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 00:00:32,603 --> 00:00:33,460 HERBERT GROSS: Hi. 9 00:00:33,460 --> 00:00:35,170 In our last two lectures, we've been 10 00:00:35,170 --> 00:00:37,540 talking quite theoretically about what 11 00:00:37,540 --> 00:00:41,260 we mean by spanning vectors, dimension, vector spaces. 12 00:00:41,260 --> 00:00:45,310 Today, what we'd like to do is add some computational know-how 13 00:00:45,310 --> 00:00:47,320 to our bag of knowledge. 14 00:00:47,320 --> 00:00:51,370 And let me say at the outset by way of a reinforcement of what 15 00:00:51,370 --> 00:00:53,170 we've already said in the study guide, 16 00:00:53,170 --> 00:00:55,720 much of the material in this block 17 00:00:55,720 --> 00:00:58,720 is very difficult if you're not used to it. 18 00:00:58,720 --> 00:01:00,340 And we have already reminded you, 19 00:01:00,340 --> 00:01:03,490 but let's do it again, watch the lectures, 20 00:01:03,490 --> 00:01:05,980 get as much out of them as you can, 21 00:01:05,980 --> 00:01:09,130 do the exercises, which hopefully will cement down 22 00:01:09,130 --> 00:01:11,320 many aspects of the lecture. 23 00:01:11,320 --> 00:01:14,260 And then if certain things still aren't crystal clear, 24 00:01:14,260 --> 00:01:16,560 watch the lecture a second time. 25 00:01:16,560 --> 00:01:19,000 Hopefully, you may not even have to watch the lecture 26 00:01:19,000 --> 00:01:21,550 the second time, but try not to read 27 00:01:21,550 --> 00:01:22,970 too much into the lectures. 28 00:01:22,970 --> 00:01:25,210 This is material, which I think that once you 29 00:01:25,210 --> 00:01:29,530 get the first insight, you must then really work hard and do 30 00:01:29,530 --> 00:01:30,730 a lot of exercises. 31 00:01:30,730 --> 00:01:34,300 But at any rate, let's go into today's lesson, which I call, 32 00:01:34,300 --> 00:01:36,490 Constructing Bases. 33 00:01:36,490 --> 00:01:39,160 And by way of review, let's see what theory 34 00:01:39,160 --> 00:01:41,740 we've had going into today's lesson. 35 00:01:41,740 --> 00:01:44,740 First of all, given a vector space, v, 36 00:01:44,740 --> 00:01:49,120 with certain elements, alpha 1 up to alpha n, the set of n 37 00:01:49,120 --> 00:01:53,400 elements, alpha 1 up to alpha n, is called a basis for v, 38 00:01:53,400 --> 00:01:55,460 provided two things happen. 39 00:01:55,460 --> 00:01:59,530 First, the alphas must span v. And secondly, the alphas 40 00:01:59,530 --> 00:02:03,010 must be linearly independent. 41 00:02:03,010 --> 00:02:05,230 And, again, if this gives you any trouble, 42 00:02:05,230 --> 00:02:08,080 think in terms of the naive example of the plane 43 00:02:08,080 --> 00:02:09,759 or three-dimensional space. 44 00:02:09,759 --> 00:02:11,770 For example, in three-dimensional space, 45 00:02:11,770 --> 00:02:15,580 i, j, and k span three-dimensional space. 46 00:02:15,580 --> 00:02:19,690 And they also happen to be linearly independent. 