1 00:00:00,090 --> 00:00:02,490 The following content is provided under a Creative 2 00:00:02,490 --> 00:00:04,030 Commons license. 3 00:00:04,030 --> 00:00:06,330 Your support will help MIT OpenCourseWare 4 00:00:06,330 --> 00:00:10,690 continue to offer high-quality educational resources for free. 5 00:00:10,690 --> 00:00:13,320 To make a donation or view additional materials 6 00:00:13,320 --> 00:00:16,920 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,920 --> 00:00:18,220 at ocw.mit.edu. 8 00:00:31,136 --> 00:00:32,030 HERBERT GROSS: Hi. 9 00:00:32,030 --> 00:00:33,447 Well, I guess I really should have 10 00:00:33,447 --> 00:00:37,130 said "goodbye," because this is the last lecture 11 00:00:37,130 --> 00:00:38,330 in our course-- 12 00:00:38,330 --> 00:00:42,030 not the last assignment, but the last lecture. 13 00:00:42,030 --> 00:00:45,720 The reason I said "hi" was, why quit after all this time 14 00:00:45,720 --> 00:00:46,880 with saying that? 15 00:00:46,880 --> 00:00:49,820 And we've reached the stage now where 16 00:00:49,820 --> 00:00:52,820 we should clean up the vector spaces 17 00:00:52,820 --> 00:00:55,610 to the best of our ability and recognize 18 00:00:55,610 --> 00:00:58,820 that, from this point on, much of the treatment of vector 19 00:00:58,820 --> 00:01:02,360 spaces requires specialized concentration. 20 00:01:02,360 --> 00:01:04,519 In fact, I envy the real people who 21 00:01:04,519 --> 00:01:06,140 make regular movies where they have 22 00:01:06,140 --> 00:01:08,060 stuntmen when things get tough. 23 00:01:08,060 --> 00:01:09,860 I would continue on with this course, 24 00:01:09,860 --> 00:01:11,870 except that I don't have a stuntman 25 00:01:11,870 --> 00:01:14,270 to do these hard lectures for me. 26 00:01:14,270 --> 00:01:17,467 And also, the particular topic, as I told you last time, 27 00:01:17,467 --> 00:01:19,550 that I have in mind for today-- the subject called 28 00:01:19,550 --> 00:01:21,020 eigenvectors-- 29 00:01:21,020 --> 00:01:23,400 has two approaches to it. 30 00:01:23,400 --> 00:01:28,070 One is that it does have some very elaborate practical 31 00:01:28,070 --> 00:01:30,620 applications, many of which occur 32 00:01:30,620 --> 00:01:32,600 in more advanced subjects. 33 00:01:32,600 --> 00:01:35,270 It also has a very nice framework 34 00:01:35,270 --> 00:01:38,570 within the game of mathematics idea. 35 00:01:38,570 --> 00:01:42,350 And my own feeling was that, since we started this course 36 00:01:42,350 --> 00:01:45,350 with the concept of the game of mathematics, 37 00:01:45,350 --> 00:01:47,420 mathematical structures, I thought 38 00:01:47,420 --> 00:01:50,210 that, rather than go into complicated applications, 39 00:01:50,210 --> 00:01:53,240 I would treat eigenvectors in terms 40 00:01:53,240 --> 00:01:56,880 of the structure of mathematics as a game. 41 00:01:56,880 --> 00:01:58,910 In fact, as an interesting aside, 42 00:01:58,910 --> 00:02:00,890 I'd like to share with you a very famous story 43 00:02:00,890 --> 00:02:05,300 in mathematics, that, when the first book on matrix algebra 44 00:02:05,300 --> 00:02:09,949 was written by Hamilton, he inscribed the book with, 45 00:02:09,949 --> 00:02:12,890 "Here at last is a branch of mathematics 46 00:02:12,890 --> 00:02:16,190 for which there will never be found practical application." 47 00:02:16,190 --> 00:02:19,760 He did not invent the subject to solve difficult physics 48 00:02:19,760 --> 00:02:20,510 problems. 49 00:02:20,510 --> 00:02:22,940 He invented the subject because it was 50 00:02:22,940 --> 00:02:25,280 an elegant mathematical device. 51 00:02:25,280 --> 00:02:27,900 And so I thought that maybe, for our last lecture, 52 00:02:27,900 --> 00:02:29,960 we should end on that vein. 53 00:02:29,960 --> 00:02:34,760 At any rate, the subject for today is called eigenvectors. 54 00:02:34,760 --> 00:02:39,600 And from a purely game point of view, the idea is this. 55 00:02:39,600 --> 00:02:42,650 Let's suppose that we have a vector space V 56 00:02:42,650 --> 00:02:45,710 and a linear mapping, a linear transformation, f, 57 00:02:45,710 --> 00:02:48,690 that maps V onto itself, say. 58 00:02:48,690 --> 00:02:52,430 And the question is, are there any vectors in V other 59 00:02:52,430 --> 00:02:55,510 than the zero vector, such that f of v 60 00:02:55,510 --> 00:02:57,620 is some scalar multiple of v? 61 00:02:57,620 --> 00:03:00,440 You see, the reason I exclude the zero vector, first of all, 62 00:03:00,440 --> 00:03:03,230 is that we already know that, for a linear transformation, 63 00:03:03,230 --> 00:03:05,720 f of 0 is 0. 64 00:03:05,720 --> 00:03:08,810 And c times 0 is 0 for all c. 65 00:03:08,810 --> 00:03:13,250 So this would be trivially true if v were the zero vector. 66 00:03:13,250 --> 00:03:15,530 So what we're really interested in-- 67 00:03:15,530 --> 00:03:18,290 given a particular linear transformation, 68 00:03:18,290 --> 00:03:20,480 are there any vectors such that, relative 69 00:03:20,480 --> 00:03:23,960 to that linear transformation, the mapping of that vector 70 00:03:23,960 --> 00:03:27,420 is just a scalar multiple of that vector itself? 71 00:03:27,420 --> 00:03:30,830 In other words, does f preserve the direction 72 00:03:30,830 --> 00:03:33,320 of any vectors in V? 73 00:03:33,320 --> 00:03:36,650 By the way, don't confuse this with conformal mapping 74 00:03:36,650 --> 00:03:39,770 that we talked about in complex variables. 75 00:03:39,770 --> 00:03:43,700 In conformal mapping, we didn't preserve any-- in general, 76 00:03:43,700 --> 00:03:46,370 we did not preserve directions of lines. 77 00:03:46,370 --> 00:03:47,630 We preserved angles. 78 00:03:47,630 --> 00:03:49,670 In other words, an angle might have 79 00:03:49,670 --> 00:03:52,160 been rotated so that the direction of the two sides 80 00:03:52,160 --> 00:03:53,180 may have changed. 81 00:03:53,180 --> 00:03:54,770 It was the angle that was preserved 82 00:03:54,770 --> 00:03:56,240 in the conformal mapping. 83 00:03:56,240 --> 00:03:59,210 What we're asking now is, given a linear transformation, 84 00:03:59,210 --> 00:04:01,370 does it preserve any directions? 85 00:04:01,370 --> 00:04:04,230 And to illustrate this in terms of an example, 86 00:04:04,230 --> 00:04:06,140 let's suppose we think of mapping 87 00:04:06,140 --> 00:04:10,190 the xy-plane into the uv-plane under the linear mapping f 88 00:04:10,190 --> 00:04:17,839 bar, where f bar maps x, y into the 2-tuple x plus 4y comma 89 00:04:17,839 --> 00:04:19,130 x plus y. 90 00:04:19,130 --> 00:04:22,700 In other words, in terms of mapping the xy-plane 91 00:04:22,700 --> 00:04:25,640 into the uv-plane, this is the mapping-- 92 00:04:25,640 --> 00:04:29,150 u equals x plus 4y, v equals x plus y. 93 00:04:29,150 --> 00:04:31,190 It's understood here, when I'm referring 94 00:04:31,190 --> 00:04:35,890 to the typical xy-plane, that my basis vectors are i and j. 95 00:04:35,890 --> 00:04:39,370 Notice, by the way, that if I look at the vector i, 96 00:04:39,370 --> 00:04:42,380 i is the 2-tuple 1 comma 0. 