,, a nn, Let A M n n (K be the n n matrix whose (i, j-entry is a ij (So A is the matrix whose j-th column is the the vector v j, written as a column Then B {v 1,, v n } is a basis of K n if and only if A is an invertible matrix; ie if and only if det A 0 in K 10

11 Proof: We begin by noting the following Let x 1,, x n be any elements of K Then x 1 v x n v n (a 11 x 1 + +a 1n x n,, a i1 x a in x n,, a n1 x a nn x n Thus if w (b 1,, b n K, the equation x 1 v x n v n w is equivalent to the system of equations a 11 x a 1n x n b 1 a i1 x a in x n b i a n1 x a nn x n b n This system, in turn, can be expressed as the matrix equation x 1 b 1 x 2 b 2 A x n b n w (2 By definition, B is a basis if and only if this system has a unique solution for every w V By first-year linear algebra, the system always has a unique solution if and only if the matrix A is invertible Remark 312 Of course, when A is invertible the unique solution of the system (2is x 1 b 1 x 2 b 2 x n A 1 This tells us how to express w (b 1,, b n as a linear combination of v 1,, v n Example 313 Consider the vectors v 1 (1, 2, v 2 (3, 4 in R 2 The associated matrix is ( 1 3 A 2 4 Since det A 2 0, B {v 1, v 2 } {(1, 2, (3, 4} is a basis for R 2 b n (3 11

12 Example 314 Consider the vectors u (1, 1, 1, v (1, 1, 1, w (1, 1, 0 Q 3 The associated matrix is A Since det A 2, B {u, v, w} is a basis of the Q-vector space Q 3 By the same computation, B is also a basis of the real vector space R 3 and of the complex vector space C 3 Example 315 Let B {u, v, w} be the basis of R 3 (or Q 3 in the last example Let s find how to express each of the vectors (1, 0, 0 and (1, 2, 1 as a linear combination of u, v and w: By our calculations above, to express a vector (a, b, c as a linear combination xu + yv + zv is equivalent to solving the system x a A y b z c where Thus we obtain Now A x y z A A 1 a b c If we let (a, b, c (1, 0, 0 this gives (x, y, z ( 1/2, 1/2, 1 So (1, 0, u v + w If we let (a, b, c (1, 2, 1 this gives (x, y, z (3/2, 1/2, 0 So (1, 2, u 1 2 v Example 316 Consider the vectors u (1, 1, 1, v (1, 1, 1, w (0, 3, 0 Q 3 The associated matrix is B Then det B 0, so {u, v, w } is not a basis of Q 3 In fact u, v, w are linearly dependent Find a dependence relation among them 12

13 Proposition 317 Let V be a nonzero vector space and let S V be a finite set with m elements If span{s} V then S contains a basis of V ; ie there is a subset B of S which is a basis of V Proof: Let B be a linearly independent subset of S of maximal size I Claim that S span{b}: Let s S If s B then certainly s span{b} Suppose s B Then B {s} is linearly dependent (by the maximality propoerty of B It follows that s span{b} by Proposition 212 This proves the Claim By the Claim, S span{b} and hence V span{s} span{b} So B is a linearly independent spanning subset of V ; ie it is a basis of V The next crucial technical result is often called The Exchange Lemma Proposition 318 (Exchange Lemma Let V be a vector space over the field K Let S be a spanning subset of V with m elements Let {w 1,, w k } be any linearly independent subset of V Then there exist k distinct vectors v 1,, v k S such that span{s k } V, where S k (S \ {v 1,, v k } {w 1,, w k } is the set obtained by exchanging w 1,, w k for v 1,, v k in S In particular, it follows that k m Proof: We will prove this by induction on k 1 If k 1, then w 1 0 and since w 1 span{s}, w 1 can be written as a linear combination of the vectors in S where not all coefficients are zero Thus there is a vector, v 1 say, in S whose coefficient a in this linear combination is nonzero We thus have w 1 av 1 + u where u span{s \ {v 1 }} It follows that v 1 1 a w 1 + ( 1 u span{s 1 } a where S 1 : (S \ {v 1 } {w 1 } Since v 1 span{s 1 } we have so that span{s 1 } V span{s 1 } span{s 1 {v 1 }} span{s} V Now suppose that k 2 and the result has been proven for 1,, k 1 So there exist v 1,, v k 1 S such that span{s k 1 } V where S k 1 (S \ {v 1,, v k 1 } {w 1,, w k 1 } Then 0 w k span{s k 1 } So w k can be written as a linear combination of the vectors in S k 1 where not all coefficients are zero Since w 1,, w k 1, w k are linearly independent, there must be a vector v k S k 1, not equal to any of w 1,, w k 1, whose coefficient, b say, in this linear combination is not zero As for the case k 1 we deduce that v k span{s k } where S k (S k 1 \ {v k } {w k } and that hence span{s k } V as required 13