47 00:02:19,690 --> 00:02:22,090 At any rate, what we then proved was, 48 00:02:22,090 --> 00:02:25,960 that if alpha 1 up to alpha n is a basis for v-- 49 00:02:25,960 --> 00:02:27,820 and don't be blinded by the alphas, now. 50 00:02:27,820 --> 00:02:29,560 What we're really saying here is, 51 00:02:29,560 --> 00:02:34,120 that if you have found a basis for v, which has n elements, 52 00:02:34,120 --> 00:02:36,570 then the following things must be true. 53 00:02:36,570 --> 00:02:39,490 One, any set of more than n elements, 54 00:02:39,490 --> 00:02:42,110 for example, any n plus 1 elements of v, 55 00:02:42,110 --> 00:02:43,720 are linearly dependent. 56 00:02:43,720 --> 00:02:46,750 Again, in terms of three-dimensional space, what 57 00:02:46,750 --> 00:02:50,200 we're saying is, if you have four or more vectors 58 00:02:50,200 --> 00:02:54,430 in three space, those vectors may or may not span v. 59 00:02:54,430 --> 00:02:56,380 But the thing that we're sure of is that they 60 00:02:56,380 --> 00:02:58,570 must be linearly dependent. 61 00:02:58,570 --> 00:03:01,750 You can't have more than three linearly-independent vectors 62 00:03:01,750 --> 00:03:04,060 in three-dimensional space. 63 00:03:04,060 --> 00:03:09,820 Secondly, fewer than n elements cannot span v. Again, 64 00:03:09,820 --> 00:03:12,640 in terms of our three-dimensional example, 65 00:03:12,640 --> 00:03:16,570 no two vectors in three space, no two vectors in three space, 66 00:03:16,570 --> 00:03:18,760 can span three space. 67 00:03:18,760 --> 00:03:22,150 In other words, then, a basis has two properties. 68 00:03:22,150 --> 00:03:27,920 If you have too many vectors, you cannot have linear 69 00:03:27,920 --> 00:03:30,290 independence, and if you have too few, 70 00:03:30,290 --> 00:03:32,340 you can't span the space. 71 00:03:32,340 --> 00:03:35,810 So what is it, then, that a basis must have? 72 00:03:35,810 --> 00:03:38,960 If one basis has n elements, every basis 73 00:03:38,960 --> 00:03:40,370 must have n elements. 74 00:03:40,370 --> 00:03:42,410 Again, by way of review, why? 75 00:03:42,410 --> 00:03:45,830 More than n couldn't be linearly independent. 76 00:03:45,830 --> 00:03:49,340 Fewer than n could not span the space. 77 00:03:49,340 --> 00:03:52,400 So every basis for v has n elements 78 00:03:52,400 --> 00:03:56,300 once we know that one basis has n elements. 79 00:03:56,300 --> 00:04:01,160 Now, this, of course, does not mean that any set of n elements 80 00:04:01,160 --> 00:04:05,540 will be a basis, again, in terms of three-dimensional space. 81 00:04:05,540 --> 00:04:07,970 Suppose, for example, that I take the following three 82 00:04:07,970 --> 00:04:10,200 vectors of three-dimensional space. 83 00:04:10,200 --> 00:04:14,810 Oh, heck, let's say i, j, and i plus j. 84 00:04:14,810 --> 00:04:17,510 Those are three vectors, i, j, and i plus j, 85 00:04:17,510 --> 00:04:20,339 but they only span two-dimensional space. 86 00:04:20,339 --> 00:04:22,190 The trouble with i, j, and i plus j 87 00:04:22,190 --> 00:04:24,440 is that they were linearly dependent. 88 00:04:24,440 --> 00:04:27,470 You see, all we know is that every basis for v 89 00:04:27,470 --> 00:04:28,580 has n elements. 