97 00:04:42,380 --> 00:04:48,170 Notice that when x is 1 and y is 0, u is 1 and v is one. 98 00:04:48,170 --> 00:04:52,850 So f bar maps 1 comma 0 into 1 comma 1. 99 00:04:52,850 --> 00:04:55,280 Again, in terms of i and j vectors, 100 00:04:55,280 --> 00:04:58,820 f bar maps i into i plus j. 101 00:04:58,820 --> 00:05:02,930 And i plus j is certainly not a scalar multiple of i. 102 00:05:02,930 --> 00:05:05,990 Similarly, what does f bar do to j? 103 00:05:05,990 --> 00:05:10,190 j, relative to the basis i and j, is written as 0 comma 1. 104 00:05:10,190 --> 00:05:15,890 When x is 0 and y is 1, we obtain that u is 4 and v is 1. 105 00:05:15,890 --> 00:05:20,720 So under f bar, 0 comma 1 is mapped into 4 comma 1. 106 00:05:20,720 --> 00:05:23,540 Or in the language of i and j components, f 107 00:05:23,540 --> 00:05:27,440 bar maps j into 4i plus j. 108 00:05:27,440 --> 00:05:31,550 And that certainly is not a scalar multiple of j. 109 00:05:31,550 --> 00:05:36,050 In other words, 4i plus j is not parallel to j. 110 00:05:36,050 --> 00:05:39,090 On the other hand, let me pull this one out of the hat. 111 00:05:39,090 --> 00:05:41,510 Let's take the vector 2i plus j-- 112 00:05:41,510 --> 00:05:44,930 in other words, the 2-tuple 2 comma 1. 113 00:05:44,930 --> 00:05:50,200 When x is 2 and y is 1, we see that x plus 4y is 6 114 00:05:50,200 --> 00:05:52,220 and x plus y is 3. 115 00:05:52,220 --> 00:05:57,170 So f bar maps 2 comma 1 into 6 comma 3. 116 00:05:57,170 --> 00:06:00,290 And that certainly is 3 times 2 comma 117 00:06:00,290 --> 00:06:02,390 1-- see, by our rule of scalar multiplication. 118 00:06:02,390 --> 00:06:06,410 In other words, what this says is that the vector 2i plus j 119 00:06:06,410 --> 00:06:11,240 is mapped into the vector which has the same sense 120 00:06:11,240 --> 00:06:17,480 and direction as 2i plus j, but is 3 times as long. 121 00:06:17,480 --> 00:06:21,140 So you see, sometimes the linear transformation 122 00:06:21,140 --> 00:06:24,020 will map a vector into a scalar multiple of itself. 123 00:06:24,020 --> 00:06:25,340 Sometimes it won't. 124 00:06:25,340 --> 00:06:26,750 Sometimes there'll be no vectors, 125 00:06:26,750 --> 00:06:30,680 other than the zero vector, that they're mapped into-- 126 00:06:30,680 --> 00:06:34,150 scalar multiples of themselves, and things of this type. 127 00:06:34,150 --> 00:06:35,630 But that's not important right now. 128 00:06:35,630 --> 00:06:38,300 In terms of a game, what we're trying to do is what? 129 00:06:38,300 --> 00:06:40,430 Solve the equation [INAUDIBLE]---- well, 130 00:06:40,430 --> 00:06:42,140 let's give it in terms of a definition. 131 00:06:42,140 --> 00:06:45,470 First of all, if we have a vector space V, 132 00:06:45,470 --> 00:06:51,350 and f is a linear transformation mapping V into itself, 133 00:06:51,350 --> 00:06:55,760 if little v is any non-zero element of the vector space V, 134 00:06:55,760 --> 00:07:00,590 and if f of v equals c times v for some scalar-- 135 00:07:00,590 --> 00:07:02,150 for some number c-- 136 00:07:02,150 --> 00:07:08,750 then v is called an eigenvector and c is called an eigenvalue. 137 00:07:08,750 --> 00:07:12,320 In other words, if a vector has its direction preserved, 138 00:07:12,320 --> 00:07:14,150 geometrically speaking, all we're saying 139 00:07:14,150 --> 00:07:16,610 is that if the direction doesn't change, 140 00:07:16,610 --> 00:07:18,980 the vector is called an eigenvector. 141 00:07:18,980 --> 00:07:20,132 And the scaling factor-- 142 00:07:20,132 --> 00:07:20,840 which means what? 143 00:07:20,840 --> 00:07:23,210 Even though the direction doesn't change, 144 00:07:23,210 --> 00:07:25,880 the image may have a different magnitude, 145 00:07:25,880 --> 00:07:28,820 because the scalar c here doesn't have to be 1. 146 00:07:28,820 --> 00:07:31,820 That scalar is called an eigenvalue. 147 00:07:31,820 --> 00:07:35,240 And I'll give you more on this in the exercises, 148 00:07:35,240 --> 00:07:37,130 and perhaps even in supplementary notes 149 00:07:37,130 --> 00:07:39,380 if the exercises seem to get too sticky. 150 00:07:39,380 --> 00:07:41,300 But we'll see how things work out. 151 00:07:41,300 --> 00:07:43,610 For the time being, all I care about 152 00:07:43,610 --> 00:07:46,130 is that you understand what an eigenvector means 153 00:07:46,130 --> 00:07:48,260 and what an eigenvalue is. 154 00:07:48,260 --> 00:07:52,910 Quickly summarized, if f maps a vector space into itself, 155 00:07:52,910 --> 00:07:56,060 an eigenvector is any non-zero vector 156 00:07:56,060 --> 00:08:00,140 which has its direction preserved under the mapping f-- 157 00:08:00,140 --> 00:08:04,130 that f maps it into a scalar multiple of itself. 158 00:08:04,130 --> 00:08:07,910 There is a matrix approach for finding eigenvectors, 159 00:08:07,910 --> 00:08:09,980 and the matrix approach also gives us 160 00:08:09,980 --> 00:08:13,040 a very nice review of many of the techniques that we've 161 00:08:13,040 --> 00:08:15,350 used previously in our course. 162 00:08:15,350 --> 00:08:17,330 For the sake of argument, let's suppose 163 00:08:17,330 --> 00:08:19,730 that V is an n-dimensional vector space, 164 00:08:19,730 --> 00:08:23,660 and that we've again chosen a particular basis, u1 up to un, 165 00:08:23,660 --> 00:08:26,810 to represent V. Suppose, also, that f 166 00:08:26,810 --> 00:08:30,470 is a linear mapping carrying V into V. 167 00:08:30,470 --> 00:08:33,049 Remember how we used the matrix approach here? 168 00:08:33,049 --> 00:08:35,030 What we said was, look at. 169 00:08:35,030 --> 00:08:39,530 The vectors u1 up to un are carried into f of u1 170 00:08:39,530 --> 00:08:41,179 up to f of un. 171 00:08:41,179 --> 00:08:43,549 And that determines the linear transformation 172 00:08:43,549 --> 00:08:48,170 f because of the fact of the linearity properties. 173 00:08:48,170 --> 00:08:53,240 In other words, when you take f of a1 u1 plus a2 u2, 174 00:08:53,240 --> 00:08:56,840 it's just a1 f of u1 plus a2 f of u2. 175 00:08:56,840 --> 00:08:59,720 So once you know what happens to the basis vectors, 176 00:08:59,720 --> 00:09:02,150 you know what happens to everything. 177 00:09:02,150 --> 00:09:06,870 But since we're expressing V in terms of the basis u1 up to un, 178 00:09:06,870 --> 00:09:10,340 that means that f of u1, et cetera, f of un 179 00:09:10,340 --> 00:09:14,220 may all be expressed as linear combinations of u1 up to un. 180 00:09:14,220 --> 00:09:18,540 And that's precisely what I've written over here. 181 00:09:18,540 --> 00:09:20,420 We also know that the vector v that we're 182 00:09:20,420 --> 00:09:21,920 trying to find-- see, remember we're 183 00:09:21,920 --> 00:09:23,540 trying to find eigenvectors. 