Elementary Linear Algebra with Applications Bernard Kolman David Hill Ninth Edition

Elementary Linear Algebra with Applications Bernard Kolman David Hill Ninth Edition

Pearson Education Limited Edinburgh Gate Harlow Essex CM JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk Pearson Education Limited 4 All rights reserved. No part of this publication may be reproduced stored in a retrieval system or transmitted in any form or by any means electronic mechanical photocopying recording or otherwise without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd Saffron House 6 Kirby Street London ECN 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners. ISBN : -9-65- ISBN : 978--9-65-6 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America

6 Chapter 4 Real Vector Spaces. Does the set S = {[ ] [ ] [ ] [ ]} span M?. Find a set of vectors spanning the solution space of Ax = where A =.. Find a set of vectors spanning the null space of A = 6. 4. The set W of all matrices A with trace equal to zero is a subspace of M. Let {[ ] [ ] [ ]} S =. Show that span S = W. 4. The set W of all matrices A with trace equal to zero is a subspace of M. (See Exercise in Section 4..) Determine a subset S of W so that span S = W. 5. The set W of all matrices of the form a b c d e is a subspace of M. (See Exercise in Section 4..) Determine a subset S of W so that span S = W. 6. Let T be the set of all matrices of the form AB BA where A and B are n n matrices. Show that span T is not M nn.(hint: Use properties of trace.) 7. Determine whether your software has a command for finding the null space (see Example in Section 4.) of a matrix A. If it does use it on the matrix A in Example and compare the command s output with the results in Example. To experiment further use Exercises and. 4.5 Linear Independence In Section 4.4 we developed the notion of the span of a set of vectors together with spanning sets of a vector space or subspace. Spanning sets S provide vectors so that any vector in the space can be expressed as a linear combination of the members of S. We remarked that a vector space can have many different spanning sets and that spanning sets for the same space need not have the same number of vectors. We illustrate this in Example. EXAMPLE In Example 5 of Section 4. we showed that the set W of all vectors of the form a b a + b where a and b are any real numbers is a subspace of R. Each of the following sets is a spanning set for W (verify): S = 5 S = S = 86

4.5 Linear Independence 7 We observe that the set S is a more efficient spanning set since each vector of W is a linear combination of two vectors compared with three vectors when using S and four vectors when using S. If we can determine a spanning set for a vector space V that is minimal in the sense that it contains the fewest number of vectors then we have an efficient way to describe every vector in V. In Example since the vectors in S span W and S is a subset of S and S it follows that the vector 5 in S must be a linear combination of the vectors in S and similarly both vectors and in S must be linear combinations of the vectors in S. Observe that + 5 + +. In addition for set S we observe that + 5 and for set S we observe that + +. It follows that if span S = V and there is a linear combination of the vectors in S with coefficients not all zero that gives the zero vector then some subset of S is also a spanning set for V. 87

8 Chapter 4 Real Vector Spaces Remark The preceding discussion motivates the next definition. In the preceding discussion which is based on Example we used the observation that S was a subset of S and of S. However that observation is a special case which need not apply when comparing two spanning sets for a vector space. DEFINITION 4.9 The vectors v v...v k in a vector space V are said to be linearly dependent if there exist constants a a...a k not all zero such that k a j v j = a v + a v + +a k v k =. () j= Otherwise v v...v k are called linearly independent. That is v v...v k are linearly independent if whenever a v + a v + +a k v k = a = a = =a k =. If S ={v v...v k } then we also say that the set S is linearly dependent or linearly independent if the vectors have the corresponding property. It should be emphasized that for any vectors v v...v k Equation () always holds if we choose all the scalars a a...a k equal to zero. The important point in this definition is whether it is possible to satisfy () with at least one of the scalars different from zero. Remark Definition 4.9 is stated for a finite set of vectors but it also applies to an infinite set S of a vector space using corresponding notation for infinite sums. Remark We connect Definition 4.9 to efficient spanning sets in Section 4.6. To determine whether a set of vectors is linearly independent or linearly dependent we use Equation (). Regardless of the form of the vectors Equation () yields a homogeneous linear system of equations. It is always consistent since a = a = =a k = is a solution. However the main idea from Definition 4.9 is whether there is a nontrivial solution. EXAMPLE Determine whether the vectors v = v = v = are linearly independent. Solution Forming Equation () a + a + a 88

4.5 Linear Independence 9 we obtain the homogeneous system (verify) EXAMPLE a + a a = a + a + a = a a =. The corresponding augmented matrix is whose reduced row echelon form is (verify). Thus there is a nontrivial solution k k k = (verify) k so the vectors are linearly dependent. Are the vectors v = [ ] v = [ ] and v = [ ] in R 4 linearly dependent or linearly independent? Solution We form Equation () a v + a v + a v = and solve for a a and a. The resulting homogeneous system is (verify) a + a = a + a = a + a + a = a + a + a =. The corresponding augmented matrix is (verify) and its reduced row echelon form is (verify). 89

Chapter 4 Real Vector Spaces Thus the only solution is the trivial solution a = a = a = so the vectors are linearly independent. EXAMPLE 4 Are the vectors v = [ ] v = [ ] v = [ ] in M linearly independent? Solution We form Equation () a [ ] + a [ ] [ ] + a = [ ] and solve for a a and a. Performing the scalar multiplications and adding the resulting matrices gives [ ] [ ] a + a a + a a =. a a a + a Using the definition for equal matrices we have the linear system a + a = a + a a = a a = a + a =. The corresponding augmented matrix is and its reduced row echelon form is (verify). Thus there is a nontrivial solution k k k k = (verify) so the vectors are linearly dependent. 9