LINEAR ALGEBRA EXAMPLES C-2 FFRREEEE SSTTUUDDYY BBOOOOKKSS UNDERSTANDING GEOMETRICAL VECTORS, VECTOR SPACES AND COMPUTER SIMULATION LINEAR MAPS RRLEOOIGGFEE MRR EMMJLCCBHHRAAONNEEYY FREE STUDY BOOKS WWW.BOOKBOON.COM Leif Mejlbro Linear Algebra Examples c-2 Geometrical Vectors, Vector spaces and Linear Maps Download free books at BookBooN.com Linear Algebra Examples c-2 – Geometrical Vectors, Vector Spaces and Linear Maps © 2009 Leif Mejlbro og Ventus Publishing Aps ISBN 978-87-7681-507-3 Download free books at BookBooN.com Linear Algebra Examples c-2 Content Indholdsfortegnelse Introduction 5 1. Geometrical vectors 6 2. Vector spaces 23 3. Linear maps 46 Index 126 what‘s missing in this equation? rt e v d a e h k t c cli e s You could be one of our future talents a e Pl maeRsK inteRnationaL teChnoLogY & sCienCe PRogRamme Are you about to graduate as an engineer or geoscientist? Or have you already graduated? If so, there may be an exciting future for you with A.P. Moller - Maersk. www.maersk.com/mitas Download free books at BookBooN.com 4 Linear Algebra Examples c-2 Introduction Introduction Here we collect all tables of contents of all the books on mathematics I have written so far for the publisher. In the rst list the topics are grouped according to their headlines, so the reader quickly can get an idea of where to search for a given topic.In order not to make the titles too long I have in the numbering added a for a compendium b for practical solution procedures (standard methods etc.) c for examples. The ideal situation would of course be that all major topics were supplied with all three forms of books, but this would be too much for a single man to write within a limited time. After the rst short review follows a more detailed review of the contents of each book. Only Linear Algebra has been supplied with a short index. The plan in the future is also to make indices of every other book as well, possibly supplied by an index of all books. This cannot be done for obvious reasons during the rst couple of years, because this work is very big, indeed. It is my hope that the present list can help the reader to navigate through this rather big collection of books. Finally, since this list from time to time will be updated, one should always check when this introduction has been signed. If a mathematical topic is not on this list, it still could be published, so the reader should also check for possible new books, which have not been included in this list yet. Unfortunately errors cannot be avoided in a rst edition of a work of this type. However, the author has tried to put them on a minimum, hoping that the reader will meet with sympathy the errors which do occur in the text. Leif Mejlbro 5th October 2008 Download free books at BookBooN.com 5 Linear Algebra Examples c-2 1. Geometrical vectors 1 Geometrical vectors Example 1.1 Given A1A2−·−·−·→A8 a regular octogon of midpoint A0. How many different vectors are there among the 81 vectors A A , where i and j belong to the set {0,1,2,...,8}? i j Remark 1.1 Thereshouldhavebeenafigurehere, butneitherLATEXnorMAPLEwillproduceitfor me properly, so it is left to the reader. ♦ This problem is a typical combinatorial problem. −−−→ Clearly, the 9 possibilities A A all represent the 0 vector, so this will giver us 1 possibility. i i From a geometrical point of view A0 is not typical. We can form 16 vector where A0 is the initial or final point. These can, however, be paired. For instance −−−→ −−−→ A1A0 =A0A5 and analogously. In