184 00:09:23,540 --> 00:09:27,476 The vector v, relative to the basis u1 up to un, 185 00:09:27,476 --> 00:09:31,723 can be written as the n-tuple x1 up to xn. 186 00:09:31,723 --> 00:09:33,890 And now I won't bother writing this, because I hope, 187 00:09:33,890 --> 00:09:35,570 by this time, you understand this. 188 00:09:35,570 --> 00:09:37,430 This is an abbreviation for saying what? 189 00:09:37,430 --> 00:09:41,930 The vector v is x1 u1 plus et cetera xn un, 190 00:09:41,930 --> 00:09:45,590 because I am always referring to the specific basis 191 00:09:45,590 --> 00:09:49,670 when I write n-tuples without any other qualifications. 192 00:09:49,670 --> 00:09:56,930 Now, the question was, how did the statement f of v equal cv 193 00:09:56,930 --> 00:10:00,260 translate in the language of matrices? 194 00:10:00,260 --> 00:10:02,870 Remember that we took this particular matrix 195 00:10:02,870 --> 00:10:05,420 of coefficients and wrote-- 196 00:10:05,420 --> 00:10:07,490 well, we transposed it. 197 00:10:07,490 --> 00:10:08,540 Remember what we said? 198 00:10:08,540 --> 00:10:10,610 We said that the matrix A would be 199 00:10:10,610 --> 00:10:14,630 the matrix whose first column would be the components of f 200 00:10:14,630 --> 00:10:19,640 of u1 and whose nth column would be the components of f of un. 201 00:10:19,640 --> 00:10:24,620 In other words, what we did was, is we said that to take f of v, 202 00:10:24,620 --> 00:10:26,990 we would just take the matrix-- 203 00:10:26,990 --> 00:10:33,870 notice how I've written it, now-- not a1,1, a1,2, a1,1, 204 00:10:33,870 --> 00:10:34,710 you see? 205 00:10:34,710 --> 00:10:37,380 a2,1, a3,1, et cetera-- 206 00:10:37,380 --> 00:10:41,250 you see, these make up the components of f of u1. 207 00:10:41,250 --> 00:10:43,950 These make up the components of f of un. 208 00:10:43,950 --> 00:10:47,082 That's what was called the matrix A. 209 00:10:47,082 --> 00:10:53,040 v itself was the n-tuple x1 up to xn, which became the column 210 00:10:53,040 --> 00:10:58,210 matrix X when we wrote it as a column vector. 211 00:10:58,210 --> 00:10:59,410 Remember that. 212 00:10:59,410 --> 00:11:02,070 And then what we said was the transpose 213 00:11:02,070 --> 00:11:05,340 of that would be c times this n-tuple. 214 00:11:05,340 --> 00:11:08,310 But if I write this n-tuple as a column vector, 215 00:11:08,310 --> 00:11:10,530 I don't need the transpose in here. 216 00:11:10,530 --> 00:11:13,110 In other words, in matrix language, 217 00:11:13,110 --> 00:11:16,080 with A being this matrix and X being 218 00:11:16,080 --> 00:11:19,530 this column matrix, this translates into the matrix 219 00:11:19,530 --> 00:11:24,000 equation A times X equals c times X, 220 00:11:24,000 --> 00:11:27,810 where we recall that the matrix A is what's given, 221 00:11:27,810 --> 00:11:30,930 and what we're trying to find is, first of all, 222 00:11:30,930 --> 00:11:35,900 are there any vectors X that are mapped into a scalar times X? 223 00:11:35,900 --> 00:11:38,010 In other words, are any column matrices 224 00:11:38,010 --> 00:11:41,100 that are mapped into a scalar times that column matrix 225 00:11:41,100 --> 00:11:42,525 with respect to A? 226 00:11:42,525 --> 00:11:46,950 And secondly, if there are such column matrices, what 227 00:11:46,950 --> 00:11:49,380 values of c correspond to that? 228 00:11:49,380 --> 00:11:52,410 Well, notice, by ordinary algebraic techniques-- 229 00:11:52,410 --> 00:11:57,180 because matrices do obey many of the ordinary rules of algebra-- 230 00:11:57,180 --> 00:12:04,530 AX equals cX is the same as saying AX minus cX is 0. 231 00:12:04,530 --> 00:12:06,960 Notice that we already know that matrices 232 00:12:06,960 --> 00:12:09,720 obey the distributive rule. 233 00:12:09,720 --> 00:12:12,780 In other words, I could factor out the matrix X from here. 234 00:12:12,780 --> 00:12:14,700 Of course, I have to be very, very careful. 235 00:12:14,700 --> 00:12:17,130 Notice that capital A is a matrix. 236 00:12:17,130 --> 00:12:18,990 Little c is a scalar. 237 00:12:18,990 --> 00:12:22,510 And to have A minus c wouldn't make much sense. 238 00:12:22,510 --> 00:12:25,110 In other words, since A is an n-by-n matrix, 239 00:12:25,110 --> 00:12:27,060 I want whatever I'm subtracting from it 240 00:12:27,060 --> 00:12:29,620 to also be an n-by-n matrix. 241 00:12:29,620 --> 00:12:33,360 So what I do is the little cute device of remembering 242 00:12:33,360 --> 00:12:36,390 the property of the identity matrix I sub n. 243 00:12:36,390 --> 00:12:41,350 I simply replace X by I sub n times X-- in other words, 244 00:12:41,350 --> 00:12:44,900 the identity matrix times X. That now says what? 245 00:12:44,900 --> 00:12:49,350 AX minus c, identity matrix n-by-n-- 246 00:12:49,350 --> 00:12:51,780 identity matrix times X equals 0. 247 00:12:51,780 --> 00:12:55,260 Now I can factor out the X. And I have what? 248 00:12:55,260 --> 00:12:59,220 The matrix A minus c times the identity matrix 249 00:12:59,220 --> 00:13:03,030 times the column matrix X equals 0. 250 00:13:03,030 --> 00:13:04,920 Now, remember, back when we were first 251 00:13:04,920 --> 00:13:08,160 talking about matrix algebra, we pointed out 252 00:13:08,160 --> 00:13:11,730 that matrices obey the same structure. 253 00:13:11,730 --> 00:13:16,200 Matrices obey the same structure with certain small reservations 254 00:13:16,200 --> 00:13:17,460 that numbers obey. 255 00:13:17,460 --> 00:13:22,890 For example, we saw that if A was not the zero matrix, 256 00:13:22,890 --> 00:13:27,690 AX equals zero did not imply that X equals zero like it 257 00:13:27,690 --> 00:13:29,220 did in ordinary arithmetic. 258 00:13:29,220 --> 00:13:32,590 But it did if A happened to be a non-singular matrix. 259 00:13:32,590 --> 00:13:34,410 In other words, what we did show was 260 00:13:34,410 --> 00:13:38,730 that if this particular matrix had an inverse, then, 261 00:13:38,730 --> 00:13:41,310 multiplying both sides of this equation 262 00:13:41,310 --> 00:13:43,470 by the inverse of this, the inverse 263 00:13:43,470 --> 00:13:47,640 would cancel this factor, and we'd be left with X equals 0. 264 00:13:47,640 --> 00:13:52,980 In other words, if A minus cI inverse exists, 265 00:13:52,980 --> 00:13:56,260 then X must be the zero column matrix. 266 00:13:56,260 --> 00:13:57,600 Now, remember what X is. 267 00:13:57,600 --> 00:14:01,515 X is the column matrix whose entries 268 00:14:01,515 --> 00:14:05,610 are the components of the v that we're looking for over here. 269 00:14:05,610 --> 00:14:08,700 Keep in mind that we were looking for a v which 270 00:14:08,700 --> 00:14:10,830 was unequal to zero. 271 00:14:10,830 --> 00:14:13,650 If v is unequal to 0, in particular, at least one 272 00:14:13,650 --> 00:14:16,560 of its components must be different from 0. 273 00:14:16,560 --> 00:14:19,320 So what we're saying is, if A-- 274 00:14:19,320 --> 00:14:23,250 if this matrix here, with its inverse, exists, 275 00:14:23,250 --> 00:14:26,190 then X must be the zero column matrix, 276 00:14:26,190 --> 00:14:29,050 which is the solution that we don't want. 277 00:14:29,050 --> 00:14:32,340 In other words, we won't get non-zero solutions. 