90 00:04:28,580 --> 00:04:31,010 It does not mean that every set of n elements 91 00:04:31,010 --> 00:04:33,470 will be a basis for v. However, what 92 00:04:33,470 --> 00:04:35,750 is true is, that if we have a set 93 00:04:35,750 --> 00:04:39,125 of n linearly-independent elements, or else 94 00:04:39,125 --> 00:04:42,770 a set of n elements which span v, then 95 00:04:42,770 --> 00:04:46,490 if either of those two criteria are obeyed, 96 00:04:46,490 --> 00:04:49,640 then the set of n elements will be a basis for v. Again, 97 00:04:49,640 --> 00:04:51,650 in three-dimensional space, if you have 98 00:04:51,650 --> 00:04:54,650 three linearly-independent vectors in three space, 99 00:04:54,650 --> 00:04:55,950 they will be a basis. 100 00:04:55,950 --> 00:04:58,110 They will automatically span the space. 101 00:04:58,110 --> 00:05:01,040 And if you have three vectors which 102 00:05:01,040 --> 00:05:03,080 span the space in three space, they 103 00:05:03,080 --> 00:05:05,980 must be linearly independent. 104 00:05:05,980 --> 00:05:06,980 That's all we're saying. 105 00:05:06,980 --> 00:05:10,910 In other words, to say that the dimensions of v equals n 106 00:05:10,910 --> 00:05:12,080 is unambiguous. 107 00:05:12,080 --> 00:05:15,740 In other words, if we define the dimension of a space 108 00:05:15,740 --> 00:05:20,090 to be the number of elements in any basis, 109 00:05:20,090 --> 00:05:24,100 then what we're saying is that that is unambiguous. 110 00:05:24,100 --> 00:05:27,040 That once one basis has n elements, 111 00:05:27,040 --> 00:05:30,940 all bases will have n elements. 112 00:05:30,940 --> 00:05:33,580 And now, with respect to a particular basis, 113 00:05:33,580 --> 00:05:34,750 what we're saying this. 114 00:05:34,750 --> 00:05:37,420 Suppose we know the dimensions of v is n, 115 00:05:37,420 --> 00:05:40,030 and we have a particular basis for v, 116 00:05:40,030 --> 00:05:43,870 say, u1 up to un, a particular basis that we single out, 117 00:05:43,870 --> 00:05:48,610 then with respect to that basis, each vector, v, in the vector 118 00:05:48,610 --> 00:05:51,850 space, v, has a unique representation 119 00:05:51,850 --> 00:05:54,380 as a linear combination of the us. 120 00:05:54,380 --> 00:05:57,930 In other words, if the us are a basis for v, 121 00:05:57,930 --> 00:06:01,180 a particular vector in v can be written 122 00:06:01,180 --> 00:06:06,550 in one and only one way, as a linear combination of the us. 123 00:06:06,550 --> 00:06:10,030 Which means, you see, that relative to the us, 124 00:06:10,030 --> 00:06:15,870 v may be written unambiguously as the n-tuple c1 up to cn, 125 00:06:15,870 --> 00:06:19,190 where that's an abbreviation for c1, u1, plus, et 126 00:06:19,190 --> 00:06:21,430 cetera, c1, un. 127 00:06:21,430 --> 00:06:26,110 And in this case, we also write that v is equal to-- 128 00:06:26,110 --> 00:06:28,120 and to indicate that we're talking 129 00:06:28,120 --> 00:06:31,150 about the n-dimensional space, v, with respect 130 00:06:31,150 --> 00:06:33,910 to a particular basis, u1 up to un, 131 00:06:33,910 --> 00:06:37,750 we simply enclose the set of basis vectors 132 00:06:37,750 --> 00:06:39,130 in square brackets. 133 00:06:39,130 --> 00:06:41,350 In other words, this expression is read, 134 00:06:41,350 --> 00:06:45,310 v is being viewed as the n-dimensional space that 135 00:06:45,310 --> 00:06:48,280 has u1 up to un as its basis. 