Worksheet for Lecture 15 (due October 23) Section 4.3 Linearly Independent Sets; Bases

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1 Worksheet for Lecture 5 (due October 23) Name: Section 4.3 Linearly Independent Sets; Bases Definition An indexed set {v,..., v n } in a vector space V is linearly dependent if there is a linear relation among them: c v + + c n v n =, with c,..., c n not all zero. Otherwise, say that {v,..., v n } is linearly independent Theorem An indexed set {v,..., v n } of two or more vectors, with v, is linearly dependent if and only if some v j (with j > ) is a linear combination of the preceding vectors v,..., v j. Example For p (t) =, p 2 (t) = t, and p 3 (t) = 4 t. Then {p, p 2, p 3 } is linearly dependent in P because p 3 = 4p p 2. Example 2 In the vector space of functions on R, the set {sin t, cos t} is linearly independent, as sin t is not a multiple of cos t as a function. the set {sin 2t, sin t cos t} because sin 2t = 2 sin t cos t. Big Concept Let H be a subspace of a vector space V. B = {v,..., v n } in V is a BASIS of H if () B is a linearly independent set, and (2) we have H = Span{v,..., v n } Example R n has a basis: e =., e 2 =.,..., e n =. {e,..., e n } is a standard basis of R n. An indexed set of vectors Example 2 In general, vectors v,..., v n R n forms a basis if and only if the matrix A = v v 2 v n is invertible It is linearly independent by Inverse Matrix Theorem; and it is spanned by v,..., v n again by Inverse Matrix Theorem. Example 3 Let P n denote the vector space of polynomials of degree n. Let S =, t, t 2,..., t n }. Then S forms a basis of S.

2 2 Spanning set theorem Let S = {v,..., v n } be a set in V, and let H = Span{v,..., v n }. (a) If one of the vectors in S, say v k, is a linear combination of the remaining vectors in S, then the set formed from S by removing v k still spans H. (b) If H {}, some subset of S is a basis of H. Illustration by example Find a basis for Col(A) with A = 4 2 v v 2 v 3 v 4 v 5 = Solution Observe that v 4 = 2v v 3. So any linear combination can be rewritten as c v + c 2 v 2 + c 3 v 3 + c 4 v 4 + c 5 v 5 Therefore, c v + c 2 v 2 + c 3 v 3 + c 4 (2v v 3 ) + c 5 v 5 =(c + 2c 4 )v + c 2 v 2 + (c 3 c 4 )v 3 + c 5 v 5. Col(A) = Span{v,..., v 5 } = Span{v, v 2, v 3, v 5 }. In a similar way, v 2 = 4v, so we can remove v 2 from the list and thus, Note that Col(A) = Span{v, v 3, v 5 }. v v 3 v 5 = The columns of this matrix is linearly independent. So {v, v 3, v 5 } is a basis of Col(A). Theorem Say A be an echelon form as above, and B is a matrix that is row equivalent to A. Say B = ( w w 2 w 3 w 4, w 5 ). Then Col(A) = Span{v, v 3, v 5 } Col(B) = Span{w, w 3, w 5 } In general, the column space of B is spanned by the columns of B indexed by the pivot columns of A Explanation: The solution set of Ax = is the same as the solution set of Bx =. So the linear relations among the columns of A are the same linear relations among the columns of B. In our example, this means that if B = ( w w 2 w 3 w 4 w 5 ), then w 2 = 4w and w 4 = 2w w 3 ; so we may remove w 2 and w 4 from the list and Col(B) = {w, w 3, w 5 }.

3 Example Find a basis of Col(A) for A = Solution We perform a row reduction to the matrix A (and we don t need to add the extra column of s) So the first, second and the fourth columns are the pivot columns. Thus { Col(A) = Span 2 2, 4 6, 4 } A good philosophy to remember: A basis of H is a linearly independent subset of H that is as large as possible; and it is at the same time a spanning set of H that is as small as possible. Section 4.4 Coordinate System Theorem (The Unique Representation Theorem) Let B = {b,..., b n } be a basis for a vector space V. Then for each x in V, there exists a unique set of scalars c,..., c n such that x = c b + c 2 b c n b n. Example For V = R 4 and the standard basis e =, e 2 =, e 3 =, e 4 =. a We can write each vector b c uniquely as ae + be 2 + ce 3 + de 4. d Definition Suppose that B = {b,..., b n } is a basis for V and x is in V. The coordinates of x relative to the basis B (or the B-coordinate of x) are the weights c,..., c n such that x = c b + + c n b n. We write [ x ] B = c. c n