this particular case we get 8 vectors. Then we consider the indices modulo 8, i.e. if an index is larger than 8 or smaller than 1, we subtract or add some multiple of 8, such that the resulting index lies in the set {1,2,...,8}. Thus e.g. 9=1+8≡1( mod 8). −−−−→ Then we h−−a−v−e−8→different vect−o−r−s→of th−e−f−o→rm AiAi+1, and these can always be paired with a vector of the form AjAj−1. Thus e.g. A1A2 =A6A5. Hence the 16 possibilities of this type will only give os 8 different vectors. −−−−→ −−−−−→ −−−−→ T−−h−e−−s→ame is true for AiAi+2 and AjAj−2 (16 possibilities and only 8 vectors), and for AiAi+3 and AjAj−3 (again 16 possibilities and 8 vectors). −−−−→ Finally, we see that we have for AiAi+4 8 possibilities, which all represent a diameter. None of these diameters can be paired with any other, so we obtain another 8 vectors. Summing up, # possibilities # vectors 0 vector 9 1 A−−0−i−s→one of the points 16 8 A−−iA−−i±→1 16 8 A−−iA−−i±→2 16 8 A−−i−A−i−±→3 16 8 A1Ai+4 8 8 I alt 81 41 By counting we find 41 different vectors among the 81 possible combinations. Download free books at BookBooN.com 6 Linear Algebra Examples c-2 1. Geometrical vectors Example 1.2 Given a point set G consisting of n points G={A1,A2,...,An}. Denoting by O the point which is chosen as origo of the vectors, prove that the point M given by the equation (cid:2) (cid:3) −−→ 1 −−→ −−→ −−→ OM = n OA1+OA2+···+OAn , does not depend on the choice of the origo O. The point M is called the midpoint or the geometrical barycenter of the point set G. Prove that the point M satisfies the equation −−−→ −−−→ −−−→ MA1+MA2+···+MAn =(cid:2)0, and that M is the only point fulfilling this equation. Let (cid:2) (cid:3) −−→ 1 −−→ −−→ −−→ OM = n OA1+OA2+···+OAn and (cid:2) (cid:3) −−−→ 1 −−−→ −−−→ −−−→ O1M1 = n O1A1+O1A2+···+O1An . www.job.oticon.dk Download free books at BookBooN.com 7 Linear Algebra Examples c-2 1. Geometrical vectors Then (cid:2) (cid:3) −−−→ −−→ −−→ −−→ 1 −−→ −−→ −−→ O1M = O1O+OM =O1O+ n OA1+OA2+···+OAn (cid:4)(cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3)(cid:5) 1 −−→ −−→ −−→ −−→ −−→ −−→ = n O1O+OA1 + O1O+OA2 +···+ O1O+OAn (cid:2) (cid:3) 1 −−−→ −−−→ −−−→ −−−→ = n O1A1+O1A2+···+O1An =O1M1, from which we conclude that M1 =M. Now choose in particular O =M. Then (cid:2) (cid:3) −−−→ 1 −−−→ −−−→ −−−→ MM =(cid:2)0= n MA1+MA2+···+MAn , thus −−−→ −−−→ −−−→ MA1+MA2+···+MAn =(cid:2)0. On the other hand, the uniqueness proved above shows that M is the only point, for which this is true. Example 1.3 Prove that if a point set G={A1,A2,...,An} has a centrum of symmetry M, then the midpoint of the set (the geometrical barycenter) lie in M. If A and A are symmetric with respect to M, then i j −−−→ −−−→ MA +MA =(cid:2)0. i j Since every point is symmetric to precisely one other point with respect to M, we get −−−→ −−−→ −−−→ MA1+MA2+···+MAn =(cid:2)0, which according to Example 1.2 means that M is also the geometrical barycenter of the set. Example 1.4 Prove that if a point set G = {A1,A2,...,An} has an axis of symmetry (cid:3), then the midpoint of the set (the geometrical barycenter) lies on (cid:3). −−→ −−→ Every point A can be paired with an A , such that OA +OA lies on (cid:3), and such that G\{A ,A } i j i j i j still has the axis of symmetry (cid:3). Remark 1.2 The problem is here that A , contrary to Example 1.3 is not uniquely determined. ♦ j Continue in this way by selecting pairs, until there are no more points left. Then the midpoints of all pairs will lie on (cid:3). Since (cid:3) is