14 Remark 319 Note that it follows that in any vector space V, if S is any spanning set and E is any (possibly completely unrelated linearly independent set then E S From Proposition 318 we immediately deduce the following theorem, of central importance in linear algebra Theorem 320 Let V be a vector space and let B 1, B 2 be (finite bases of V Then B 1 B 2 That is, any two bases of a vector space have the same number of elements Proof: Since B 1 spans and since B 2 is linearly independent, Proposition 318 tells us that B 2 B 1 Reversing the roles of B 1 and B 2 in this argument, we also have B 1 B 2 This theorem is the basis for the following definition(s: V is said to be finite- Definition 321 Let V be a vector space over a field K dimensional over K if V has a finite basis If V does not have a finite basis it is sais to be infinite-dimensional over K If V is a finite-dimensional vector space, then the dimension of V (over K is defined to be the number of elements in any basis The dimension of V is denoted dim K (V Remark 322 By Proposition 317, any vector space V which has a finite spanning set has a finite basis, and so is finite-dimensional Example 323 let K be any field For any n 1, the vector K n is finite-dimensional of dimension n, since the standard basis {e 1,, e n } has n elements In particular, taking n 1, we have that K is a one-dimensional vector space over itself Example 324 Let P be the plane 2x 2y z 0 in R 3 We have seen that {v (1, 0, 2, w (2, 2, 0} is a basis of P Thus dim R (P 2 since one (and hence all bases have two elements Example 325 Let W be the real vector space of solutions to the differential equation y + y 0 Then dim R (W 2 since {sin x, cos x} is a basis with two elements Example 326 For n 0, let R[x] n denote the set of real polynomials of degree at most n Since this set is nonempty and closed under addition and multiplication by scalars, it is a subspace of R[x] Thus R[x] n is a real vector space in its own right By definition, every element of R[x] n can be expressed in one and only one way as a linear combination a 0 + a 1 x + + a n x n of 1, x,, x n It follows that the set {1, x,, x n } is a basis of R[x] n Thus R[x] n is a finite-dimensional real vector space of dimension n

15 Example 327 Consider the vector space M m n (K For any 1 i m, 1 j n, let E i,j denote the m n matrix whose (i, j-entry is 1 and all of whose other entries are 0 Let A M m n (K have (i, j-entry a i,j Then A i,j a i,je i,j Thus every element of M m n (K can be expressed uniquely as a linear combination of the E i,j It follows that E {E i,j } m,n i1,j1 is a basis of M m n(k called the standard basis and that dim(m m n (K E mn Example 328 Let V be the real vector space R[x] of all real polynomial functions a 0 + a 1 x + + a d x d Observe that any n 1 the set {1, x,, x n } is linearly independent over R; for a linear combination of 1,, x n is just a polynomial a a 1 x + + a n x n of degree at most n If such an function is 0 (that is, equal to the 0 function then necessarily a 0 a 1 a n Thus {1, x,, x n } is linearly independent by Proposition 211 It follows from Proposition 318 that the vector space R[x] has no finite spanning set In particular, R[x] has no finite basis and thus is an infinite-dimensional vector space over R Example 329 Consider R as a vector space over Q We have already seen in Example 27 that R contains linearly independent sets of any size we want As in the last example, it follows from Proposition 318 that R does not have a finite basis as a vector space over Q So R is an infinite-dimensional vector space over Q (Of course, R is a 1-dimensional vector space over R Example 330 Let C be the real vector space of all real continuous functions Let D be the subspace of all real differentiable functions Then both of these spaces contain R[x] as a subspace Since R[x] contains linearly independent sets of any given size, so do the spaces D and C Thus the real vector spaces D and C are infinite-dimensional Here is another important consequence of Proposition 318: Proposition 331 Let V be a finite-dimensional vector space over a field K Let S be any linearly independent subset of V Then there exists a basis B of V with S B That is, any linearly independent subset of V can be extended to a basis Proof: Let E be any basis of V Since E spans V and S is linearly independent, Proposition 318 tells us that there is a subset T B with T S and such that B : (E \ T S spans V 15