4 4 and call it the coordinate vector of x (relative to B). Then the map x [ x ] is the B coordinate mapping V R n. { Example Consider a basis B = b =, b 2 = 2, b 3 = 4 } of R 3. What s 3 9 the B-coordinate of a vector x = 5? 7 Solution We solve the equation x b + x 2 b 2 + x 3 b 3 = x. In other words: So [ x ] B = 5 7. Remark You are supposed to be always get a unique solution. Coordinate mapping Theorem Let B = {b,..., b n } be a basis for a vector space V. Then the coordinate mapping x [ x ] B is a one-to-one linear transformation from V onto Rn. (We say that V is isomorphic to R n.) Example Let B = {, t, t 2, t 3 } be the standard basis of the space P 3 of polynomials of degree 3. Then the coordinate mapping is given by p(t) = a + a t + a 2 t 2 + a 3 t 3 [ p(t) ] B = Example 2 With the same notation as above, B = { + t, t, t 2, t 3 + t 2 } is also a basis of P 3, then the coordinate mapping with respect to B is given by p(t) = a +a t+a 2 t 2 +a 3 t 3 = a (+t)+(a a )t+(a 2 a 3 )t 2 +a 3 (t 2 +t 3 ) [ p(t) ] = a a B a 2 a 3 Upshot: The vector space P 3 is isomorphic to R 4. But there are many ways to write this linear isomorphism, and this depends on a choice of basis. It is an important concept in linear algebra to allow working with different basis of P 3, and therefore different ways to realize P 3 as a more concrete vector space R 4. a a a 2 a 3 a a 3

5 True/False Questions () If H = Span{b,..., b n }, then {b,..., b n } is a basis for H. False. For {b,..., b n } to be a basis for H, we also need it to be linearly independent. (2) The columns of an invertible matrix form a basis for R n. True. This is because the columns of an invertible basis spans R n and are linearly independent. (3) A linearly independent set in a subspace H is a basis for H. False. A linearly independent set in H need not span H in general. (4) If a finite set S of nonzero vectors spans a vector space V, then some subset of S is a basis for V. True. This is the Spanning Set Theorem. (5) A basis is a linearly independent set that is as large as possible. True. (6) If B is an echelon form of a matrix A, then the pivot columns of B form a basis for Col(A). False. A basis of Col(A) is given by the columns of A whose indices are the pivot columns of B. (7) If x is in V and if B contains n vectors, then the B-coordinate vector of x is in R n. True. (8) If B is the standard basis for R n, then the B-coordinate vector of an x R n is x itself. True. 5 Exercise Consider a basis B = { b =, b 2 = 3 3 4, b 3 = } of R 3. 4 What s the B-coordinate of a vector x = 8 9? 6 Solution We solve the equation x b + x 2 b 2 + x 3 b 3 = x. In other words:

6 6 So [ x ] B =. 3 Exercise 2 The set B = { + t 2, t + t 2, + 2t + t 2 } is a basis for P 2. Find the coordinate vector of p(t) = + 4t + 7t 2 relative to B. Solution The coordinate [ p(t) ] B = c satisfies c 2 c 3 p(t) = + 4t + 7t 2 = c ( + t 2 ) + c 2 (t + t 2 ) + c 3 ( + 2t + t 2 ) Then we have c + +c 3 = c 2 + 2c 3 = 4 c + c 2 + c 3 = So [ p(t) ] B = 2 6. Exercise 3 Find a basis for Col(A) and for Nul(A) with 3 2 A = Solution We first find a basis for Col(A). For this we do a row reduction on A The pivot columns are the first, second, and the fourth. So { } Col(A) = Span 3, 2,

7 To compute Nul(A), we need to solve [ A ] We have already done lots of row operations above; so we continue with that (but adding the last column of zeros): So So we have x x 2 x 3 x 4 x 5 3x 3 4x = 4x 3 + 5x 5 x 3 2x 5 = x x x { 4 Nul(A) = Span, 5 }. 2 7