a straight line, the midpoint of all points in G will also lie on (cid:3). Download free books at BookBooN.com 8 Linear Algebra Examples c-2 1. Geometrical vectors Example 1.5 Given a regular−h−→exagon of the vertices A1, A2, ..., A6. Denote the center of the hexagon by O. Find the vector OM from O to the midpoint (the geometrical barycenter) M of 1. the point set {A1,A2,A3,A4,A5}, 2. the point set {A1,A2,A3}. Remark 1.3 Again a figure would have been very useful and again neither LATEXnor MAPLE will produce it properly. The drawing is therefore left to the reader. ♦ 1. It follows from −−→ −−→ −−→ −−→ −−→ −−→ OA1+OA2+OA3+OA4+OA5+OA6 =(cid:2)0, by adding something and then subtracting it again that (cid:4) (cid:5) −−→ 1 −−→ −−→ −−→ −−→ −−→ OM = OA1+OA2+OA3+OA4+OA5 5 (cid:4)(cid:2) (cid:3) (cid:5) 1 −−→ −−→ −−→ −−→ −−→ −−→ −−→ = OA1+OA2+OA3+OA4+OA5+OA6 −OA6 5 1−−→ 1−−→ = − OA6 = OA3. 5 5 −−→ −−→ −−→ 2. Since OA1+OA3 =OA2 (follows fromthemissingfigure, whichthereaderofcoursehasdrawn already), we get (cid:4) (cid:5) −−→ 1 −−→ −−→ −−→ 2−−→ OM = OA1+OA2+OA3 = OA2. 3 3 Example 1.6 Prove by vector calculus that the medians of a triangle pass through the same point and that they cut each other in the proportion 1:2. Remark 1.4 In this case there would be a theoretical possibility of sketching a figure in LATEX. It will, however, be very small, and the benefit of if will be too small for all the troubles in creating the figure. LATEXis not suited for figures. ♦ Let O denote the reference point. Let M denote the midpoint of BC and analogously of the others. A Then the median from A is given by the line segment AM , and analogously. A It follows from the definition of M that A −−−→ 1 −−→ −−→ OM = (OB+OC), A 2 −−−→ 1 −→ −−→ OM = (OA+OC), B 2 Download free books at BookBooN.com 9 Linear Algebra Examples c-2 1. Geometrical vectors −−−→ 1 −→ −−→ OM = (OA+OB). C 2 Then we conclude that 1 −→ −−→ −−→ 1−→ −−−→ 1−−→ −−−→ 1−−−→ (OA+OB+OC)= OA+OM = OB+OM = OM . 2 2 A 2 B 2 C −−→ −−→ −−→ Choose O =M, such that MA+MB+MC =(cid:2)0, i.e. M is the geometrical barycenter. Then we get by multiplying by 2 that −−→ −−−−→ −−→ −−−−→ −−→ −−−−→ (cid:2)0=MA+2MM =MB+2MM =MC+2MM , A B C whichprovesthatM liesonallthreelinesAM , BM andCM , andthatM cutseachoftheseline A B C segments in the proportion 2:1. Example 1.7 We define the median from a vertex A of a tetrahedron ABCD as the line segment from A to the point of intersection of the medians of the triangle BCD. Prove by vector calculus that the four medians of a tetrahedron all pass through the same point and cut each other in the proportion 1:3. Furthermore, prove that the point mentioned above is the common midpoint of the line segments which connect the midpoints of opposite edges of the tetrahedron. Remark 1.5 It is again left to the reader to sketch a figure of a tetrahedron. ♦ It follows from Example 1.6 that M is the geometrical barycentrum of (cid:4)BCD, i.e. A (cid:2) (cid:3) −−−→ 1 −−→ −−→ −−→ OM = OB+OC+OD , A 3 and analogously. Thus (cid:2) (cid:3) 1 −→ −−→ −−→ −−→ 1−→ −−−→ 1−−→ −−−→ 1−−→ −−−→ OA+OB+OC+OD = OA+OM = OB+OM = OC+OM 3 3 A 3 B 3 C 1−−→ −−−−−→ = OD+ONM . 3 D By choosing O =M as the geometrical barycenter of A, B, C and D, i.e. −−→ −−→ −−→ −−→ MA+MB+MC+MD =(cid:2)0, we get 1−−→ −−−−→ 1−−→ −−−−→ 1−−→ −−−−→ 1−−→ −−−−→ MA+MM = MB+MM = MC+MM = MD+MM , 3 A 3 B 3 C 3 D so we conclude as in Example 1.6 that the four medians all pass through M, and that M divides each median in the proportion 3:1. Download free books at BookBooN.com 10