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 --> 00:00:17,270 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,270 --> 00:00:18,200 at ocw.mit.edu. 8 00:00:32,082 --> 00:00:33,070 HERBERT GROSS: Hi. 9 00:00:33,070 --> 00:00:35,740 As I was getting myself prepared for the lecture, 10 00:00:35,740 --> 00:00:38,320 an old shaggy dog science story came 11 00:00:38,320 --> 00:00:41,380 to mind that's usually told as a tribute 12 00:00:41,380 --> 00:00:44,110 to the German scientist's thoroughness. 13 00:00:44,110 --> 00:00:47,110 This was the story of when all the scientists of the world 14 00:00:47,110 --> 00:00:49,840 got together and decided for an annual project. 15 00:00:49,840 --> 00:00:53,650 Each country would study exhaustively a different animal 16 00:00:53,650 --> 00:00:56,140 and report at the end of the year 17 00:00:56,140 --> 00:00:57,490 as to what they had found out. 18 00:00:57,490 --> 00:01:00,430 And at the end of the year, every country but Germany 19 00:01:00,430 --> 00:01:01,810 had been heard from. 20 00:01:01,810 --> 00:01:04,330 No one knew what had happened to the German scientists. 21 00:01:04,330 --> 00:01:09,250 And five years later, the German report came in, a huge epitome, 22 00:01:09,250 --> 00:01:13,860 and it was entitled, "Handbook of the Elephant, Volume 1." 23 00:01:13,860 --> 00:01:16,820 And the reason that this story comes to mind is twofold. 24 00:01:16,820 --> 00:01:19,750 First of all, the study of vector spaces 25 00:01:19,750 --> 00:01:25,390 has so many ramifications, that to study the subject thoroughly 26 00:01:25,390 --> 00:01:28,210 should be a full year's course by itself. 27 00:01:28,210 --> 00:01:32,170 And secondly, with respect to the particular topic of basis 28 00:01:32,170 --> 00:01:35,800 vectors and spanning vectors, and whether subelements 29 00:01:35,800 --> 00:01:38,050 are linearly independent, whereas we've 30 00:01:38,050 --> 00:01:41,950 invented a rather nice row reduced matrix technique 31 00:01:41,950 --> 00:01:47,330 for finding out what space is spanned by a set of vectors 32 00:01:47,330 --> 00:01:49,550 and what a basis for that space is, 33 00:01:49,550 --> 00:01:52,830 frequently we don't want that much information. 34 00:01:52,830 --> 00:01:55,760 And what I'm leading up to is a topic which you've all 35 00:01:55,760 --> 00:01:57,560 had in a previous context. 36 00:01:57,560 --> 00:01:59,000 It's called determinants. 37 00:01:59,000 --> 00:02:00,830 And today, what I would like to do 38 00:02:00,830 --> 00:02:06,470 is to study determinants within the framework of vector spaces. 39 00:02:06,470 --> 00:02:09,000 And the way it comes up is as follows. 40 00:02:09,000 --> 00:02:12,020 Suppose we have an n-dimensional vector space, v, 41 00:02:12,020 --> 00:02:14,730 and suppose that we pick a particular basis, 42 00:02:14,730 --> 00:02:18,270 u1 up to un, for v itself. 43 00:02:18,270 --> 00:02:20,010 Now, the idea is this. 44 00:02:20,010 --> 00:02:22,080 Knowing that the dimension of v is n, 45 00:02:22,080 --> 00:02:25,140 we immediately know that more than n vectors 46 00:02:25,140 --> 00:02:27,210 can't be a basis for v, because they 47 00:02:27,210 --> 00:02:29,370 will be linearly dependent rather than linearly 48 00:02:29,370 --> 00:02:31,080 independent. 49 00:02:31,080 --> 00:02:35,700 Fewer than n vectors of v cannot be a basis, 50 00:02:35,700 --> 00:02:39,180 because fewer than n vectors, since the dimension is n, 51 00:02:39,180 --> 00:02:43,950 cannot span v. Consequently, the only point of interest is what 52 00:02:43,950 --> 00:02:48,030 happens when we're given a set of n vectors and v. Namely, 53 00:02:48,030 --> 00:02:52,620 given the n vectors, alpha 1 up to alpha n and v, 54 00:02:52,620 --> 00:02:55,770 is this set a basis or isn't it? 55 00:02:55,770 --> 00:02:59,610 And rather than to use the row reduced matrix technique, here 56 00:02:59,610 --> 00:03:02,080 what we're saying is we don't want, 57 00:03:02,080 --> 00:03:05,190 for example, in many cases, to know what the betas look 58 00:03:05,190 --> 00:03:06,900 like that we talked about in our lecture 59 00:03:06,900 --> 00:03:08,670 on spanning vectors and the like. 60 00:03:08,670 --> 00:03:11,280 All we want to know is, are these vectors linearly 61 00:03:11,280 --> 00:03:13,560 independent or aren't they? 62 00:03:13,560 --> 00:03:16,440 And so what we do is we essentially 63 00:03:16,440 --> 00:03:18,247 invent a function machine. 64 00:03:18,247 --> 00:03:19,830 In other words, what we're going to do 65 00:03:19,830 --> 00:03:22,350 is to construct a function in which, 66 00:03:22,350 --> 00:03:26,010 given that the dimension of our space is n-dimensional, 67 00:03:26,010 --> 00:03:29,730 the input of our function machine 68 00:03:29,730 --> 00:03:33,420 will be any set of n vectors from that space. 69 00:03:33,420 --> 00:03:36,630 And the machine will be programmed to give 0 70 00:03:36,630 --> 00:03:40,030 as an output if the vectors are linearly dependent, 71 00:03:40,030 --> 00:03:43,730 in other words, if they are not a basis, and non-zero-- 72 00:03:43,730 --> 00:03:46,000 and I'll explain in more detail what non-zero-- 73 00:03:46,000 --> 00:03:49,650 why I picked non-zero rather than a specific non-zero 74 00:03:49,650 --> 00:03:51,900 number-- 75 00:03:51,900 --> 00:03:54,160 if the n vectors do form a basis. 76 00:03:54,160 --> 00:03:56,550 In other words, I am going to invent 77 00:03:56,550 --> 00:04:00,180 a function, D, capital D, to indicate the word determinant 78 00:04:00,180 --> 00:04:06,360 here, such that D of the n vectors, alpha 1 up to alpha n, 79 00:04:06,360 --> 00:04:10,770 will be 0 if and only if the set of n vectors 80 00:04:10,770 --> 00:04:12,990 are linearly dependent. 81 00:04:12,990 --> 00:04:14,880 Now, obviously, a function machine 82 00:04:14,880 --> 00:04:16,649 can do nothing by itself. 83 00:04:16,649 --> 00:04:17,839 We have to give-- 84 00:04:17,839 --> 00:04:21,160 and I hope this is well-known by now in our course-- 85 00:04:21,160 --> 00:04:24,390 we have to endow anything that we're 86 00:04:24,390 --> 00:04:27,190 working with with a particular structure 87 00:04:27,190 --> 00:04:31,140 so that is feels free to work logically for us. 88 00:04:31,140 --> 00:04:33,000 Well, among other things, what do we 89 00:04:33,000 --> 00:04:35,970 know for sure is a set of n vectors 90 00:04:35,970 --> 00:04:38,550 which are a basis for v? 91 00:04:38,550 --> 00:04:43,740 Since we were given that the specific basis that v is being 92 00:04:43,740 --> 00:04:48,296 referred to with respect to are u1 up tp un, 93 00:04:48,296 --> 00:04:53,550 then certainly, if u1 up to un is the input of the D machine, 94 00:04:53,550 --> 00:04:55,740 we want a non-zero output. 95 00:04:55,740 --> 00:04:59,340 For the sake of just normalizing things, 96 00:04:59,340 --> 00:05:03,160 let's program the machine so that D of u1 up to un 97 00:05:03,160 --> 00:05:05,750 will be 1. 98 00:05:05,750 --> 00:05:08,290 Now, notice, by the way, this only 99 00:05:08,290 --> 00:05:10,480 tells us one special basis. 