136 00:06:48,280 --> 00:06:51,820 And unless otherwise specified, every time we 137 00:06:51,820 --> 00:06:55,030 write an n-tuple in v, it's understood 138 00:06:55,030 --> 00:06:58,030 that that n-tuple is being abbreviated 139 00:06:58,030 --> 00:07:01,770 relative to the basis u1 up to un. 140 00:07:01,770 --> 00:07:04,090 And I think, again, the best way to illustrate 141 00:07:04,090 --> 00:07:06,490 this is by means of an example. 142 00:07:06,490 --> 00:07:09,310 And, in fact, the remainder of today's lesson 143 00:07:09,310 --> 00:07:13,930 will be one example wherein we will take one problem 144 00:07:13,930 --> 00:07:16,840 and try to look at it from as many different points of view 145 00:07:16,840 --> 00:07:20,230 as possible so that we get a broad spectrum as to what's 146 00:07:20,230 --> 00:07:24,040 involved in constructing bases, looking at n-tuples, 147 00:07:24,040 --> 00:07:27,520 and how the n-tuple depends on the basis, et cetera. 148 00:07:27,520 --> 00:07:30,518 So the example I have in mind is the following. 149 00:07:30,518 --> 00:07:31,310 Let's suppose that. 150 00:07:31,310 --> 00:07:35,090 All I know is that I have a four-dimensional vector space. 151 00:07:35,090 --> 00:07:36,700 What does that mean again? 152 00:07:36,700 --> 00:07:39,700 It means if the dimensions is four, 153 00:07:39,700 --> 00:07:42,190 that any basis has four elements. 154 00:07:42,190 --> 00:07:44,740 Let me now pick a particular basis, which 155 00:07:44,740 --> 00:07:48,040 I will call u1, u2, u3 and u4. 156 00:07:48,040 --> 00:07:49,690 And then notice that notation now. 157 00:07:49,690 --> 00:07:53,290 I'm saying that v is the vector space, which has u1, u2, u3, 158 00:07:53,290 --> 00:07:55,330 and u4 as a basis. 159 00:07:55,330 --> 00:07:57,520 What that means is, if I now elect 160 00:07:57,520 --> 00:08:00,220 to use the 4-tuple notation for v, 161 00:08:00,220 --> 00:08:02,840 that unless anything is said to the contrary, 162 00:08:02,840 --> 00:08:07,060 it means that that 4-tuple is relative to the vectors u1, u2, 163 00:08:07,060 --> 00:08:08,750 u3, and u4. 164 00:08:08,750 --> 00:08:11,860 For example, let's suppose I was to say to you, 165 00:08:11,860 --> 00:08:16,290 describe the subspace of v spanned by the four 166 00:08:16,290 --> 00:08:20,800 vectors alpha 1, alpha 2, alpha 3, alpha 4, where alpha 1 is 167 00:08:20,800 --> 00:08:25,840 the 4-tuple 1, 1, 2, 3, alpha 2 is the 4-tuple 2, 3, 4, 168 00:08:25,840 --> 00:08:30,640 5, alpha 3 is the 4-tuple 3, 7, 6, 5, and alpha 4 169 00:08:30,640 --> 00:08:34,294 is the 4-tuple 4, 5, 9, 9. 170 00:08:34,294 --> 00:08:35,919 Notice that what I'm really saying here 171 00:08:35,919 --> 00:08:44,560 is that alpha 1 is 1u1 plus 1u2 plus 2u3 plus 3u4, alpha 2 172 00:08:44,560 --> 00:08:50,080 is 2u1 plus 3u2 plus 4u3 plus 5u4. 173 00:08:50,080 --> 00:08:52,150 In essence, what I'm saying then is what? 174 00:08:52,150 --> 00:08:55,000 Once I've specified a particular basis, 175 00:08:55,000 --> 00:08:59,140 every 4-tuple notation is relative to that basis. 176 00:08:59,140 --> 00:09:02,950 And by the way, what do I mean by being vague over here when I 177 00:09:02,950 --> 00:09:05,140 say, describe the space? 