278 00:14:32,340 --> 00:14:34,260 Or from a different point of view, 279 00:14:34,260 --> 00:14:37,800 what this says is, if we want to be able to find a column 280 00:14:37,800 --> 00:14:42,780 vector X which is not the zero column vector, in particular, 281 00:14:42,780 --> 00:14:46,750 A minus cI had better be a singular matrix. 282 00:14:46,750 --> 00:14:48,960 In other words, it's inverse doesn't exist. 283 00:14:48,960 --> 00:14:53,400 And as we saw back in block 4, when a matrix is singular, 284 00:14:53,400 --> 00:14:55,680 it means that its determinant is 0. 285 00:14:55,680 --> 00:14:58,890 Consequently, in order for there to be any chance 286 00:14:58,890 --> 00:15:03,210 that we can find non-zero solutions of this equation, 287 00:15:03,210 --> 00:15:08,190 it must be that the determinant of A minus cI must be 0. 288 00:15:08,190 --> 00:15:11,560 And by the way, what does A minus cI look like? 289 00:15:11,560 --> 00:15:15,430 Notice that I is the n-by-n identity matrix. 290 00:15:15,430 --> 00:15:19,860 When you multiply a matrix by a scalar, 291 00:15:19,860 --> 00:15:24,240 you multiply each entry of that matrix by that scalar. 292 00:15:24,240 --> 00:15:29,250 Since all of the entries off the diagonal are 0, c times 0 293 00:15:29,250 --> 00:15:30,520 will still be 0. 294 00:15:30,520 --> 00:15:35,140 So notice that c times the n-by-n identity matrix is just 295 00:15:35,140 --> 00:15:41,050 the n-by-n diagonal matrix, each of whose diagonal elements 296 00:15:41,050 --> 00:15:43,330 is c. 297 00:15:43,330 --> 00:15:46,870 And now, remembering what A is, and remembering 298 00:15:46,870 --> 00:15:50,170 how we subtract two matrices, we subtract 299 00:15:50,170 --> 00:15:52,840 them component-by-component. 300 00:15:52,840 --> 00:15:57,640 Notice that the only non-zero components of cI are the c's 301 00:15:57,640 --> 00:15:58,990 down the diagonal. 302 00:15:58,990 --> 00:16:04,480 What this says is, if we take our matrix A, A minus c In is 303 00:16:04,480 --> 00:16:07,210 simply, from a manipulative point of view, 304 00:16:07,210 --> 00:16:11,590 obtained by subtracting c from each of the diagonal elements. 305 00:16:11,590 --> 00:16:17,230 You see, it's a1,1 minus c, a2,2 minus c. 306 00:16:17,230 --> 00:16:19,600 But every place else, you're subtracting 0, 307 00:16:19,600 --> 00:16:21,940 because the entry here is 0. 308 00:16:21,940 --> 00:16:24,130 So this is what this matrix looks like. 309 00:16:24,130 --> 00:16:26,530 I want its determinant to be 0. 310 00:16:26,530 --> 00:16:30,490 Last time, we showed how we computed a determinant. 311 00:16:30,490 --> 00:16:33,400 Notice that the a's are given numbers. 312 00:16:33,400 --> 00:16:35,500 c is the only unknown. 313 00:16:35,500 --> 00:16:39,130 If we expand this determinant and equate it to 0, 314 00:16:39,130 --> 00:16:42,700 we get an nth degree polynomial in c. 315 00:16:42,700 --> 00:16:46,130 I'm not going to go into much detail about that now. 316 00:16:46,130 --> 00:16:49,090 In fact, I'm not going to go into any detail about this now. 317 00:16:49,090 --> 00:16:51,250 I will save that for the exercises. 318 00:16:51,250 --> 00:16:54,490 But what I will do is take the very simple case, where 319 00:16:54,490 --> 00:16:56,500 we have a two-dimensional vector space, 320 00:16:56,500 --> 00:16:59,710 and apply this theory to the two-by-two case. 321 00:16:59,710 --> 00:17:04,000 And I think that the easiest example to pick, in this case, 322 00:17:04,000 --> 00:17:08,290 is the same example as we started with-- example 1-- 323 00:17:08,290 --> 00:17:09,819 and revisit it. 324 00:17:09,819 --> 00:17:12,250 In other words, this is called "example 1 revisited." 325 00:17:12,250 --> 00:17:13,089 That doesn't sound-- 326 00:17:13,089 --> 00:17:14,920 Let's just call it example 2. 327 00:17:14,920 --> 00:17:18,460 In example 2, you were thinking of a two-dimensional space 328 00:17:18,460 --> 00:17:20,290 relative to a particular basis. 329 00:17:20,290 --> 00:17:26,470 The 2-tuple x comma y got mapped into x plus 4y comma x plus y. 330 00:17:26,470 --> 00:17:30,780 In particular, 1 comma 0 got mapped into 1 comma 1. 331 00:17:30,780 --> 00:17:33,640 0 comma 1 got mapped into 4 comma 1. 332 00:17:33,640 --> 00:17:35,410 The only reason I've left the bar off 333 00:17:35,410 --> 00:17:37,300 here is so that you don't get the feeling 334 00:17:37,300 --> 00:17:39,610 that this has to be interpreted geometrically. 335 00:17:39,610 --> 00:17:41,740 This could be any two-dimensional space 336 00:17:41,740 --> 00:17:43,960 relative to any basis. 337 00:17:43,960 --> 00:17:46,510 But at any rate, using this example, 338 00:17:46,510 --> 00:17:50,470 remembering how the matrix A is obtained, the first column of A 339 00:17:50,470 --> 00:17:51,850 are these components. 340 00:17:51,850 --> 00:17:52,990 That's 1, 1. 341 00:17:52,990 --> 00:17:56,110 The second column of A are these components-- 342 00:17:56,110 --> 00:17:57,040 4, 1. 343 00:17:57,040 --> 00:18:00,760 So the matrix A associated with f 344 00:18:00,760 --> 00:18:04,900 relative to our given basis that represents the 2-tuples here 345 00:18:04,900 --> 00:18:07,390 is 1, 4, 1, 1. 346 00:18:07,390 --> 00:18:11,180 If I want to now look at A minus c I2-- 347 00:18:11,180 --> 00:18:13,390 see, n is 2, in this case-- the two-by-two identity 348 00:18:13,390 --> 00:18:14,750 matrix-- what do I do? 349 00:18:14,750 --> 00:18:20,470 I just subtract c from each diagonal element this way 350 00:18:20,470 --> 00:18:23,620 so the determinant of that is this determinant. 351 00:18:23,620 --> 00:18:26,120 We already know how to expand a two-by-two determinant. 352 00:18:26,120 --> 00:18:29,320 It's this times this minus this times this. 353 00:18:29,320 --> 00:18:32,500 This is what? c squared minus 2c. 354 00:18:32,500 --> 00:18:35,050 Plus 1 minus 4 is minus 3. 355 00:18:35,050 --> 00:18:37,780 This is c squared minus 2c minus 3. 356 00:18:37,780 --> 00:18:41,140 And therefore, the only way this determinant can be 0 357 00:18:41,140 --> 00:18:42,580 is if this is 0. 358 00:18:42,580 --> 00:18:46,040 This factors into c minus 3 times c plus 1. 359 00:18:46,040 --> 00:18:50,240 Therefore, it must be that c is 3 or c is minus 1. 360 00:18:50,240 --> 00:18:52,630 In other words, the only possible characteristic-- or, 361 00:18:52,630 --> 00:18:57,970 eigenvalues, in this problem, are 3 and minus 1. 362 00:18:57,970 --> 00:19:00,700 And we'll see what that means in a moment in terms of looking 363 00:19:00,700 --> 00:19:02,000 at these cases separately. 364 00:19:02,000 --> 00:19:04,000 You may have noticed I made a slip of the tongue 365 00:19:04,000 --> 00:19:07,720 and said characteristic values instead of eigenvalues. 366 00:19:07,720 --> 00:19:12,130 You will find, in many textbooks, the word eigenvalue. 367 00:19:12,130 --> 00:19:14,980 In other books, you'll find characteristic value. 368 00:19:14,980 --> 00:19:17,650 These terms are used interchangeably. 369 00:19:17,650 --> 00:19:20,590 Eigenvalue was the German translation 370 00:19:20,590 --> 00:19:22,150 of characteristic value. 