100 00:05:10,480 --> 00:05:13,450 There are many bases that I could have chosen for v, 101 00:05:13,450 --> 00:05:15,760 and certainly condition one by itself 102 00:05:15,760 --> 00:05:19,210 isn't going to tell me any-- has no information programmed 103 00:05:19,210 --> 00:05:21,550 in it to tell me anything other than what 104 00:05:21,550 --> 00:05:24,220 it does to u1 up to un. 105 00:05:24,220 --> 00:05:28,360 As a second input to my D machine for programming it, 106 00:05:28,360 --> 00:05:31,450 I certainly know that given the n vectors, 107 00:05:31,450 --> 00:05:34,600 if any two of the vectors happen to be equal, then 108 00:05:34,600 --> 00:05:37,210 those vectors are linearly dependent. 109 00:05:37,210 --> 00:05:40,330 Consequently, I instruct my D machine, 110 00:05:40,330 --> 00:05:45,030 that if the input is a set of vectors alpha 1 up to alpha n, 111 00:05:45,030 --> 00:05:47,830 and at least two of the alphas are equal-- 112 00:05:47,830 --> 00:05:49,750 and the way we say that mathematically 113 00:05:49,750 --> 00:05:55,060 is that alpha i equals alpha j for a sum i unequal to j-- 114 00:05:55,060 --> 00:05:57,520 that if two of the vectors in the set are equal, 115 00:05:57,520 --> 00:06:00,580 we tell the machine to grind out 0 as an output. 116 00:06:00,580 --> 00:06:02,470 And by the way, at this stage we should 117 00:06:02,470 --> 00:06:05,590 be very careful to recognize that this is not 118 00:06:05,590 --> 00:06:07,720 the only way in which a set of vectors 119 00:06:07,720 --> 00:06:09,170 can be linearly dependent. 120 00:06:09,170 --> 00:06:11,230 There are many ways in which a set of vectors 121 00:06:11,230 --> 00:06:14,290 can be linearly dependent, even if no two of the vectors 122 00:06:14,290 --> 00:06:15,560 are equal. 123 00:06:15,560 --> 00:06:18,490 So consequently, given a linearly dependent set 124 00:06:18,490 --> 00:06:21,230 of vectors in which no two are equal, 125 00:06:21,230 --> 00:06:25,720 notice that condition two here will not work or do 126 00:06:25,720 --> 00:06:26,660 anything for me. 127 00:06:26,660 --> 00:06:30,160 So all I have done by conditions one and two 128 00:06:30,160 --> 00:06:33,700 is I have endowed my D machine with two standards. 129 00:06:33,700 --> 00:06:37,150 One particular case in which the output of the D machine 130 00:06:37,150 --> 00:06:40,780 will be the number 1, and a particular situation 131 00:06:40,780 --> 00:06:43,990 in which the output of the D machine will be 0. 132 00:06:43,990 --> 00:06:45,850 But I have not solved the problem 133 00:06:45,850 --> 00:06:49,630 that I wanted so far, namely, to make sure 134 00:06:49,630 --> 00:06:53,740 that D grinds out 0 if and only if the set of n vectors 135 00:06:53,740 --> 00:06:56,320 are linearly dependent. 136 00:06:56,320 --> 00:07:01,020 Let me now endow D with a third property. 137 00:07:01,020 --> 00:07:04,360 It is a property that will suggest linearity to you, 138 00:07:04,360 --> 00:07:07,540 and it's one which is amazingly powerful. 139 00:07:07,540 --> 00:07:10,420 And what I mean by amazingly powerful hopefully 140 00:07:10,420 --> 00:07:12,580 will become clear in a few moments. 141 00:07:12,580 --> 00:07:14,920 But the third property is this. 142 00:07:14,920 --> 00:07:16,840 And rather than work with n-tuples 143 00:07:16,840 --> 00:07:19,060 here, to make things easier to read, 144 00:07:19,060 --> 00:07:21,350 let me just take a 3-tuple over here. 145 00:07:21,350 --> 00:07:24,040 Suppose I have a three-dimensional vector space, 146 00:07:24,040 --> 00:07:26,530 and I pick three vectors in that space. 147 00:07:26,530 --> 00:07:28,240 Suppose one of those vectors-- the way 148 00:07:28,240 --> 00:07:29,770 I've written that here, it happens 149 00:07:29,770 --> 00:07:31,090 to be the second vector. 150 00:07:31,090 --> 00:07:34,240 Suppose one of those vectors is itself written 151 00:07:34,240 --> 00:07:37,180 as a sum of two other vectors in the space. 152 00:07:37,180 --> 00:07:41,500 Then I tell the D machine to compute this as follows. 153 00:07:41,500 --> 00:07:44,800 Linearize this, in other words, compute this 154 00:07:44,800 --> 00:07:49,810 as if it were the sum of two separate determinants, one 155 00:07:49,810 --> 00:07:53,080 of which had the alpha 2 missing, and the other of which 156 00:07:53,080 --> 00:07:54,670 had the beta 2 missing. 157 00:07:54,670 --> 00:07:56,650 In other words, what I do is is I 158 00:07:56,650 --> 00:07:58,660 endow the machine with the property 159 00:07:58,660 --> 00:08:04,210 that D of alpha 1 comma, alpha 2 plus beta 2 comma, alpha 3 160 00:08:04,210 --> 00:08:09,850 will be D of alpha 1, alpha 2, alpha 3 plus D 161 00:08:09,850 --> 00:08:13,410 of alpha 1, beta 2, alpha 3. 162 00:08:16,420 --> 00:08:18,875 And thirdly-- well, thirdly. 163 00:08:18,875 --> 00:08:19,750 I don't mean thirdly. 164 00:08:19,750 --> 00:08:21,790 I mean the other lineal property. 165 00:08:21,790 --> 00:08:27,880 If one of the input vectors is multiplied by a scalar, 166 00:08:27,880 --> 00:08:29,690 I can factor the scalar out. 167 00:08:29,690 --> 00:08:33,190 In other words, if the three vectors that are being tested 168 00:08:33,190 --> 00:08:36,790 in my three-dimensional space have the form alpha 1, 169 00:08:36,790 --> 00:08:42,549 c alpha 2, and alpha 3, D of alpha 1 comma, c alpha 2 comma, 170 00:08:42,549 --> 00:08:44,340 alpha 3 will be-- 171 00:08:44,340 --> 00:08:45,970 see, factor of the c out. 172 00:08:45,970 --> 00:08:50,200 c times D of alpha 1, alpha 2, alpha 3. 173 00:08:50,200 --> 00:08:53,950 And it's extremely important to notice that the c does not 174 00:08:53,950 --> 00:08:58,360 have to be a scalar multiple of all three vectors in the input. 175 00:08:58,360 --> 00:09:00,610 Notice, that the fact that the c was 176 00:09:00,610 --> 00:09:05,230 multiplying one of the vectors is enough to factor the c out. 177 00:09:05,230 --> 00:09:07,660 Without belaboring this point, all I wanted to say 178 00:09:07,660 --> 00:09:10,950 was, that if the c were multiplying each of the alphas, 179 00:09:10,950 --> 00:09:14,230 in other words, if this were D of c alpha 1 comma, 180 00:09:14,230 --> 00:09:19,900 c alpha 2 comma, c alpha 3, this would equal c cubed D alpha 1, 181 00:09:19,900 --> 00:09:21,880 alpha 2, alpha 3, because we would 182 00:09:21,880 --> 00:09:24,490 fact about the c each time-- 183 00:09:24,490 --> 00:09:26,860 one for each vector. 184 00:09:26,860 --> 00:09:30,270 But let me illustrate this in terms of a 2 by 2-- 185 00:09:30,270 --> 00:09:32,770 a two-dimensional example for you. 186 00:09:32,770 --> 00:09:35,170 Suppose I'm dealing in two-dimensional space, 187 00:09:35,170 --> 00:09:39,170 and I'm dealing with respect to a particular basis, u1 and u2. 188 00:09:39,170 --> 00:09:42,460 And let's suppose that relative to that basis, u1 and u2, 189 00:09:42,460 --> 00:09:45,040 alpha 1 is the vector 3 comma, 1, 190 00:09:45,040 --> 00:09:49,570 beta 1 is the vector 6 comma, 7, and beta 2 191 00:09:49,570 --> 00:09:53,930 is the vector 4 common, 5. 192 00:09:53,930 --> 00:09:57,460 Let's recall the fact that, one way or another, 193 00:09:57,460 --> 00:10:00,190 we already know how to expand 2 by 2 determinants, 194 00:10:00,190 --> 00:10:02,350 even though we may not know rigorously 195 00:10:02,350 --> 00:10:04,000 why the rules were chosen. 196 00:10:04,000 --> 00:10:07,510 The idea is, what would alpha 1 plus beta 1 be? 197 00:10:07,510 --> 00:10:11,950 It would be the 2-tuple whose first entry was 3 plus 6 198 00:10:11,950 --> 00:10:15,190 and whose second entry was 1 plus 7. 