178 00:09:05,140 --> 00:09:09,370 What I mean is, for example, is the subspace all of v? 179 00:09:09,370 --> 00:09:13,120 In other words, we know that the dimension of v is four. 180 00:09:13,120 --> 00:09:14,650 I have four vectors. 181 00:09:14,650 --> 00:09:17,590 If those four vectors are linearly independent, 182 00:09:17,590 --> 00:09:21,400 they will span all of v. I guess an equivalent question would 183 00:09:21,400 --> 00:09:25,570 be, therefore, are the alphas linearly independent? 184 00:09:25,570 --> 00:09:27,010 I can make this even more vague. 185 00:09:27,010 --> 00:09:31,385 I can say, if w doesn't equal v, what does w equal? 186 00:09:31,385 --> 00:09:35,770 In other words, what subspace is spanned by the vectors alpha 1, 187 00:09:35,770 --> 00:09:38,630 alpha 2, alpha 3, and alpha 4? 188 00:09:38,630 --> 00:09:41,020 Now, what I'd like to do is to introduce to you 189 00:09:41,020 --> 00:09:44,580 at this time a rather nice computational technique. 190 00:09:44,580 --> 00:09:46,850 It's nice for two reasons. 191 00:09:46,850 --> 00:09:49,990 The first reason is that it gets the job done very nicely, 192 00:09:49,990 --> 00:09:52,210 which is perhaps the best reason of all. 193 00:09:52,210 --> 00:09:54,670 The second reason, which is also very nice, 194 00:09:54,670 --> 00:09:57,670 is the fact that we've already had the technique before. 195 00:09:57,670 --> 00:10:01,720 Keep in mind the following property of spanning vectors. 196 00:10:01,720 --> 00:10:05,500 If we have a set of vectors and we talk about the space spanned 197 00:10:05,500 --> 00:10:08,680 by those vectors, notice from our previous homework 198 00:10:08,680 --> 00:10:12,100 exercises that we have shown, that if you replace 199 00:10:12,100 --> 00:10:15,370 any member of that spanning set by itself 200 00:10:15,370 --> 00:10:19,130 plus a multiple of any other member of the spanning set, 201 00:10:19,130 --> 00:10:23,280 you do not change the space spanned by those vectors. 202 00:10:23,280 --> 00:10:25,290 Oh, again, by means of an example-- 203 00:10:25,290 --> 00:10:27,360 because as I say, this sounds very 204 00:10:27,360 --> 00:10:29,490 difficult from an abstract point of view-- 205 00:10:29,490 --> 00:10:32,640 but if you use three-dimensional space as an analogy, 206 00:10:32,640 --> 00:10:34,290 as a reference point, I think you'll 207 00:10:34,290 --> 00:10:36,330 see more clearly what these things mean. 208 00:10:36,330 --> 00:10:39,420 For example, i, j, and k are a basis 209 00:10:39,420 --> 00:10:42,300 for three-dimensional space. 210 00:10:42,300 --> 00:10:45,030 i, j, and k span three-dimensional space. 211 00:10:45,030 --> 00:10:50,700 Suppose I replace i, or for the sake of argument, by i plus 3j. 212 00:10:50,700 --> 00:10:52,110 Look at the three vectors-- 213 00:10:52,110 --> 00:10:55,950 i plus 3j, j, and k. 214 00:10:55,950 --> 00:10:58,400 Those three vectors span the same space 215 00:10:58,400 --> 00:11:01,500 as i, j, and k, namely, all of three-dimensional space. 216 00:11:01,500 --> 00:11:06,180 In other words, i plus 3j and j are non-parallel vectors 217 00:11:06,180 --> 00:11:07,620 in the x, y plane. 218 00:11:07,620 --> 00:11:10,740 Consequently, they spend the entire x, y plane. 219 00:11:10,740 --> 00:11:14,250 And then with the k vector, all of three space is spanned. 