371 00:19:22,150 --> 00:19:25,120 So use these terms interchangeably, all right? 372 00:19:25,120 --> 00:19:27,010 But let's take a look at what happens 373 00:19:27,010 --> 00:19:30,640 in this particular example, in the case that c equals 3. 374 00:19:30,640 --> 00:19:34,810 If c equals 3, this tells me that my vector v 375 00:19:34,810 --> 00:19:38,050 is determined by f of v is 3 times v, where 376 00:19:38,050 --> 00:19:40,920 v is the 2-tuple x comma y. 377 00:19:40,920 --> 00:19:43,420 Writing out what this means in terms of my matrix now, 378 00:19:43,420 --> 00:19:45,985 my matrix A is 1, 4, 1, 1. 379 00:19:45,985 --> 00:19:48,340 v written as a column vector is this. 380 00:19:48,340 --> 00:19:51,640 And 3 times this column vector is this. 381 00:19:51,640 --> 00:19:53,980 Remembering that to compare two matrices to be equal, 382 00:19:53,980 --> 00:19:56,230 they must be equal entry by entry, 383 00:19:56,230 --> 00:19:58,870 the first entry of this product is what? 384 00:19:58,870 --> 00:20:01,450 It's x plus 4y. 385 00:20:01,450 --> 00:20:04,750 The second entry is x plus y. 386 00:20:04,750 --> 00:20:08,440 Therefore, it must be that x plus 4y equals 3x, 387 00:20:08,440 --> 00:20:11,020 and x plus y equals 3y. 388 00:20:11,020 --> 00:20:14,470 And both of these two conditions together say, quite simply, 389 00:20:14,470 --> 00:20:16,440 that x equals 2y. 390 00:20:16,440 --> 00:20:18,230 And what does that tell us? 391 00:20:18,230 --> 00:20:23,120 It says that if you take any 2-tuple of the form x comma y, 392 00:20:23,120 --> 00:20:24,820 where x is twice y-- 393 00:20:24,820 --> 00:20:30,230 in other words, if you take the set of all 2-tuples 2y comma y, 394 00:20:30,230 --> 00:20:32,570 these are eigenvectors. 395 00:20:32,570 --> 00:20:36,690 And they correspond to the eigenvalue 3. 396 00:20:36,690 --> 00:20:39,740 And I'm going to show you that pictorially in a few moments. 397 00:20:39,740 --> 00:20:41,690 I just want you to get used to the computation 398 00:20:41,690 --> 00:20:43,130 here for the time being. 399 00:20:43,130 --> 00:20:46,010 Secondly, if you want to think of this geometrically, 400 00:20:46,010 --> 00:20:50,840 x equals 2y doesn't have to be viewed as a set of vectors. 401 00:20:50,840 --> 00:20:54,500 It can be viewed as a line in the plane. 402 00:20:54,500 --> 00:20:58,190 And what this says is that f preserves the direction 403 00:20:58,190 --> 00:21:00,800 of the line x equals 2y. 404 00:21:00,800 --> 00:21:03,980 Oh, just as a quick check over here-- 405 00:21:03,980 --> 00:21:05,780 whichever interpretation you want-- 406 00:21:05,780 --> 00:21:09,020 notice that if you replace x by 2y-- 407 00:21:09,020 --> 00:21:11,930 remember what the definition of f was? 408 00:21:11,930 --> 00:21:13,190 f was what? 409 00:21:13,190 --> 00:21:17,330 f of x, y was x plus 4y comma x plus y. 410 00:21:17,330 --> 00:21:22,190 So if x is 2y, this becomes 6y comma 3y. 411 00:21:22,190 --> 00:21:26,660 In other words, f of 2y comma y is 6y comma 3y. 412 00:21:26,660 --> 00:21:30,200 That's the same as 3 times 2y comma y. 413 00:21:30,200 --> 00:21:32,640 And by the way, notice that the special case 414 00:21:32,640 --> 00:21:36,860 y equals 1 corresponded to part of our example number 415 00:21:36,860 --> 00:21:39,800 1, when we showed that f bar of 2 comma 416 00:21:39,800 --> 00:21:43,220 1 was three times 2 comma 1. 417 00:21:43,220 --> 00:21:45,980 In a similar way, c equals minus 1 418 00:21:45,980 --> 00:21:48,230 is the other characteristic value. 419 00:21:48,230 --> 00:21:52,520 Namely, if c equals minus 1, the equation f of v 420 00:21:52,520 --> 00:21:56,720 equals cv becomes f of v equals minus v. 421 00:21:56,720 --> 00:22:00,710 And in matrix language, that's AX equals minus X. 422 00:22:00,710 --> 00:22:03,710 Recalling what A and X are from before, 423 00:22:03,710 --> 00:22:08,390 remember, A times X will be the column matrix x plus 4y, 424 00:22:08,390 --> 00:22:10,070 x plus y. 425 00:22:10,070 --> 00:22:13,970 Minus X is the column matrix minus x minus y. 426 00:22:13,970 --> 00:22:18,020 Equating corresponding entries, we get this pair of equations. 427 00:22:18,020 --> 00:22:19,730 And notice that both of these equations 428 00:22:19,730 --> 00:22:22,670 say that x must equal minus 2y. 429 00:22:22,670 --> 00:22:26,180 In other words, if the x component is minus twice the y 430 00:22:26,180 --> 00:22:28,430 component, relative to the given basis 431 00:22:28,430 --> 00:22:30,530 that we're talking about here, notice 432 00:22:30,530 --> 00:22:35,090 that the set of all 2-tuples of the form minus 2y comma y 433 00:22:35,090 --> 00:22:37,400 are eigenvectors, in this case, and 434 00:22:37,400 --> 00:22:40,420 the corresponding eigenvalue is minus 1, 435 00:22:40,420 --> 00:22:43,490 and that f preserves the direction 436 00:22:43,490 --> 00:22:46,990 of the line x equals minus 2y. 437 00:22:46,990 --> 00:22:49,010 And I think, now, the time has come 438 00:22:49,010 --> 00:22:50,450 to show you what this thing means 439 00:22:50,450 --> 00:22:53,270 in terms of a simple geometric interpretation. 440 00:22:53,270 --> 00:22:55,400 Keep in mind, many of the applications 441 00:22:55,400 --> 00:22:58,070 of eigenvalues and eigenvectors come up 442 00:22:58,070 --> 00:23:02,010 in boundary value problems of partial differential equations. 443 00:23:02,010 --> 00:23:04,130 I will show you, in one of our exercises, 444 00:23:04,130 --> 00:23:07,730 that even our linear homogeneous differential equations may be 445 00:23:07,730 --> 00:23:09,710 viewed as eigenvector problems. 446 00:23:09,710 --> 00:23:11,330 They come up in many applications. 447 00:23:11,330 --> 00:23:13,730 But I'm saying, in terms of the spirit of a game, 448 00:23:13,730 --> 00:23:16,310 let's take the simplest physical interpretation. 449 00:23:16,310 --> 00:23:19,970 And that's simply the mapping of the xy-plane into the uv-plane. 450 00:23:19,970 --> 00:23:22,970 And all we're saying is that the mapping f bar 451 00:23:22,970 --> 00:23:25,550 that we're talking about-- what mapping are we talking about? 452 00:23:25,550 --> 00:23:28,130 The mapping that carries x, y into-- 453 00:23:28,130 --> 00:23:32,720 what was it that's written down here? x plus 4y comma x plus y. 454 00:23:32,720 --> 00:23:35,720 What that mapping does is it changes the direction 455 00:23:35,720 --> 00:23:37,430 of most lines in the plane. 456 00:23:37,430 --> 00:23:40,880 But there are two lines that it leaves alone in direction. 457 00:23:40,880 --> 00:23:44,870 Namely, the line x equal 2y gets mapped into the line u 458 00:23:44,870 --> 00:23:48,170 equals 2v, and the line x equals minus 2y 459 00:23:48,170 --> 00:23:51,200 gets mapped into the line u equals minus 2v. 460 00:23:51,200 --> 00:23:54,770 By the way, notice we are not saying that the points remain 461 00:23:54,770 --> 00:23:55,790 fixed here. 462 00:23:55,790 --> 00:23:59,630 Remember that the characteristic value corresponding 463 00:23:59,630 --> 00:24:01,955 to this eigenvector was 3. 