199 00:10:15,190 --> 00:10:18,985 In other words, the first row of the determinant 200 00:10:18,985 --> 00:10:22,170 that I'm going to be talking about has as its entries 201 00:10:22,170 --> 00:10:24,940 3 plus 6 and 1 plus 7. 202 00:10:24,940 --> 00:10:28,960 Beta 2, which will make up the second row with my determinant, 203 00:10:28,960 --> 00:10:31,580 has as its components 4 and 5. 204 00:10:31,580 --> 00:10:34,840 So the second row of my determinant is 4, 5. 205 00:10:34,840 --> 00:10:37,720 And using the usual rule for multiplying determinants, 206 00:10:37,720 --> 00:10:43,300 this is 5 times 9 minus 4 times 8, which is 13. 207 00:10:43,300 --> 00:10:45,790 And by the way, just to refresh your memories here, 208 00:10:45,790 --> 00:10:49,840 what if I computed in terms of my D language and the alpha 1, 209 00:10:49,840 --> 00:10:52,090 alpha 2, and beta 2 is used here? 210 00:10:52,090 --> 00:11:00,370 This is D of alpha 1 plus beta 1 comma, beta 2. 211 00:11:00,370 --> 00:11:05,140 On the other hand, what would D of alpha 1, beta 2 be? 212 00:11:05,140 --> 00:11:07,960 D of alpha 1 beta 2 would be the 2 213 00:11:07,960 --> 00:11:11,800 by 2 determinant, whose first row was 3, 1, 214 00:11:11,800 --> 00:11:14,970 and whose second row was 4, 5. 215 00:11:14,970 --> 00:11:20,320 D of alpha 1, beta 2 would be the determinant 216 00:11:20,320 --> 00:11:23,670 whose first row-- 217 00:11:23,670 --> 00:11:24,230 I'm sorry. 218 00:11:24,230 --> 00:11:27,960 D of beta 1 comma, beta 2, would be the determinant 219 00:11:27,960 --> 00:11:32,590 whose first row was 6, 7 and whose second row was 4, 5. 220 00:11:32,590 --> 00:11:36,240 In other words, these matrices here, what is this again? 221 00:11:36,240 --> 00:11:42,510 This is D of alpha 1, beta 2. 222 00:11:42,510 --> 00:11:50,300 And this one here is D of beta 1, beta 2. 223 00:11:50,300 --> 00:11:51,830 And notice, that this determinant 224 00:11:51,830 --> 00:11:54,440 by the traditional way of expanding is 11. 225 00:11:54,440 --> 00:11:57,380 This determinant is 30 minus 28, which is 2. 226 00:11:57,380 --> 00:11:59,300 11 plus 2 is 13. 227 00:11:59,300 --> 00:12:02,300 And we see that, at least in this case, D of alpha 1 228 00:12:02,300 --> 00:12:05,380 plus beta 1 comma, beta 2, is the same as D 229 00:12:05,380 --> 00:12:10,070 of alpha 1 comma, beta 2 plus D of beta 1 comma, beta 2. 230 00:12:10,070 --> 00:12:13,640 So at least the first part of property three 231 00:12:13,640 --> 00:12:16,460 is obeyed in this particular example. 232 00:12:16,460 --> 00:12:18,710 And to show what the second property is, 233 00:12:18,710 --> 00:12:20,540 let's suppose alpha 1 now-- 234 00:12:20,540 --> 00:12:21,590 I pick a new alpha 1. 235 00:12:21,590 --> 00:12:23,570 We'll call that 2 comma, 6. 236 00:12:23,570 --> 00:12:26,300 Suppose alpha 2 is 3 comma, 4. 237 00:12:26,300 --> 00:12:30,650 Then the determinant of alpha 1 and alpha 2 would be what? 238 00:12:30,650 --> 00:12:33,583 2, 6, 3, 4. 239 00:12:33,583 --> 00:12:34,250 And that's what? 240 00:12:34,250 --> 00:12:37,130 8 minus 18, which is minus 10. 241 00:12:37,130 --> 00:12:40,580 On the other hand, notice that a common-- 242 00:12:40,580 --> 00:12:42,680 that this vector here could have been 243 00:12:42,680 --> 00:12:45,890 written as twice 1 comma, 3. 244 00:12:45,890 --> 00:12:47,900 I can, therefore, factor out the 2. 245 00:12:47,900 --> 00:12:49,670 See, notice this very carefully. 246 00:12:49,670 --> 00:12:54,950 Notice that the alphas, which are written in a row here-- 247 00:12:54,950 --> 00:12:56,450 one row this way-- 248 00:12:56,450 --> 00:13:01,080 each alpha forms a row when I use the matrix interpretation. 249 00:13:01,080 --> 00:13:06,930 So what I'm saying is, that 2 comma, 6 is one vector. 250 00:13:06,930 --> 00:13:10,070 It's a 2-tuple with respect to the basis u1 and u2. 251 00:13:10,070 --> 00:13:13,490 All I'm saying is factor out the 2 from the first row, 252 00:13:13,490 --> 00:13:14,750 from the first vector. 253 00:13:14,750 --> 00:13:19,670 That leaves me with 2 outside, 1, 3, 3, 4 inside. 254 00:13:19,670 --> 00:13:21,880 And notice, that 4 times-- 255 00:13:21,880 --> 00:13:26,220 4 minus 9 is minus 5 times 2 is also minus 10. 256 00:13:26,220 --> 00:13:29,090 Notice that these, indeed, are equal, and notice 257 00:13:29,090 --> 00:13:33,140 that I could factor out the 2 simply by virtue of the fact 258 00:13:33,140 --> 00:13:35,600 that it was a common factor in one row. 259 00:13:35,600 --> 00:13:39,020 It did not have to be a common factor in both rows. 260 00:13:39,020 --> 00:13:41,810 In fact, if it had been a common factor in both rows, 261 00:13:41,810 --> 00:13:44,540 I would have had to factor out a 2 twice, 262 00:13:44,540 --> 00:13:47,240 in other words, a 4 over here. 263 00:13:47,240 --> 00:13:49,740 Now, I can go on with things like this, 264 00:13:49,740 --> 00:13:52,790 but, again, I want to stress the overview. 265 00:13:52,790 --> 00:13:56,480 And the key point is that these three simple properties, which 266 00:13:56,480 --> 00:13:59,210 I've just called simply one, two, and three, those three 267 00:13:59,210 --> 00:14:03,740 properties that I've programmed the D machine with 268 00:14:03,740 --> 00:14:06,650 are enough to completely determine 269 00:14:06,650 --> 00:14:09,980 D. In other words, if I wanted to be dramatic over here, 270 00:14:09,980 --> 00:14:14,660 the key point is that one, two, and three completely determine 271 00:14:14,660 --> 00:14:17,240 D. I'll put an exclamation point down there. 272 00:14:17,240 --> 00:14:20,480 Because what I claim is, that once these three properties are 273 00:14:20,480 --> 00:14:23,690 obeyed, there is no possible way for D 274 00:14:23,690 --> 00:14:28,450 to behave other than in a very unique well-defined way. 275 00:14:28,450 --> 00:14:31,580 That D now has a perfectly well-defined structure. 276 00:14:31,580 --> 00:14:34,130 Let me give you some examples of that structure. 277 00:14:34,130 --> 00:14:35,990 I'll just prove a couple of theorems. 278 00:14:35,990 --> 00:14:38,120 I'll prove them in two-dimensional space, 279 00:14:38,120 --> 00:14:40,580 leaving it for the exercises to work 280 00:14:40,580 --> 00:14:44,120 on higher dimensional space, but this doesn't get too cluttered. 281 00:14:44,120 --> 00:14:46,760 First of all, what I claim is that the D machine 282 00:14:46,760 --> 00:14:49,190 is so finicky, that if you interchange 283 00:14:49,190 --> 00:14:53,780 the order in which the two vectors are given, 284 00:14:53,780 --> 00:14:58,700 you change the sign of the output. 285 00:14:58,700 --> 00:15:01,820 In other words, D of alpha 1, alpha 2, in that order, 286 00:15:01,820 --> 00:15:05,360 is the negative of D of alpha 2, alpha 1. 287 00:15:05,360 --> 00:15:07,910 And the proof, again, follows very nicely 288 00:15:07,910 --> 00:15:10,040 in terms of our game structure. 289 00:15:10,040 --> 00:15:13,190 The gimmick to begin with is that we must be clever enough 290 00:15:13,190 --> 00:15:16,610 to decide that we'll compute the determinant of alpha 1 291 00:15:16,610 --> 00:15:20,210 plus alpha 2 comma, alpha 1 plus alpha 2. 