220 00:11:14,250 --> 00:11:19,260 So the idea is, that if ever we replace a vector by itself 221 00:11:19,260 --> 00:11:23,460 plus a multiple of any other factor in the spanning space, 222 00:11:23,460 --> 00:11:26,370 we do not change the space spanned by the vectors. 223 00:11:26,370 --> 00:11:30,180 And my claim is that this suggests the row reduced 224 00:11:30,180 --> 00:11:31,680 matrix technique. 225 00:11:31,680 --> 00:11:33,060 Now, why is that? 226 00:11:33,060 --> 00:11:35,730 Well, the best way to see this is let's 227 00:11:35,730 --> 00:11:40,660 write, as a coding system, the alphas as a matrix. 228 00:11:40,660 --> 00:11:42,870 In other words, what we'll do is is we 229 00:11:42,870 --> 00:11:47,760 will label the columns of our matrix, u1, u2, u3, and then 230 00:11:47,760 --> 00:11:50,060 the rows of our matrix will simply be 1, 231 00:11:50,060 --> 00:11:56,010 1, 2, 3, 2, 3, 4, 5, 3, 7, 6, 5, 4, 5, 9, 9. 232 00:11:56,010 --> 00:11:58,980 In other words, in terms of our code, we have what? 233 00:11:58,980 --> 00:12:04,795 Alpha 1 is 1u1 plus 1u2 plus 2u3 plus u4, et cetera. 234 00:12:04,795 --> 00:12:07,770 So this is my coding matrix. 235 00:12:07,770 --> 00:12:10,710 Now, what did we do when we row reduced? 236 00:12:10,710 --> 00:12:14,010 For example, to get a 0 in this entry-- 237 00:12:14,010 --> 00:12:15,330 remember our old technique? 238 00:12:15,330 --> 00:12:18,540 We said something like, let's take the second row 239 00:12:18,540 --> 00:12:22,080 and replace it by the second minus twice the first. 240 00:12:22,080 --> 00:12:24,320 In terms of our coding system, this 241 00:12:24,320 --> 00:12:27,870 says, let's take the spanning vectors, alpha 1, alpha 2, 242 00:12:27,870 --> 00:12:31,440 alpha 3, alpha 4, and let's replace alpha 2 243 00:12:31,440 --> 00:12:36,120 by alpha 2 minus twice alpha 1. 244 00:12:36,120 --> 00:12:39,690 You see, we are replacing alpha 2 by alpha 2 245 00:12:39,690 --> 00:12:42,450 plus a scalar multiple-- the scalar happens to be minus 2, 246 00:12:42,450 --> 00:12:46,470 but that's irrelevant-- plus a scalar multiple of alpha 1. 247 00:12:46,470 --> 00:12:49,960 So the space spanned will not have changed. 248 00:12:49,960 --> 00:12:52,558 In other words, if I now go through this row reduction, 249 00:12:52,558 --> 00:12:53,100 what do I do? 250 00:12:53,100 --> 00:12:54,820 I get 0s every place here. 251 00:12:54,820 --> 00:12:56,280 How do I do that? 252 00:12:56,280 --> 00:12:58,350 I replace the second row by the second minus 253 00:12:58,350 --> 00:13:01,350 twice the first, the third row by the third minus 3 254 00:13:01,350 --> 00:13:05,040 times the first, the fourth row by the fourth minus 4 times 255 00:13:05,040 --> 00:13:05,730 the first. 256 00:13:05,730 --> 00:13:08,230 In terms of the coding system, what am I doing? 257 00:13:08,230 --> 00:13:12,030 I'm replacing alpha 2 by alpha 2 minus twice alpha 1. 258 00:13:12,030 --> 00:13:16,660 I'm replacing alpha 3 by alpha 3 minus 3 times alpha 1. 259 00:13:16,660 --> 00:13:21,510 I'm replacing alpha 4 by alpha 4 minus 4 times alpha 1. 260 00:13:21,510 --> 00:13:25,700 When I do that, my resulting matrix looks like this. 261 00:13:25,700 --> 00:13:28,350 See, notice now what's happened is these columns 262 00:13:28,350 --> 00:13:31,440 are still u1, u2, u3, and u4. 