464 00:24:01,955 --> 00:24:05,780 In other words, notice that 2 comma 1 doesn't get 465 00:24:05,780 --> 00:24:07,670 mapped into 2 comma 1 here. 466 00:24:07,670 --> 00:24:10,340 It got mapped into 6 comma 3-- 467 00:24:10,340 --> 00:24:12,950 that the characteristic value tells you, 468 00:24:12,950 --> 00:24:17,660 once you know what directions are preserved, how much 469 00:24:17,660 --> 00:24:19,230 the vector was stretched out. 470 00:24:19,230 --> 00:24:25,250 In other words, 2i plus j get stretched out into 6i plus 3j. 471 00:24:25,250 --> 00:24:27,500 Well, I'm not going to go into that in any more detail 472 00:24:27,500 --> 00:24:28,300 right now. 473 00:24:28,300 --> 00:24:31,220 All I do want to observe is that, if I 474 00:24:31,220 --> 00:24:34,230 was studying the particular mapping f bar, 475 00:24:34,230 --> 00:24:38,810 notice that the lines x equal 2y and x equal minus 2y are, 476 00:24:38,810 --> 00:24:44,000 in a way, a better coordinate system than the axes x and y, 477 00:24:44,000 --> 00:24:46,880 because notice that the x-axis and the y-axis 478 00:24:46,880 --> 00:24:49,880 have their directions changed under this mapping. 479 00:24:49,880 --> 00:24:53,420 But x equals 2y and x equals minus 2y don't 480 00:24:53,420 --> 00:24:55,260 have their directions changed. 481 00:24:55,260 --> 00:24:57,230 In fact, to look at this a different way, let's 482 00:24:57,230 --> 00:24:59,630 pick a representative vector from this line 483 00:24:59,630 --> 00:25:01,610 and a representative vector from this line. 484 00:25:01,610 --> 00:25:03,230 Let's take y to be 1. 485 00:25:03,230 --> 00:25:05,270 In this case, that would say x is 2. 486 00:25:05,270 --> 00:25:07,355 In this case, it says x is minus 2. 487 00:25:07,355 --> 00:25:12,010 Let's pick, as two new vectors, alpha 1 to be 2i plus j 488 00:25:12,010 --> 00:25:15,230 and alpha 2 to be minus 2i plus j. 489 00:25:15,230 --> 00:25:16,750 And my claim is-- 490 00:25:16,750 --> 00:25:18,380 I'll write them with arrows here, 491 00:25:18,380 --> 00:25:20,630 as long as we are going to think of this is a mapping. 492 00:25:20,630 --> 00:25:22,610 My claim is that alpha 1 and alpha 2 493 00:25:22,610 --> 00:25:26,000 is a very nice basis for E2 with respect 494 00:25:26,000 --> 00:25:28,560 to the linear transformation f. 495 00:25:28,560 --> 00:25:29,710 Well, y? 496 00:25:29,710 --> 00:25:32,110 Well, what do we already know about alpha 1? 497 00:25:32,110 --> 00:25:36,320 Alpha 1 is an eigenvector with characteristic value 3. 498 00:25:36,320 --> 00:25:40,720 In other words, f of alpha 1 is 3 alpha 1. 499 00:25:40,720 --> 00:25:43,870 Alpha 2 is also an eigenvector with characteristic value 500 00:25:43,870 --> 00:25:45,020 minus 1. 501 00:25:45,020 --> 00:25:48,360 So f of alpha 2 is minus alpha 2. 502 00:25:48,360 --> 00:25:51,520 Notice, then, therefore, from an algebraic point of view, 503 00:25:51,520 --> 00:25:55,270 if I pick alpha 1 and alpha 2 as my bases for-- well, 504 00:25:55,270 --> 00:25:56,770 I should be consistent here. 505 00:25:56,770 --> 00:26:01,600 I called this V. I suppose it should have been E2, 506 00:26:01,600 --> 00:26:03,520 simply to match the notation here. 507 00:26:03,520 --> 00:26:04,690 But that's not important. 508 00:26:04,690 --> 00:26:08,170 Suppose I pick alpha 1 and alpha 2 to be my new bases. 509 00:26:08,170 --> 00:26:15,316 Notice, you see, that f of alpha 1 is 3 alpha 1 plus 0 alpha 2. 510 00:26:15,316 --> 00:26:21,900 f of alpha 2 is 0 alpha 1 minus 1 alpha 2. 511 00:26:21,900 --> 00:26:27,590 So my matrix of f, relative to alpha 1 and alpha 2 as a basis, 512 00:26:27,590 --> 00:26:30,470 would have its first column being 3 and 0. 513 00:26:30,470 --> 00:26:34,230 It would have its second column being 0 and minus 1. 514 00:26:34,230 --> 00:26:39,020 In other words, the matrix now is a diagonal matrix, 3, 0, 0, 515 00:26:39,020 --> 00:26:40,340 minus 1. 516 00:26:40,340 --> 00:26:44,390 It's not only a diagonal matrix, but the diagonal elements 517 00:26:44,390 --> 00:26:47,660 themselves yield the eigenvalues. 518 00:26:47,660 --> 00:26:53,550 Notice how easy this matrix is to use for computing-- 519 00:26:53,550 --> 00:26:58,530 A times X-- if X happens to be written relative to the alphas, 520 00:26:58,530 --> 00:27:01,770 because the easiest type of matrix to multiply by 521 00:27:01,770 --> 00:27:03,900 is a diagonal matrix. 522 00:27:03,900 --> 00:27:08,520 And I'm not going to go through this here, but when you write-- 523 00:27:08,520 --> 00:27:12,260 when you pick the basis consisting of eigenvalues, 524 00:27:12,260 --> 00:27:15,960 eigenvectors, and write this diagonal matrix, 525 00:27:15,960 --> 00:27:18,600 the resulting diagonal matrix gives you 526 00:27:18,600 --> 00:27:21,240 a tremendous amount of insight as to what 527 00:27:21,240 --> 00:27:22,710 the space looks like. 528 00:27:22,710 --> 00:27:24,900 And I'll bring that out in the exercises. 529 00:27:24,900 --> 00:27:27,390 All I want you to get out of this overview 530 00:27:27,390 --> 00:27:31,510 is what eigenvectors are and how we compute them. 531 00:27:31,510 --> 00:27:33,510 And I thought that, to finish up with, 532 00:27:33,510 --> 00:27:37,980 I would like to give you a very, very profound result, which 533 00:27:37,980 --> 00:27:41,700 I won't prove for you, but which I will state-- 534 00:27:41,700 --> 00:27:43,500 has, also, a profound name. 535 00:27:43,500 --> 00:27:45,120 But I'll get to that in a moment. 536 00:27:45,120 --> 00:27:46,950 I call this an important aside. 537 00:27:46,950 --> 00:27:48,540 It really isn't an aside. 538 00:27:48,540 --> 00:27:53,200 It's the backbone of much of advanced matrix algebra. 539 00:27:53,200 --> 00:27:54,810 But the interesting thing is this. 540 00:27:54,810 --> 00:27:57,540 Remember, given an n-by-n matrix A, 541 00:27:57,540 --> 00:27:59,700 we were fooling around with looking 542 00:27:59,700 --> 00:28:05,040 at the determinant of A minus cI equaling 0 543 00:28:05,040 --> 00:28:08,037 and trying to find a scalar c that would do this for us. 544 00:28:08,037 --> 00:28:09,870 That's how we get the characteristic values. 545 00:28:09,870 --> 00:28:11,520 A was the given matrix. 546 00:28:11,520 --> 00:28:13,560 I was the given identity matrix. 547 00:28:13,560 --> 00:28:15,870 c was a scalar whose value we were 548 00:28:15,870 --> 00:28:21,270 trying to get the determinant of this matrix to be 0. 549 00:28:21,270 --> 00:28:25,980 The amazing point is that if you substitute the matrix 550 00:28:25,980 --> 00:28:30,885 A for c, in this equation, it will satisfy this equation. 551 00:28:30,885 --> 00:28:32,790 And what do I mean by that? 552 00:28:32,790 --> 00:28:38,790 Just replace c by A over here, and this equation is satisfied. 553 00:28:38,790 --> 00:28:40,470 By the way, that may look trivial. 554 00:28:40,470 --> 00:28:42,510 You may say to me, gee, whiz. 555 00:28:42,510 --> 00:28:43,440 What a big deal. 556 00:28:43,440 --> 00:28:45,720 If I take c and replace it by A, this 557 00:28:45,720 --> 00:28:47,760 is A times the identity matrix, which 558 00:28:47,760 --> 00:28:51,050 is still A. A minus A is the zero matrix. 