292 00:15:20,210 --> 00:15:22,380 On the one hand, by our second property, 293 00:15:22,380 --> 00:15:26,390 since two of the vectors making up the set being tested 294 00:15:26,390 --> 00:15:29,960 are equal, it means that that determinant must be 0. 295 00:15:29,960 --> 00:15:31,820 On the other hand, by property three, 296 00:15:31,820 --> 00:15:35,780 by linearity, splitting this up as two terms, 297 00:15:35,780 --> 00:15:40,730 notice that this is D of alpha 1 comma, alpha 1 plus alpha 2 298 00:15:40,730 --> 00:15:45,420 plus D of alpha 2 comma, alpha 1 plus alpha 2. 299 00:15:45,420 --> 00:15:47,390 In other words, this result here. 300 00:15:47,390 --> 00:15:51,470 Now, in turn, noticing that these form a sum, 301 00:15:51,470 --> 00:15:54,080 I can rewrite this one as what? 302 00:15:54,080 --> 00:16:00,620 D of alpha 1 comma, alpha 1 plus D of alpha 1 comma, alpha 2. 303 00:16:00,620 --> 00:16:04,310 This one is D of alpha 2 comma, alpha 1 304 00:16:04,310 --> 00:16:08,090 plus D of alpha 2 comma, alpha 2. 305 00:16:08,090 --> 00:16:10,850 I've just written this whole thing out on this next line. 306 00:16:10,850 --> 00:16:16,500 By property two, D of alpha 1 comma, alpha 1 must be 0. 307 00:16:16,500 --> 00:16:19,130 D of alpha 2 comma, alpha 2 must be 0, 308 00:16:19,130 --> 00:16:24,380 because after all, you see, two vectors are equal in this. 309 00:16:24,380 --> 00:16:25,850 So what do I have left? 310 00:16:25,850 --> 00:16:30,950 Comparing this with this, I have the D of alpha 1 comma, alpha 2 311 00:16:30,950 --> 00:16:34,520 plus D of alpha 2 comma, alpha 1 is 0. 312 00:16:34,520 --> 00:16:37,130 And that means that the number-- and keep that in mind, 313 00:16:37,130 --> 00:16:38,510 the determinant is a number. 314 00:16:38,510 --> 00:16:40,790 It maps the n vectors into a number, which 315 00:16:40,790 --> 00:16:42,380 is either 0 or non-zero. 316 00:16:42,380 --> 00:16:45,260 But the determinant of alpha 1 comma, alpha 2, 317 00:16:45,260 --> 00:16:48,890 therefore, must be the negative of the term of alpha 2, 318 00:16:48,890 --> 00:16:52,010 alpha 1, because their sum is 0. 319 00:16:52,010 --> 00:16:54,170 The second theorem-- and this is the one 320 00:16:54,170 --> 00:16:58,280 that relates determinants to matrices, a very fantastic 321 00:16:58,280 --> 00:17:01,850 result, one that has tremendous practical application that 322 00:17:01,850 --> 00:17:04,319 will also talk about later in the lecture. 323 00:17:04,319 --> 00:17:08,359 And that is, that if I take any one of my input vectors 324 00:17:08,359 --> 00:17:13,490 and replace it by itself, plus a scalar multiple of another, 325 00:17:13,490 --> 00:17:15,740 I do not change the determinant. 326 00:17:15,740 --> 00:17:19,910 In other words, for example, if I replace alpha 1 by alpha 1 327 00:17:19,910 --> 00:17:24,550 plus a scalar multiple of alpha 2, and I leave alpha 2 alone, 328 00:17:24,550 --> 00:17:27,530 the determinant of alpha 1 and alpha 2 329 00:17:27,530 --> 00:17:30,170 will be the same as the determinant of alpha 1 330 00:17:30,170 --> 00:17:33,140 plus c alpha 2 comma, alpha 2. 331 00:17:33,140 --> 00:17:35,570 And again, the proof is very easy, 332 00:17:35,570 --> 00:17:38,870 namely, look at the expression D of alpha 1 333 00:17:38,870 --> 00:17:41,390 plus c alpha 2 comma, alpha 2. 334 00:17:41,390 --> 00:17:44,120 Use the linearity property that this 335 00:17:44,120 --> 00:17:51,050 is D of alpha 1 comma, alpha 2 plus D of c alpha 2 comma, 336 00:17:51,050 --> 00:17:51,770 alpha 2. 337 00:17:54,390 --> 00:17:57,530 Then use the second part of the linearity property 338 00:17:57,530 --> 00:18:00,780 that the constant factor, c, can be taken outside here. 339 00:18:00,780 --> 00:18:03,410 So this becomes D of alpha 1, alpha 2 340 00:18:03,410 --> 00:18:07,520 plus c times D of alpha 2 comma, alpha 2. 341 00:18:07,520 --> 00:18:11,780 And notice, that since alpha 2 is repeated here, 342 00:18:11,780 --> 00:18:13,310 this determinant is 0. 343 00:18:13,310 --> 00:18:14,930 And consequently, what we've proven 344 00:18:14,930 --> 00:18:18,302 is that this is equal to this. 345 00:18:18,302 --> 00:18:20,010 Now, I've just proven those two theorems, 346 00:18:20,010 --> 00:18:21,620 because that's about all I need. 347 00:18:21,620 --> 00:18:23,780 Let me show you now, that if I had never 348 00:18:23,780 --> 00:18:27,140 seen that shortcut method of expanding a 2 349 00:18:27,140 --> 00:18:29,780 by 2 determinant, as learned in high school, 350 00:18:29,780 --> 00:18:32,630 how these three properties uniquely 351 00:18:32,630 --> 00:18:34,265 determine what D has to mean. 352 00:18:34,265 --> 00:18:36,890 In other words, why I said that these three properties uniquely 353 00:18:36,890 --> 00:18:38,630 determine D. 354 00:18:38,630 --> 00:18:42,020 As an example, let me take a two-dimensional vector space 355 00:18:42,020 --> 00:18:42,950 again. 356 00:18:42,950 --> 00:18:47,150 Let me pick u1 and u2 as a specific basis, 357 00:18:47,150 --> 00:18:48,260 and that's important. 358 00:18:48,260 --> 00:18:51,200 I've picked a specific basis, u1, u2. 359 00:18:51,200 --> 00:18:53,000 Now, I pick any two vectors. 360 00:18:53,000 --> 00:18:56,990 Since v is a two-dimensional space, I pick any two vectors-- 361 00:18:56,990 --> 00:18:59,210 alpha 1 and alpha 2. 362 00:18:59,210 --> 00:19:01,790 And because of this notation, this 363 00:19:01,790 --> 00:19:05,360 means that alpha 1 and alpha 2 are both linear combinations 364 00:19:05,360 --> 00:19:06,830 of u1 and u2. 365 00:19:06,830 --> 00:19:10,130 Say alpha 1 is this, and alpha 2 is this. 366 00:19:10,130 --> 00:19:13,520 Therefore, what is D of alpha 1 comma, alpha 2? 367 00:19:13,520 --> 00:19:20,460 Well, by direct substitution, D of alpha 1, alpha 2, just 368 00:19:20,460 --> 00:19:23,430 replace alpha 1 by what it's equal to here, alpha 2 369 00:19:23,430 --> 00:19:25,020 by what it's equal to here. 370 00:19:25,020 --> 00:19:29,290 And we get that D of alpha 1, alpha 2 is this expression. 371 00:19:29,290 --> 00:19:31,800 Now, we use the linearity property. 372 00:19:31,800 --> 00:19:34,230 We treat this as one number for the time being 373 00:19:34,230 --> 00:19:36,090 and split this up as a sum. 374 00:19:36,090 --> 00:19:38,040 In other words, it's going to be D 375 00:19:38,040 --> 00:19:44,940 of this term comma, this whole term plus D of this term 376 00:19:44,940 --> 00:19:47,020 comma, this whole term. 377 00:19:47,020 --> 00:19:51,150 In other words, if I do that, this breaks down to this. 378 00:19:51,150 --> 00:19:54,300 Now, I notice that each of these two 379 00:19:54,300 --> 00:19:59,760 is a 2-tuple in which the second entry is a sum of two terms. 380 00:19:59,760 --> 00:20:02,730 So I now split this up into what? 381 00:20:02,730 --> 00:20:09,456 D of this comma, this plus d of this comma, this. 382 00:20:09,456 --> 00:20:15,180 This term becomes D of this comma, this term plus D 383 00:20:15,180 --> 00:20:17,940 of this comma, this term. 384 00:20:17,940 --> 00:20:20,130 And I hope it doesn't sound boorish 385 00:20:20,130 --> 00:20:23,370 on my part to keep saying "D of this comma, this." 386 00:20:23,370 --> 00:20:25,800 I prefer to say that for you so that you 387 00:20:25,800 --> 00:20:27,600 can watch what I'm doing, and then 388 00:20:27,600 --> 00:20:33,000 have you just be able afterwards to read what these things mean. 389 00:20:33,000 --> 00:20:35,250 Now, when I'm down to here, notice 390 00:20:35,250 --> 00:20:37,740 that by my second part of linearity 391 00:20:37,740 --> 00:20:41,520 I can factor out the a11 from here. 392 00:20:41,520 --> 00:20:44,340 I can factor out the a21, because it's 393 00:20:44,340 --> 00:20:46,060 a common factor of this term. 394 00:20:46,060 --> 00:20:48,510 In other words, I can write this term 395 00:20:48,510 --> 00:20:53,220 as a11, a21, D of u1 comma, u2. 