263 00:13:31,440 --> 00:13:35,790 But notice now, that whereas this vector here is still 264 00:13:35,790 --> 00:13:40,240 alpha 1, the vector 0, 1, 0, minus 1, in other words, 265 00:13:40,240 --> 00:13:44,160 what, u2 minus u4, is no longer alpha 2. 266 00:13:44,160 --> 00:13:45,660 In fact, what is it? 267 00:13:45,660 --> 00:13:46,860 How do we get this row? 268 00:13:46,860 --> 00:13:50,880 It was alpha 2 minus twice alpha 1. 269 00:13:50,880 --> 00:13:54,354 In a similar way, this would be what? 270 00:13:54,354 --> 00:13:56,760 0, 4, 0, minus 4. 271 00:13:56,760 --> 00:13:58,110 But it's also what? 272 00:13:58,110 --> 00:14:04,530 Alpha 3 minus 3 alpha 1, and the vector 0, 1, 1, minus 3 273 00:14:04,530 --> 00:14:08,220 is alpha 4 minus 4 alpha 4. 274 00:14:08,220 --> 00:14:11,010 What we're saying is these four vectors 275 00:14:11,010 --> 00:14:13,740 may not look the same as our original vectors, 276 00:14:13,740 --> 00:14:15,990 but the thing that we're sure of is 277 00:14:15,990 --> 00:14:17,820 that they span the same space. 278 00:14:17,820 --> 00:14:20,130 Now, at this point, let me pause for a moment 279 00:14:20,130 --> 00:14:22,710 and try to give you some pseudo motivation as to why 280 00:14:22,710 --> 00:14:24,360 I'm doing this. 281 00:14:24,360 --> 00:14:26,940 You see, in the same way that we solved systems 282 00:14:26,940 --> 00:14:28,980 of linear equations, when we said 283 00:14:28,980 --> 00:14:32,790 that the same system of equations 284 00:14:32,790 --> 00:14:36,390 could be represented in different ways, in other words, 285 00:14:36,390 --> 00:14:40,260 two different systems could have the same solution set, 286 00:14:40,260 --> 00:14:43,260 but one of the systems was easier to solve than the other. 287 00:14:43,260 --> 00:14:46,380 See, one thing I'm saying over here is that, somehow or other, 288 00:14:46,380 --> 00:14:50,610 these four vectors look perhaps a bit simpler 289 00:14:50,610 --> 00:14:52,560 than these four because of the fact 290 00:14:52,560 --> 00:14:54,780 that I have 0s appearing over here. 291 00:14:54,780 --> 00:14:58,090 Well, I'll go into that in more detail later. 292 00:14:58,090 --> 00:14:59,640 The other thing is this. 293 00:14:59,640 --> 00:15:02,610 Suppose, by row reducing this way, 294 00:15:02,610 --> 00:15:05,640 suppose I were to eventually wind up with the 4 295 00:15:05,640 --> 00:15:07,660 by 4 identity matrix. 296 00:15:07,660 --> 00:15:14,670 In other words, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 297 00:15:14,670 --> 00:15:15,450 1-- 298 00:15:15,450 --> 00:15:17,520 what would that matrix represent? 299 00:15:17,520 --> 00:15:21,255 Notice, in terms of this coding system that 1, 0, 0, 300 00:15:21,255 --> 00:15:24,080 0 would just be u1. 301 00:15:24,080 --> 00:15:27,900 0, 1, 0, 0 would just be u2, et cetera. 302 00:15:27,900 --> 00:15:30,420 In other words, if by row reducing this 303 00:15:30,420 --> 00:15:33,798 I could ultimately wind up with the identity matrix, 304 00:15:33,798 --> 00:15:35,340 that would say that the space spanned 305 00:15:35,340 --> 00:15:38,400 by alpha 1, alpha 2, alpha 3, and alpha 4 306 00:15:38,400 --> 00:15:42,720 is the same as the space spanned by u1, u2, u3, and u4. 