559 00:28:51,050 --> 00:28:54,000 And the determinant of the zero matrix is clearly 0. 560 00:28:54,000 --> 00:28:59,930 The metaphysical thing here is, notice that c is a number. 561 00:28:59,930 --> 00:29:01,380 It's a scalar. 562 00:29:01,380 --> 00:29:04,650 And A is a matrix. 563 00:29:04,650 --> 00:29:07,620 Structurally, you cannot let c equal A. 564 00:29:07,620 --> 00:29:10,620 All we're saying is a remarkable result, 565 00:29:10,620 --> 00:29:13,830 that if you mechanically replace c by A, 566 00:29:13,830 --> 00:29:16,290 this equation is satisfied. 567 00:29:16,290 --> 00:29:19,680 And before I illustrate that for you, I've made a big decision. 568 00:29:19,680 --> 00:29:22,328 I'm going to tell you what this theorem is called. 569 00:29:22,328 --> 00:29:24,120 I wasn't originally going to tell you that. 570 00:29:24,120 --> 00:29:26,700 It's called the Cayley-Hamilton theorem. 571 00:29:26,700 --> 00:29:30,270 And by my telling you this name, you now know as much 572 00:29:30,270 --> 00:29:31,740 about the subject as I do. 573 00:29:31,740 --> 00:29:33,390 That's why I didn't want to tell you what the name was, 574 00:29:33,390 --> 00:29:35,910 so I'd still know something more than you did about it. 575 00:29:35,910 --> 00:29:37,770 But that's not too important. 576 00:29:37,770 --> 00:29:39,840 Let me illustrate how this thing works. 577 00:29:39,840 --> 00:29:45,120 Let's go back to our matrix 1, 4, 1, 1, all right? 578 00:29:45,120 --> 00:29:49,920 The determinant of A minus cI, we already saw, 579 00:29:49,920 --> 00:29:54,770 was c squared minus 2c minus 3. 580 00:29:54,770 --> 00:29:58,070 My claim is, if I replace c by A in here, 581 00:29:58,070 --> 00:30:01,410 this will still be obeyed, only with one slight modification. 582 00:30:01,410 --> 00:30:02,480 See, this becomes what? 583 00:30:02,480 --> 00:30:05,030 A squared minus 2A. 584 00:30:05,030 --> 00:30:08,900 And I can't write minus 3, because 3 is a number, not 585 00:30:08,900 --> 00:30:09,740 a matrix. 586 00:30:09,740 --> 00:30:11,390 It's always understood, when you're 587 00:30:11,390 --> 00:30:16,440 converting to matrix form, that the I is over here. 588 00:30:16,440 --> 00:30:17,900 And if you want to see why, you can 589 00:30:17,900 --> 00:30:23,060 think of this as being A to the 0, and think of the number 1 590 00:30:23,060 --> 00:30:25,070 as being c to the 0. 591 00:30:25,070 --> 00:30:26,690 In other words, structurally, this 592 00:30:26,690 --> 00:30:30,840 is A squared minus 2A minus 3A to the 0 power. 593 00:30:30,840 --> 00:30:35,000 This is c squared minus 2c minus 3c to the 0. 594 00:30:35,000 --> 00:30:37,790 And my claim is that this equation 595 00:30:37,790 --> 00:30:40,130 will be obeyed by the matrix A. 596 00:30:40,130 --> 00:30:42,240 Let's just check it out and see if it's true. 597 00:30:42,240 --> 00:30:45,180 Remember that A was the matrix 1, 4, 1, 1. 598 00:30:45,180 --> 00:30:47,660 To square it means multiply it by itself. 599 00:30:47,660 --> 00:30:50,960 If I go through the usual recipe for multiplying 600 00:30:50,960 --> 00:30:53,570 two two-by-two matrices, I very quickly 601 00:30:53,570 --> 00:30:56,460 see that the product is 5, 8, 2, 5. 602 00:30:56,460 --> 00:31:01,490 Notice that, since A is 1, 4, 1, 1, multiplying by minus 2 603 00:31:01,490 --> 00:31:03,840 multiplies each entry by minus 2. 604 00:31:03,840 --> 00:31:06,540 So minus 2A is this matrix. 605 00:31:06,540 --> 00:31:09,620 Notice that minus 3 times the identity matrix 606 00:31:09,620 --> 00:31:13,210 is a diagonal matrix that has minus 3 607 00:31:13,210 --> 00:31:15,580 as each main diagonal element-- 608 00:31:15,580 --> 00:31:17,040 in other words, this matrix here. 609 00:31:17,040 --> 00:31:19,550 And notice, now, just for the sake of argument, 610 00:31:19,550 --> 00:31:22,250 if I add these up, what do I get? 611 00:31:22,250 --> 00:31:26,720 5 minus 2 minus 3, which is 0, 8 minus 8 plus 0, 612 00:31:26,720 --> 00:31:31,820 which is 0, 2 minus 2 plus 0, which is 0, 5 minus 2 minus 3, 613 00:31:31,820 --> 00:31:32,720 which is 0. 614 00:31:32,720 --> 00:31:37,910 In other words, this sum is the zero matrix, not 615 00:31:37,910 --> 00:31:39,710 the zero number. 616 00:31:39,710 --> 00:31:43,190 You see, technically speaking here, in this equation here, 617 00:31:43,190 --> 00:31:46,190 the 0 refers to a number, because the determinant 618 00:31:46,190 --> 00:31:47,400 is a number. 619 00:31:47,400 --> 00:31:49,010 But here, we're talking about, it 620 00:31:49,010 --> 00:31:52,100 satisfied in matrix language. 621 00:31:52,100 --> 00:31:56,910 And what this means is that matrices can now 622 00:31:56,910 --> 00:31:59,035 be reduced by long division. 623 00:31:59,035 --> 00:32:00,660 So I'll give you a very simple example. 624 00:32:00,660 --> 00:32:02,880 But what the main impact of this is, 625 00:32:02,880 --> 00:32:06,710 I can now invent power series of matrices. 626 00:32:06,710 --> 00:32:10,410 In other words, I can define e to the x, where 627 00:32:10,410 --> 00:32:14,520 x is a matrix, to be 1 plus x plus x squared over 628 00:32:14,520 --> 00:32:18,600 2 factorial plus x cubed over 3 factorial, et cetera, the same 629 00:32:18,600 --> 00:32:20,220 as we did in scalar cases. 630 00:32:20,220 --> 00:32:24,940 And the main reason is that, once I'm given a matrix, 631 00:32:24,940 --> 00:32:29,520 and I find the basic polynomial equation that it satisfies, 632 00:32:29,520 --> 00:32:31,800 I can reduce every matrix to that. 633 00:32:31,800 --> 00:32:33,360 Let me give you an example. 634 00:32:33,360 --> 00:32:35,040 The key thing I want you to keep in mind 635 00:32:35,040 --> 00:32:38,400 here is that we already know, for this particular matrix A, 636 00:32:38,400 --> 00:32:43,140 that A squared minus 2A minus 3I is the zero matrix. 637 00:32:43,140 --> 00:32:45,747 Suppose, now, I wanted to compute A cubed. 638 00:32:45,747 --> 00:32:47,330 Now, in this assignment here I'm going 639 00:32:47,330 --> 00:32:51,480 to show you how I can reduce matrices by long division. 640 00:32:51,480 --> 00:32:53,820 And in the exercises, I'll actually 641 00:32:53,820 --> 00:32:55,740 do the long division for you. 642 00:32:55,740 --> 00:32:59,250 But what is long division in factoring form? 643 00:32:59,250 --> 00:33:03,180 What I'm saying is, I know that A squared minus 2A minus 3I 644 00:33:03,180 --> 00:33:04,380 is 0. 645 00:33:04,380 --> 00:33:06,930 So I would like to write A cubed in such a way 646 00:33:06,930 --> 00:33:10,350 that I can factor out an A squared minus 2A minus 3I. 647 00:33:10,350 --> 00:33:13,950 The way I do that is I notice that I must multiply A squared 648 00:33:13,950 --> 00:33:16,560 by A in order to get A cubed. 