396 00:20:53,220 --> 00:20:55,540 And again, sparing you the details, 397 00:20:55,540 --> 00:21:00,870 notice I can factor out an a11, a22 from this term. 398 00:21:00,870 --> 00:21:06,210 I can factor out an a12, a21 from this term, an a12, 399 00:21:06,210 --> 00:21:08,460 a22 from this term. 400 00:21:08,460 --> 00:21:12,870 And that this now can be written in this particular way. 401 00:21:12,870 --> 00:21:16,560 Now, what properties do I know that D is endowed with? 402 00:21:16,560 --> 00:21:18,480 I know, first of all, that whenever 403 00:21:18,480 --> 00:21:22,050 D operates on a set of vectors where at least two of them 404 00:21:22,050 --> 00:21:25,360 are equal, D must give 0 as an output. 405 00:21:25,360 --> 00:21:28,620 So D of u1 comma, u1 is 0 again. 406 00:21:28,620 --> 00:21:32,640 u1 comma, u2 is the particular basis with respect 407 00:21:32,640 --> 00:21:35,400 to which I define v, you see. 408 00:21:35,400 --> 00:21:41,280 Therefore, by property one, D of u1 comma, u2 must be 1. 409 00:21:41,280 --> 00:21:46,860 D of u2 comma, u1 is just the permuted order of u1 and u2. 410 00:21:46,860 --> 00:21:52,800 Therefore, by our first theorem, that must be minus D of u1, u2. 411 00:21:52,800 --> 00:21:54,930 Therefore, it must be minus 1. 412 00:21:54,930 --> 00:21:58,260 And finally, D of u2 comma, u2 is 0. 413 00:21:58,260 --> 00:22:00,210 And if I now collect everything that I 414 00:22:00,210 --> 00:22:02,200 have left here, what do I have? 415 00:22:02,200 --> 00:22:08,020 I have the D of alpha 1 comma, alpha 2 is equal to what? 416 00:22:08,020 --> 00:22:15,460 It's equal to a11, a22, see, times 1, minus a12, a21. 417 00:22:15,460 --> 00:22:17,700 In other words, this is the determinant 418 00:22:17,700 --> 00:22:20,310 of alpha 1 comma, alpha 2. 419 00:22:20,310 --> 00:22:24,210 And let me, again, put an exclamation point here. 420 00:22:24,210 --> 00:22:26,400 Because if you now go back to the high school 421 00:22:26,400 --> 00:22:29,010 way of computing this determinant-- 422 00:22:29,010 --> 00:22:30,510 remember how we did it? 423 00:22:30,510 --> 00:22:32,060 What would we have obtained? 424 00:22:32,060 --> 00:22:34,050 And, again, let me just come back to this board for a second 425 00:22:34,050 --> 00:22:34,580 here. 426 00:22:34,580 --> 00:22:36,835 Remember how you used the matrix of coefficients? 427 00:22:36,835 --> 00:22:38,460 So take the determinant, would be what? 428 00:22:38,460 --> 00:22:44,690 a11, a22 minus a12, a21. 429 00:22:44,690 --> 00:22:47,820 So, again, notice two things that happen here. 430 00:22:47,820 --> 00:22:49,860 First of all, I get the same answer 431 00:22:49,860 --> 00:22:51,570 as I would have got the traditional way. 432 00:22:51,570 --> 00:22:53,750 And second, just as a minor aside 433 00:22:53,750 --> 00:22:56,170 that I'll emphasize more later in the lecture 434 00:22:56,170 --> 00:22:58,650 and also in the exercises, notice 435 00:22:58,650 --> 00:23:01,980 that you can begin to suspect that the actual value 436 00:23:01,980 --> 00:23:04,440 of the determinant of alpha 1 and alpha 2 437 00:23:04,440 --> 00:23:07,410 should depend on what basis was chosen. 438 00:23:07,410 --> 00:23:09,870 Because relative to a different basis, 439 00:23:09,870 --> 00:23:12,570 notice that the coefficients might very well 440 00:23:12,570 --> 00:23:15,300 be different for alpha 1 and alpha 2. 441 00:23:15,300 --> 00:23:18,090 But I don't want to mention that right now. 442 00:23:18,090 --> 00:23:20,190 I think, as I say, what the amazing result 443 00:23:20,190 --> 00:23:22,590 is, that what I meant by saying that these three 444 00:23:22,590 --> 00:23:26,250 properties completely determine D, is the fact 445 00:23:26,250 --> 00:23:27,900 that with these three properties it 446 00:23:27,900 --> 00:23:32,410 turns out that the high school definition was ironclad. 447 00:23:32,410 --> 00:23:35,350 Meaning, there was no other possible way 448 00:23:35,350 --> 00:23:37,660 to define what the determinant should 449 00:23:37,660 --> 00:23:41,800 be if you wanted these three properties to be obeyed. 450 00:23:41,800 --> 00:23:43,690 By the way, I have some quick checks. 451 00:23:43,690 --> 00:23:45,400 Notice that relative to the basis 452 00:23:45,400 --> 00:23:48,850 u1, u2, u1 can be written this way, 453 00:23:48,850 --> 00:23:50,770 u2 can be written this way. 454 00:23:50,770 --> 00:23:53,890 We know that D of u1, u2 should be 1. 455 00:23:53,890 --> 00:23:57,310 And using the old fashioned way for checking this, 456 00:23:57,310 --> 00:23:58,690 what would the determinant be? 457 00:23:58,690 --> 00:24:03,880 It would be 1 times 1 minus 0 times 0, which is 1. 458 00:24:03,880 --> 00:24:08,170 Secondly, we also know that property two should yield 0 459 00:24:08,170 --> 00:24:12,220 as a determinant if these two vectors that were making up 460 00:24:12,220 --> 00:24:13,750 the input were equal. 461 00:24:13,750 --> 00:24:17,230 Let's call the vectors au1, bu2. 462 00:24:17,230 --> 00:24:19,930 So the 2 by 2 determinant I would get in this case 463 00:24:19,930 --> 00:24:21,580 is ab, ab. 464 00:24:21,580 --> 00:24:23,890 And if I expand that determinant, it's what? 465 00:24:23,890 --> 00:24:28,970 a times b minus a times b, which is 0. 466 00:24:28,970 --> 00:24:33,450 And now, let me come back to that idea of what 467 00:24:33,450 --> 00:24:35,460 motivated the whole block of material 468 00:24:35,460 --> 00:24:39,490 here, the trouble with writing vectors as n-tuples. 469 00:24:39,490 --> 00:24:42,870 That there really is something that requires great care. 470 00:24:42,870 --> 00:24:46,690 In advanced applications, one must always be wary of this. 471 00:24:46,690 --> 00:24:49,410 I am not going to give you the advanced applications here. 472 00:24:49,410 --> 00:24:52,500 All I want you to do is to become prepared 473 00:24:52,500 --> 00:24:54,480 against the pitfalls. 474 00:24:54,480 --> 00:24:57,120 And the thing is, that in dealing with determinants, 475 00:24:57,120 --> 00:25:00,900 as I said before, you must be very, very careful about what 476 00:25:00,900 --> 00:25:04,380 basis you're referring to for a given vector space. 477 00:25:04,380 --> 00:25:07,500 Obviously, the vectors do not depend on the basis, 478 00:25:07,500 --> 00:25:09,420 but their representation does. 479 00:25:09,420 --> 00:25:11,190 Let me give you a for instance. 480 00:25:11,190 --> 00:25:17,250 Using v and u1 and u2 as an example number 1, 481 00:25:17,250 --> 00:25:21,220 suppose we let alpha 1 be the 2-tuple, 3 comma, 4. 482 00:25:21,220 --> 00:25:23,650 In other words, 3u1 plus 4u2. 483 00:25:23,650 --> 00:25:26,370 And let alpha 2 be 2 comma, 5. 