16 We will show that B is a basis: Note that B E dim(v Now B spans V By Proposition 317, B contains a basis, B say By Theorem 320, B E B Thus B B is a basis We will often use the following easy corollary: Corollary 332 Let V be an n-dimensional vector space If B is any linearly independent subset with n elements, then B is a basis (and hence spans V Proof: By Proposition 331, B is contained in a basis, A, say of V Since dim(v A B it follows that B A 4 Coordinate systems Let V be an n-dimensional vector space with (ordered basis B {v 1,, v n } Let v V Then there exists a unique list a 1,, a n of scalars such that We let [v] B denote the column vector v a 1 v a n v n [v] B : a 1 a 2 a n M n 1(K [v] B is called the coordinate vector of v with respect to the basis B Example 41 Let E {e 1, e 2, e 3 } be the standard basis of R 3 Then (2, 5, 7 2 e e e 3 Thus 2 [(2, 5, 7] E 5 7 Example 42 Let B denote the basis {1, x, x 2 } of the real vector space, R[x] 2, of polynomials of degree at most 2 Then [ ] 3 3 x + 2x 2 1 B 2 Example 43 Let P be the plane 2x 2y z 0 in R 3 We have seen that B {v (1, 0, 2, w (1, 1, 0} is a basis of P Observe that u (2, 1, 6 P Let s compute [u] B : We need to solve (2, 1, 6 sv + tw (s + t, t, 2s for s, t This gives s 3, t 1 So u 3v w and hence ( 3 [u] B 1 16

17 Example 44 Let us find the coordinate vector of v (2, 7 with respect to the basis B {(1, 2, (1, 1} of R 2 : We must find x, y R such that This amounts to solving the system where x (1, 2 + y (1, 1 (2, 7 A ( x y A ( ( 2 7 is the 2 2 matrix whose columns are the vectors in the basis Thus we find [v] B ( x y A 1 ( 2 7 ( ( 2 7 ( 5 3 Suppose, more generally, that we have two different bases A and B of a vector space V The following crucial theorem tells us how to switch between the two associated coordinate systems Theorem 45 (Change of Basis Theorem Let V be an n-dimensional vector space with bases A {v 1,, v n } and B {w 1,, w n } Let C be the n n matrix whose j-th column is [v j ] B Then, for any v V we have [v] B C [v] A Proof: Denote the (i, j-entry of C by c i,j Then by definition of C we have, for j 1,, n, n v j c i,j w i Let v V and suppose that v a 1 v a n v n n j1 a jv j Thus [v] A i1 a 1 a n 17

18 Thus we have v n a j v j j1 ( n n a j c i,j w i i1 ( n n a j c i,j w i j1 i1 j1 ( n n c i,j a j w i i1 j1 n b i w i i1 where b i : n j1 c i,ja j But n j1 c i,ja j is precisely the i-th entry in the column vector C a 1 a n Thus [v] B b 1 b n C a 1 a n C [v] A Notation: When needed, we will use the notation C(B, A for the matrix C occuring in this theorem It is called the change-of-basis matrix Thus C(B, A is the matrix whoxe j-th column is the coordinate vector with respect to B of the j-th vector in the basis A For any v V, we thus have [v] B C(B, A [v] A Remark 46 A special case is when V K n and the basis B is the standard basis E {e 1,, e n } In this case, the change-of-basis matrix C(E, A is just the matrix whose columns are the vectors of A We have met this matrix many times already Remark 47 Of course, the matrix C(A, A is just the n n identity matrix I n Problem 48 Let A, B both be n m matrices Suppose that A v B v for all column vectors v M m 1 (K Prove that A B 18

19 Proposition 49 Suppose that A, B, C are all bases of the finite-dimensional vector space V Then C(C, B C(B, A C(C, A Proof: Let A {v 1,, v n } Let v be any column vector Then v [w] A where w a 1 v a n v n Thus a 1 a n C(C, B C(B, A v C(C, B C(B, A [w] A C(C, B [w] B [w] C C(C, A [w] A C(C, A v So C(C, B C(B, A v C(C, A v for all column vectors v Hence C(C, B C(B, A C(C, A as required Corollary 410 Let A and B be bases of the vector space V Then the matrix C(B, A is invertible and C(B, A 1 C(A, B Proof: By Proposition 49 we have C(A, B C(B, A C(A, A I n Example 411 Find the coordinate vectors of v (5, 7 and w ( 4, 3 in R 2 with respect to the basis A {(1, 1, (3, 2} Solution: Let E be the standard basis We have ( 1 3 C(E, A 1 2 Thus ( 1 3 C(A, E ( Thus and [v] A [w] A ( ( ( 2 3 [v] E 1 1 ( 2 3 [w] E 1 1 ( 5 7 ( 4 3 ( ( 1 1, 19

20 Remark 412 Let A and B be two bases of the vector space K n Since the matrices C(E, A and C(E, B are easy to write down, we can use the formula to calculate C(B, A C(B, A C(B, E C(E, A C(E, B 1 C(E, A Example 413 Calculate the change of basis matrix, C(B, A for the two bases A {(1, 2, (3, 4} and B {(1, 1, ( 2, 5}: Solution: So C(E, A ( ( 1 2, C(E, B 1 5 C(B, A C(E, B 1 C(E, A ( ( (

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