649 00:33:16,560 --> 00:33:19,920 The trouble is, when I multiply, I now have a minus 2A 650 00:33:19,920 --> 00:33:22,350 squared on the right-hand side that I don't want, 651 00:33:22,350 --> 00:33:23,580 because it's not here. 652 00:33:23,580 --> 00:33:26,100 And I have a minus 3A on the right-hand side that I don't 653 00:33:26,100 --> 00:33:27,970 want, because it's not here. 654 00:33:27,970 --> 00:33:31,740 So to compensate for that, I simply add on a 2A squared, 655 00:33:31,740 --> 00:33:33,030 and I add on a 3A. 656 00:33:33,030 --> 00:33:34,530 In other words, even though this may 657 00:33:34,530 --> 00:33:36,120 look like a funny way of doing it, 658 00:33:36,120 --> 00:33:41,140 a very funny way of writing A cubed is this expression here. 659 00:33:41,140 --> 00:33:43,290 And the reason that I choose this expression is, 660 00:33:43,290 --> 00:33:49,170 notice that this being 0 means that A cubed is just 2A squared 661 00:33:49,170 --> 00:33:51,060 plus 3A. 662 00:33:51,060 --> 00:33:54,690 Moreover, notice that I can still get the structural form A 663 00:33:54,690 --> 00:33:59,760 squared minus 2A minus 3I out of this thing by writing it. 664 00:33:59,760 --> 00:34:03,730 Seeing that my first term here is going to be 2A squared-- 665 00:34:03,730 --> 00:34:05,220 so I put a 2 over here-- 666 00:34:05,220 --> 00:34:07,860 this gives me my 2A squared, which I want. 667 00:34:07,860 --> 00:34:09,929 This gives me a minus 4A. 668 00:34:09,929 --> 00:34:12,659 But I want to have a plus 3A, so I add 669 00:34:12,659 --> 00:34:15,510 on 7A to compensate for that. 670 00:34:15,510 --> 00:34:19,120 This gives me a minus 6I, which I don't have up here. 671 00:34:19,120 --> 00:34:22,050 So to wipe that out, I add on a 6I. 672 00:34:22,050 --> 00:34:24,270 In other words, another way of writing I cubed, 673 00:34:24,270 --> 00:34:27,600 therefore, is this plus this. 674 00:34:27,600 --> 00:34:32,010 And since this is 0, this says that A cubed is nothing 675 00:34:32,010 --> 00:34:34,770 more than 7A plus 6I. 676 00:34:34,770 --> 00:34:36,969 In fact, in this particular case, 677 00:34:36,969 --> 00:34:40,710 notice that any power of A can be reduced 678 00:34:40,710 --> 00:34:44,340 to a linear combination of A and I, 679 00:34:44,340 --> 00:34:48,120 because as long as I have even a quadratic power in here, 680 00:34:48,120 --> 00:34:50,330 I can continue my long division. 681 00:34:50,330 --> 00:34:52,650 It's just like finding a remainder 682 00:34:52,650 --> 00:34:54,270 in ordinary long division. 683 00:34:54,270 --> 00:34:58,620 You keep on going until the remainder is of a lower 684 00:34:58,620 --> 00:35:00,835 degree than the divisor. 685 00:35:00,835 --> 00:35:02,460 In this particular case, I've shown you 686 00:35:02,460 --> 00:35:05,070 that A cubed is 7A plus 6I. 687 00:35:05,070 --> 00:35:07,440 And I picked an easy case just so we could check it. 688 00:35:07,440 --> 00:35:12,060 Notice that we already know that A squared is 5, 8, 2, 5. 689 00:35:12,060 --> 00:35:13,920 A is 1, 4, 1, 1. 690 00:35:13,920 --> 00:35:16,200 So A cubed is this times this. 691 00:35:16,200 --> 00:35:18,450 Multiplying these two-by-two's together, 692 00:35:18,450 --> 00:35:20,640 I get this particular matrix. 693 00:35:20,640 --> 00:35:25,890 On the other hand, knowing that A is 1, 4, 1, 1, 7A 694 00:35:25,890 --> 00:35:31,230 is this matrix, 6I is this matrix-- 695 00:35:31,230 --> 00:35:33,600 and if I now add 7A and 6I-- 696 00:35:33,600 --> 00:35:34,620 how do I add? 697 00:35:34,620 --> 00:35:36,030 Component-by-component. 698 00:35:36,030 --> 00:35:43,940 I get 7 plus 6, which is 13, 28 plus 0, which is 28, 7 plus 0, 699 00:35:43,940 --> 00:35:47,480 which is 7, 7 plus 6, which is 13. 700 00:35:47,480 --> 00:35:53,130 In other words, 7A plus 6I is, indeed, A cubed, as asserted. 701 00:35:53,130 --> 00:35:56,570 And I think you can see now why I wanted to end here. 702 00:35:56,570 --> 00:36:00,950 From here on in, the course becomes a very, very technical 703 00:36:00,950 --> 00:36:03,980 subject and one that's used best in conjunction 704 00:36:03,980 --> 00:36:07,370 with advanced math courses that are using these techniques. 705 00:36:07,370 --> 00:36:09,710 So we come to the end of part 2. 706 00:36:09,710 --> 00:36:11,780 I want to tell you what an enjoyable experience 707 00:36:11,780 --> 00:36:13,550 it was teaching you all. 708 00:36:13,550 --> 00:36:15,950 If nothing else, as I told you after part 1, 709 00:36:15,950 --> 00:36:19,670 I emerge smarter, because it takes a lot of preparation 710 00:36:19,670 --> 00:36:21,680 to get these boards pre-written. 711 00:36:21,680 --> 00:36:23,720 I couldn't do it alone, and I would like-- 712 00:36:23,720 --> 00:36:27,050 in addition to the camera people, the floor people, 713 00:36:27,050 --> 00:36:30,540 there are three people who work very closely with this project 714 00:36:30,540 --> 00:36:32,090 that I would like to single out. 715 00:36:32,090 --> 00:36:34,490 I would like to thank, especially, John Fitch, who 716 00:36:34,490 --> 00:36:38,270 is the manager of our self-study project, who also doubles 717 00:36:38,270 --> 00:36:42,970 in as director and producer of the tape, the film series, 718 00:36:42,970 --> 00:36:45,320 and is also my advisor for the study guide 719 00:36:45,320 --> 00:36:46,640 and things of the like. 720 00:36:46,640 --> 00:36:49,340 I would like to thank Charles Patton, who 721 00:36:49,340 --> 00:36:53,960 is the one responsible the most for the clear pictures 722 00:36:53,960 --> 00:36:57,320 and the excellent photogenic features that you notice of me, 723 00:36:57,320 --> 00:36:59,240 the sharpness of the camera. 724 00:36:59,240 --> 00:37:03,080 I would also like to thank Elise Pelletier, who, 725 00:37:03,080 --> 00:37:05,660 in addition to being a very able secretary, 726 00:37:05,660 --> 00:37:08,930 doubles in in the master control room as the master everything, 727 00:37:08,930 --> 00:37:12,620 from running the video tape recorder to making hasty phone 728 00:37:12,620 --> 00:37:14,600 calls and things of this type. 729 00:37:14,600 --> 00:37:17,660 I would also like to thank two other colleagues, Arthur 730 00:37:17,660 --> 00:37:20,330 [INAUDIBLE] and Paul Brown, administrative offices 731 00:37:20,330 --> 00:37:23,540 at the center, who have provided me most excellent working 732 00:37:23,540 --> 00:37:26,750 conditions, and finally, Harold [INAUDIBLE],, who 733 00:37:26,750 --> 00:37:28,760 was the first director of the center 734 00:37:28,760 --> 00:37:32,540 and whose idea it was to produce Calculus Revisited Part 735 00:37:32,540 --> 00:37:34,430 1 and Part 2. 736 00:37:34,430 --> 00:37:35,910 It has been my pleasure. 737 00:37:35,910 --> 00:37:38,560 I hope that our paths cross again. 738 00:37:38,560 --> 00:37:40,880 But until such a time, God bless you all. 739 00:37:43,530 --> 00:37:45,930 Funding for the publication of this video 740 00:37:45,930 --> 00:37:50,820 was provided by the Gabriella and Paul Rosenbaum Foundation. 741 00:37:50,820 --> 00:37:54,960 Help OCW continue to provide free and open access to MIT 742 00:37:54,960 --> 00:38:00,246 courses by making a donation at ocw.mit.edu/donate.