484 00:25:26,370 --> 00:25:28,950 Then by what we've just proven, the determinant 485 00:25:28,950 --> 00:25:31,920 of alpha 1 comma, alpha 2 must be 486 00:25:31,920 --> 00:25:34,110 the determinant of what matrix? 487 00:25:34,110 --> 00:25:36,810 The one whose first row is 3 comma, 4, and whose 488 00:25:36,810 --> 00:25:38,940 second row is 2 comma, 5. 489 00:25:38,940 --> 00:25:41,730 If I compute that determinant-- and by the way, notice, 490 00:25:41,730 --> 00:25:43,290 now that I know that the shortcut 491 00:25:43,290 --> 00:25:45,582 way has to be the right answer, I 492 00:25:45,582 --> 00:25:47,040 don't do this the long way anymore. 493 00:25:47,040 --> 00:25:51,630 I just say, OK, it's 3 times 5 minus 2 times 4 15 minus 8, 494 00:25:51,630 --> 00:25:52,590 which is 7. 495 00:25:52,590 --> 00:25:56,670 So the determinant of alpha 1, alpha 2 is 7. 496 00:25:56,670 --> 00:25:57,930 So far so good. 497 00:25:57,930 --> 00:26:00,450 But let me point out the following thing. 498 00:26:00,450 --> 00:26:04,320 Notice, that alpha 1 and alpha 2 are also linearly independent. 499 00:26:04,320 --> 00:26:09,750 Namely, 3 common, 4 is not a scalar multiple of 2 comma, 5. 500 00:26:09,750 --> 00:26:12,480 Consequently, since alpha 1 and alpha 2 our linearly 501 00:26:12,480 --> 00:26:16,260 independent, and since v is a two-dimensional space, 502 00:26:16,260 --> 00:26:20,430 it means that alpha 1 and alpha 2 are themselves a basis for v. 503 00:26:20,430 --> 00:26:24,570 In other words, v is equal to the space spanned by alpha 1 504 00:26:24,570 --> 00:26:26,400 and alpha 2 as a basis. 505 00:26:26,400 --> 00:26:28,590 Notice, that relative to this new basis, 506 00:26:28,590 --> 00:26:34,140 alpha 1 comma, alpha 2, alpha 1 is 1 alpha 1 plus 0 alpha 2, 507 00:26:34,140 --> 00:26:37,740 alpha 2 is 0 alpha 1 plus alpha 2. 508 00:26:37,740 --> 00:26:40,560 And therefore, using the traditional method 509 00:26:40,560 --> 00:26:42,780 of computing the determinant, notice 510 00:26:42,780 --> 00:26:45,900 that the determinant of alpha 1 comma, alpha 2 511 00:26:45,900 --> 00:26:48,990 relative to the basis alpha 1 and alpha 2 512 00:26:48,990 --> 00:26:50,120 would be what determinant? 513 00:26:50,120 --> 00:26:51,953 It would be to the determinant of the matrix 514 00:26:51,953 --> 00:26:55,980 whose first row was 1, 0, and whose second role was 0, 1. 515 00:26:55,980 --> 00:26:57,750 And that is 1. 516 00:26:57,750 --> 00:26:59,190 This looks like a contradiction. 517 00:26:59,190 --> 00:27:01,830 You see, on the one hand, we have the D of alpha 1 518 00:27:01,830 --> 00:27:03,420 and alpha 2 is 1. 519 00:27:03,420 --> 00:27:09,210 But down here, we just saw that D of alpha 1 and alpha 2 was 7. 520 00:27:09,210 --> 00:27:11,870 Which is correct? 521 00:27:11,870 --> 00:27:14,420 Well, the answer is, they're both correct. 522 00:27:14,420 --> 00:27:17,630 That the first observation is, that if the determinant 523 00:27:17,630 --> 00:27:21,800 of alpha 1 and alpha 2 is not 0, the value 524 00:27:21,800 --> 00:27:24,090 depends on the particular basis. 525 00:27:24,090 --> 00:27:26,870 In other words, if the determinant is 0, 526 00:27:26,870 --> 00:27:29,030 it turns out that no matter what basis 527 00:27:29,030 --> 00:27:32,150 you use to represent the determinant of alpha 1 528 00:27:32,150 --> 00:27:36,005 and alpha 2, you will get 0 once the terminator is 529 00:27:36,005 --> 00:27:37,340 0 in one basis. 530 00:27:37,340 --> 00:27:40,160 And if the determinant is not 0 with respect 531 00:27:40,160 --> 00:27:43,880 to a particular basis, it will be non-zero with respect 532 00:27:43,880 --> 00:27:45,200 to all bases. 533 00:27:45,200 --> 00:27:49,620 But what non-zero number it will be does depend on the basis. 534 00:27:49,620 --> 00:27:51,950 And that was why way back at the beginning 535 00:27:51,950 --> 00:27:57,080 we told the D machine simply to give a non-zero output 536 00:27:57,080 --> 00:28:01,670 if the input was a set of n linearly-independent vectors. 537 00:28:01,670 --> 00:28:05,960 In summary, when one says that the determinant of alpha 1 538 00:28:05,960 --> 00:28:09,350 and alpha 2 is 7, it's tacitly assumed 539 00:28:09,350 --> 00:28:14,330 that D is being referred to with respect to the particular basis 540 00:28:14,330 --> 00:28:15,770 u1, u2. 541 00:28:15,770 --> 00:28:19,010 On the other hand, when one says that D of alpha 1 and alpha 2 542 00:28:19,010 --> 00:28:22,130 is 1, it's tacitly assumed that the basis 543 00:28:22,130 --> 00:28:26,810 that we're using to express v is alpha 1 and alpha 2. 544 00:28:26,810 --> 00:28:30,440 Again, I'll drill that more in the exercises. 545 00:28:30,440 --> 00:28:32,810 Let me go on now to generalize what 546 00:28:32,810 --> 00:28:35,510 happens in n-dimensional space, keeping in mind 547 00:28:35,510 --> 00:28:37,370 the fact that when we generalize what 548 00:28:37,370 --> 00:28:41,210 happens in n-dimensional space, the actual proofs become 549 00:28:41,210 --> 00:28:44,690 messier, but the theory remains very much the same. 550 00:28:44,690 --> 00:28:49,040 I prefer to leave the messy details to the exercises, 551 00:28:49,040 --> 00:28:51,080 either optional or required, depending 552 00:28:51,080 --> 00:28:53,490 on how hard the exercises may be. 553 00:28:53,490 --> 00:28:55,790 But let me summarize what the results are 554 00:28:55,790 --> 00:28:58,010 for any n-dimensional space, keeping 555 00:28:58,010 --> 00:29:00,440 in mind we've proven the results rigorously 556 00:29:00,440 --> 00:29:02,780 for the case n equals 2. 557 00:29:02,780 --> 00:29:04,490 The generalization is this. 558 00:29:04,490 --> 00:29:06,470 If I have an n-dimensional vector space, 559 00:29:06,470 --> 00:29:11,000 v, with respect to a particular basis, u1 up to un, 560 00:29:11,000 --> 00:29:16,010 and if alpha 1 up to alpha n are n vectors chosen from v so 561 00:29:16,010 --> 00:29:19,490 that they are linear combinations of the us, 562 00:29:19,490 --> 00:29:22,040 say, in the traditional way that we've written this 563 00:29:22,040 --> 00:29:24,860 all the time, it turns out the following. 564 00:29:24,860 --> 00:29:28,820 That the determinant of alpha 1 up to alpha n-- 565 00:29:28,820 --> 00:29:31,940 first of all, it's conventional and convenient to write 566 00:29:31,940 --> 00:29:34,550 the determinant as if it were a matrix, only 567 00:29:34,550 --> 00:29:37,850 replacing the square brackets by sort of absolute value 568 00:29:37,850 --> 00:29:39,560 signs-- vertical lines. 569 00:29:39,560 --> 00:29:44,650 What we do is, notice that alpha 1 is written as an n-tuple, 570 00:29:44,650 --> 00:29:47,060 a11 up to a1n. 571 00:29:47,060 --> 00:29:49,460 In other words, the first rule of the determinant 572 00:29:49,460 --> 00:29:54,470 represents alpha 1 as an n-tuple vector relative to the basis 573 00:29:54,470 --> 00:29:55,820 u1 up to un. 574 00:29:55,820 --> 00:29:59,450 And a similar thing holds for alpha 2 through alpha n. 575 00:29:59,450 --> 00:30:01,880 And you may remember, I told you what the recipe 576 00:30:01,880 --> 00:30:04,442 was when we were dealing with-- 577 00:30:04,442 --> 00:30:05,900 I told you what the recipe was when 578 00:30:05,900 --> 00:30:08,300 we were dealing with cross products and the like. 579 00:30:08,300 --> 00:30:10,130 And the idea is simply this. 580 00:30:10,130 --> 00:30:12,650 What you do is is you start in the upper left-hand corner 581 00:30:12,650 --> 00:30:14,250 writing a plus sign. 582 00:30:14,250 --> 00:30:18,110 Then you alternate going along rows and columns 583 00:30:18,110 --> 00:30:21,170 in any order you want writing plus, minus, plus, minus, 584 00:30:21,170 --> 00:30:23,690 et cetera, plus, minus, plus, minus, et cetera. 585 00:30:23,690 --> 00:30:26,150 Pick any row or column that you want. 586 00:30:26,150 --> 00:30:29,090 And then you go down that row or column factoring 587 00:30:29,090 --> 00:30:32,960 out a particular term, the plus sign telling you to take out 588 00:30:32,960 --> 00:30:37,280 the term as it is, the minus sign telling you to take out 589 00:30:37,280 --> 00:30:40,910 the term and change its sign, and you multiply that term 590 00:30:40,910 --> 00:30:44,030 by the n minus 1 by n minus 1 matrix that's 591 00:30:44,030 --> 00:30:47,390 left when you strike out the row and column in which that term 592 00:30:47,390 --> 00:30:48,210 occurs. 593 00:30:48,210 --> 00:30:50,360 Now, if that still sounds like a tongue-twister, 594 00:30:50,360 --> 00:30:54,050 let's do this in terms of a specific semi-abstract, 595 00:30:54,050 --> 00:30:56,960 semi-concrete, four-dimensional case. 596 00:30:56,960 --> 00:31:00,380 Namely, let's suppose v is a four-dimensional vector space 597 00:31:00,380 --> 00:31:04,520 relative to a particular basis, u1, u2, u3, and u4. 598 00:31:04,520 --> 00:31:09,670 Suppose alpha 1, alpha 2, alpha 3, and alpha 4 599 00:31:09,670 --> 00:31:14,360 are these four specific vectors of v. And again-- 600 00:31:14,360 --> 00:31:16,190 I can't keep emphasizing this too much, 601 00:31:16,190 --> 00:31:18,800 even though I hope some of you are bored because you know it 602 00:31:18,800 --> 00:31:19,610 so well-- 603 00:31:19,610 --> 00:31:21,860 it's always understood when I write this 604 00:31:21,860 --> 00:31:25,820 that the components are relative to u1, u2, u3, u4. 605 00:31:25,820 --> 00:31:31,340 For example, this is 3u1 plus 5u2 plus 7u3 plus 2u4. 606 00:31:31,340 --> 00:31:34,670 Now, the traditional way of expanding this determinant 607 00:31:34,670 --> 00:31:38,640 is you write down the rows as the n-tuples. 608 00:31:38,640 --> 00:31:41,610 See, you write down-- this is your first row, second row, 609 00:31:41,610 --> 00:31:43,140 third row, fourth row. 610 00:31:43,140 --> 00:31:46,910 So you notice that the determinant 611 00:31:46,910 --> 00:31:52,520 is an n by an array of numbers that you have four vectors. 612 00:31:52,520 --> 00:31:55,432 Each vector is a 4-tuple, and this is 613 00:31:55,432 --> 00:31:56,640 how you get this determinant. 614 00:31:56,640 --> 00:31:57,780 Now, what do we do next? 615 00:31:57,780 --> 00:31:59,360 Let's say, for the sake of argument, 616 00:31:59,360 --> 00:32:03,560 I elect to expand this determinant along the top row. 617 00:32:03,560 --> 00:32:05,970 I first take the 1 out. 618 00:32:05,970 --> 00:32:08,540 Because it has a plus sign, that comes out as 1. 619 00:32:08,540 --> 00:32:11,040 I multiply that by the 3 by 3 determinant 620 00:32:11,040 --> 00:32:14,790 that's left when I strike out the row and column in which 1 621 00:32:14,790 --> 00:32:16,320 appears. 622 00:32:16,320 --> 00:32:19,650 I take out 2, but make it a minus, because of the signature 623 00:32:19,650 --> 00:32:23,160 here, and multiply that by the 3 by 3 determinant that's 624 00:32:23,160 --> 00:32:26,730 left when I strike out the row and column in which 2 appears. 625 00:32:26,730 --> 00:32:30,330 I factor out 1 as it is, and multiply that by the 3 626 00:32:30,330 --> 00:32:32,280 by 3 determinant that's left when 627 00:32:32,280 --> 00:32:36,670 I strike out the row and column in which this 1 appears. 628 00:32:36,670 --> 00:32:38,475 And finally, I take out this minus 1 629 00:32:38,475 --> 00:32:41,800 and change its sign, because of this minus code over here. 630 00:32:41,800 --> 00:32:44,580 In other words, this comes out as a plus 1 multiplied by the 3 631 00:32:44,580 --> 00:32:47,160 by 3 determinant that's left when I strike out 632 00:32:47,160 --> 00:32:50,130 the row and column in which minus 1 appears in. 633 00:32:50,130 --> 00:32:52,650 And to make a long story short, all we're saying 634 00:32:52,650 --> 00:32:55,800 is that this 4 by 4 determinant can 635 00:32:55,800 --> 00:33:00,330 be written as a sum of four 3 by 3 determinants, 636 00:33:00,330 --> 00:33:04,770 namely, this, which is precisely what I was 637 00:33:04,770 --> 00:33:06,960 saying the long way up here. 638 00:33:06,960 --> 00:33:10,350 Now, what we say is, each of these 3 by 3s 639 00:33:10,350 --> 00:33:12,960 can be written as three 2 by 2s. 640 00:33:12,960 --> 00:33:14,670 Namely, I can factor out-- 641 00:33:14,670 --> 00:33:16,530 say again, I'm going across the top row. 642 00:33:16,530 --> 00:33:18,150 I can pick any row or any column, 643 00:33:18,150 --> 00:33:20,370 but let me stick to the top row, like we did in i, j, 644 00:33:20,370 --> 00:33:21,900 and k, for the time being. 645 00:33:21,900 --> 00:33:22,950 But this becomes what? 646 00:33:22,950 --> 00:33:28,410 5 times this 2 by 2 matrix, minus 3 times 647 00:33:28,410 --> 00:33:34,330 this 2 by 2 matrix, plus 1 times this 2 by 2 matrix. 648 00:33:34,330 --> 00:33:36,960 In other words, this expression here. 649 00:33:36,960 --> 00:33:39,190 I can now do that for each of these. 650 00:33:39,190 --> 00:33:43,680 So each of the 3 by 3s gives me three 2 by 2s. 651 00:33:43,680 --> 00:33:46,710 So altogether, I would have 12 2 by 2s 652 00:33:46,710 --> 00:33:49,320 and go through this whole mess to compute this thing. 653 00:33:49,320 --> 00:33:52,290 And this can be very, very difficult, sure. 654 00:33:52,290 --> 00:33:54,270 This was only a four-dimensional case. 655 00:33:54,270 --> 00:33:56,610 Imagine trying to apply this technique 656 00:33:56,610 --> 00:33:59,340 for a 10-dimensional vector space, 657 00:33:59,340 --> 00:34:00,810 for the sake of argument. 658 00:34:00,810 --> 00:34:06,460 It turns out that that recipe that we called 659 00:34:06,460 --> 00:34:11,270 theorem two gives us a tie-in between row 660 00:34:11,270 --> 00:34:15,780 reduced matrices and a quick way to compute determinants. 661 00:34:15,780 --> 00:34:17,578 For example, let's suppose we still 662 00:34:17,578 --> 00:34:19,620 wanted to compute the same determinant that we've 663 00:34:19,620 --> 00:34:21,370 written down over here. 664 00:34:21,370 --> 00:34:24,556 Let's call that-- let's write it again over here. 665 00:34:24,556 --> 00:34:26,909 What we already know is, if I were 666 00:34:26,909 --> 00:34:30,989 to replace the second vector by the second minus twice 667 00:34:30,989 --> 00:34:33,840 the first, the determinant does not change. 668 00:34:33,840 --> 00:34:35,100 That's what theorem two said. 669 00:34:35,100 --> 00:34:37,350 If you replace one vector by itself 670 00:34:37,350 --> 00:34:38,790 plus a scalar multiple of another, 671 00:34:38,790 --> 00:34:40,590 you don't change the determinant. 672 00:34:40,590 --> 00:34:43,380 Similarly, if I were to replace this vector 673 00:34:43,380 --> 00:34:46,260 by three times the first vector, I 674 00:34:46,260 --> 00:34:48,239 would still have the same determinant. 675 00:34:48,239 --> 00:34:50,730 And finally, if I were to replace this vector 676 00:34:50,730 --> 00:34:52,920 by three times the first vector, I 677 00:34:52,920 --> 00:34:55,825 would still have the same determinant. 678 00:34:55,825 --> 00:34:58,200 I don't know if you've noticed what I've been driving at. 679 00:34:58,200 --> 00:35:01,380 But wasn't what I was saying here the same computational 680 00:35:01,380 --> 00:35:05,050 steps that one goes through in row reducing a matrix? 681 00:35:05,050 --> 00:35:08,050 In other words, let me row reduce this matrix 682 00:35:08,050 --> 00:35:12,720 so I get 0s every place, say, in my first column, 683 00:35:12,720 --> 00:35:14,130 except in the upper column. 684 00:35:14,130 --> 00:35:16,440 By the way, notice over here I always 685 00:35:16,440 --> 00:35:20,730 have the habit of putting a 1 in the upper left-hand corner. 686 00:35:20,730 --> 00:35:24,250 That's simply to facilitate the arithmetic. 687 00:35:24,250 --> 00:35:26,460 I hope by this stage of the game you realize, 688 00:35:26,460 --> 00:35:29,730 if this weren't a 1, I can always divide through by it, 689 00:35:29,730 --> 00:35:32,388 or what have you, and do the arithmetic another way. 690 00:35:32,388 --> 00:35:33,930 I just do this thing for convenience. 691 00:35:33,930 --> 00:35:37,680 But the important point is that this determinant is now 692 00:35:37,680 --> 00:35:39,780 equal to this determinant. 693 00:35:39,780 --> 00:35:41,070 And be very careful. 694 00:35:41,070 --> 00:35:43,430 Notice, if these had been matrices-- in other words, 695 00:35:43,430 --> 00:35:46,440 if I had put square brackets here and had made these 696 00:35:46,440 --> 00:35:47,710 matrices-- 697 00:35:47,710 --> 00:35:53,160 we would call the two matrices row equivalent but different 698 00:35:53,160 --> 00:35:54,020 matrices. 699 00:35:54,020 --> 00:35:56,340 Notice, the determinant is a number. 700 00:35:56,340 --> 00:35:59,130 And what we're saying is that the number named 701 00:35:59,130 --> 00:36:03,330 by this determinant is equal to the number named 702 00:36:03,330 --> 00:36:06,300 by this determinant-- 703 00:36:06,300 --> 00:36:08,190 equal. 704 00:36:08,190 --> 00:36:10,110 Here's the key point. 705 00:36:10,110 --> 00:36:13,980 I now notice that I have three 0s in the first column. 706 00:36:13,980 --> 00:36:17,610 I, therefore, elect to expand this determinant 707 00:36:17,610 --> 00:36:19,440 along the first column. 708 00:36:19,440 --> 00:36:20,510 Why do I do this? 709 00:36:20,510 --> 00:36:21,510 Well, let's take a look. 710 00:36:21,510 --> 00:36:24,870 First of all, I get a 1 multiplied by this 3 711 00:36:24,870 --> 00:36:27,630 by 3 determinant that's left when I strike out the row 712 00:36:27,630 --> 00:36:29,790 and column in which 1 appears. 713 00:36:29,790 --> 00:36:33,900 The beauty is that when I form the other three determinants, 3 714 00:36:33,900 --> 00:36:37,870 by 3 determinants, they're all going to be multiplied by 0. 715 00:36:37,870 --> 00:36:39,870 For example, over here, I factor out-- well 716 00:36:39,870 --> 00:36:41,910 minus 0, which is still 0. 717 00:36:41,910 --> 00:36:43,710 What do I multiply that by? 718 00:36:43,710 --> 00:36:46,240 I multiply that by the 3 by 3 determinant 719 00:36:46,240 --> 00:36:48,090 that's left when I strike out the row 720 00:36:48,090 --> 00:36:50,670 and column that this 0 appears in, namely, 721 00:36:50,670 --> 00:36:53,610 the 3 by 3 determinant, 2, 1, minus 1, 722 00:36:53,610 --> 00:36:56,490 minus 1, 4, 5, 2, 3, 4. 723 00:36:56,490 --> 00:36:58,980 But since that's being multiplied by 0, 724 00:36:58,980 --> 00:37:00,510 the product will be 0. 725 00:37:00,510 --> 00:37:04,620 To make a long story short, to evaluate this determinant, 726 00:37:04,620 --> 00:37:05,930 it's simply what? 727 00:37:05,930 --> 00:37:09,920 Plus 1 times this 3 by 3 determinant. 728 00:37:09,920 --> 00:37:13,190 I put the 1 in parentheses here to indicate that I really 729 00:37:13,190 --> 00:37:14,570 factored that 1 out. 730 00:37:14,570 --> 00:37:18,470 But really what we're saying is that this 4 by 4 determinant 731 00:37:18,470 --> 00:37:21,860 has the same value as this 3 by 3 determinant. 732 00:37:21,860 --> 00:37:24,860 And now, I row reduce this 3 by 3 determinant. 733 00:37:24,860 --> 00:37:29,000 I replace the second row, in other words, the second vector, 734 00:37:29,000 --> 00:37:32,630 by the second plus the first, the third by the third minus 735 00:37:32,630 --> 00:37:33,800 twice the first. 736 00:37:33,800 --> 00:37:37,010 I now wind up with this 3 by 3 determinant. 737 00:37:37,010 --> 00:37:39,590 Since I don't change the value of the determinant 738 00:37:39,590 --> 00:37:41,540 when I replace one vector by itself 739 00:37:41,540 --> 00:37:43,430 plus a scalar multiple of another, 740 00:37:43,430 --> 00:37:46,550 then notice very quickly that what happens over here 741 00:37:46,550 --> 00:37:49,100 is that these two determinants are equal. 742 00:37:49,100 --> 00:37:54,830 Again, I elect to expand along the first column factoring out 743 00:37:54,830 --> 00:37:59,780 the plus 1 being left with the determinant whose entries are 744 00:37:59,780 --> 00:38:01,730 5, 8, 1, 1. 745 00:38:01,730 --> 00:38:04,610 And the other two terms contribute nothing, 746 00:38:04,610 --> 00:38:07,690 because the coefficient is 0. 747 00:38:07,690 --> 00:38:09,770 But I already know how to expand-- 748 00:38:09,770 --> 00:38:13,422 I've proven that-- how to expand the 2 by 2 matrix, determinant. 749 00:38:13,422 --> 00:38:14,630 That's just going to be what? 750 00:38:14,630 --> 00:38:17,630 5 minus 8 or minus 3. 751 00:38:17,630 --> 00:38:20,240 And that's how this shortcut row reduction works-- 752 00:38:20,240 --> 00:38:24,410 much, much more elegantly than the brute force technique. 753 00:38:24,410 --> 00:38:26,510 And by the way, this is the technique 754 00:38:26,510 --> 00:38:28,940 used in most, if not all, computers 755 00:38:28,940 --> 00:38:33,050 in determining the determinant of a set of n vectors 756 00:38:33,050 --> 00:38:35,240 in an n-dimensional vector space. 757 00:38:35,240 --> 00:38:38,040 Well, as I say, this was meant as an overview. 758 00:38:38,040 --> 00:38:40,880 I hope you now see the overall picture of what determinants 759 00:38:40,880 --> 00:38:41,870 are all about. 760 00:38:41,870 --> 00:38:44,660 And in the exercises, I will try to give you 761 00:38:44,660 --> 00:38:48,560 some more sophisticated and elaborate details. 762 00:38:48,560 --> 00:38:51,410 Next time what we will do is talk 763 00:38:51,410 --> 00:38:54,710 about an application of how determinants 764 00:38:54,710 --> 00:38:58,477 are used in certain aspects of vector spaces. 765 00:38:58,477 --> 00:39:00,560 In particular, we're going to talk about something 766 00:39:00,560 --> 00:39:04,220 called eigenvalues or eigenvectors, but more 767 00:39:04,220 --> 00:39:05,780 about that next time. 768 00:39:05,780 --> 00:39:09,790 Until next time, goodbye. 769 00:39:09,790 --> 00:39:12,190 Funding for the publication of this video 770 00:39:12,190 --> 00:39:17,050 was provided by the Gabriella and Paul Rosenbaum Foundation. 771 00:39:17,050 --> 00:39:21,220 Help OCW continue to provide free and open access to MIT 772 00:39:21,220 --> 00:39:26,655 courses by making a donation at ocw.mit.edu/donate.