307 00:15:42,720 --> 00:15:47,070 But by definition of the us, those four vectors 308 00:15:47,070 --> 00:15:50,050 span my entire four-dimensional space. 309 00:15:50,050 --> 00:15:53,130 In other words, if in row reducing my alphas 310 00:15:53,130 --> 00:15:55,350 I wind up with the identity matrix, 311 00:15:55,350 --> 00:15:58,140 that will be proof, in terms of my coding system, 312 00:15:58,140 --> 00:16:01,110 that the alphas span all of v, because they 313 00:16:01,110 --> 00:16:05,520 will span the same space as u1, u2, u3, and u4. 314 00:16:05,520 --> 00:16:08,340 So maybe that's a more immediate reason for proceeding the way 315 00:16:08,340 --> 00:16:09,180 we're going. 316 00:16:09,180 --> 00:16:11,400 At any rate, what I'm going to do now 317 00:16:11,400 --> 00:16:15,310 is to continue on with my row reduction scheme. 318 00:16:15,310 --> 00:16:18,450 In other words, what will I do now in terms of reduction? 319 00:16:18,450 --> 00:16:23,430 I will replace the first vector here by the first minus 320 00:16:23,430 --> 00:16:23,940 the second. 321 00:16:23,940 --> 00:16:26,460 In other words, I'm replacing alpha 1 322 00:16:26,460 --> 00:16:29,280 by alpha 1 minus this vector, which 323 00:16:29,280 --> 00:16:31,350 is alpha 2 minus 2 alpha 1. 324 00:16:31,350 --> 00:16:33,060 The important point being what? 325 00:16:33,060 --> 00:16:36,450 I'm replacing a spanning vector by itself 326 00:16:36,450 --> 00:16:39,040 plus a scalar multiple of another one. 327 00:16:39,040 --> 00:16:42,280 In any event, going through this operation 328 00:16:42,280 --> 00:16:47,610 what I will end up with now is the following 4 by 4 matrix. 329 00:16:47,610 --> 00:16:50,160 And you see, again, what I'm saying 330 00:16:50,160 --> 00:16:57,835 is that the 4-tuples 1, 0, 2, 4, 0, 1, 0, minus 1, 0, 0, 0, 0, 331 00:16:57,835 --> 00:17:04,170 and 0, 0, 1, minus 2 span the same space as alpha 1, 332 00:17:04,170 --> 00:17:06,119 alpha 2, alpha 3, alpha 4. 333 00:17:06,119 --> 00:17:08,010 Now let me keep emphasizing that. 334 00:17:08,010 --> 00:17:10,319 Even though I said a 4-tuple always 335 00:17:10,319 --> 00:17:14,550 means with respect to u1, u2, u3, u4, unless 336 00:17:14,550 --> 00:17:16,619 otherwise specified, I think I will always 337 00:17:16,619 --> 00:17:20,880 specify it so we make sure that we get this idea hammered home. 338 00:17:20,880 --> 00:17:23,760 Now, here's the first key point. 339 00:17:23,760 --> 00:17:27,810 My dream of having this matrix row reduced to the identity 340 00:17:27,810 --> 00:17:29,760 matrix is now shattered. 341 00:17:29,760 --> 00:17:31,080 Why is it shattered? 342 00:17:31,080 --> 00:17:34,740 The third row is already all 0s, so I can't possibly 343 00:17:34,740 --> 00:17:36,800 get a 1 into this position here. 344 00:17:36,800 --> 00:17:39,750 In other words, I can no longer hope to get the identity matrix 345 00:17:39,750 --> 00:17:40,710 here. 346 00:17:40,710 --> 00:17:43,510 But this tells me something much more than that. 347 00:17:43,510 --> 00:17:47,280 In other words, even failure is success in this situation. 348 00:17:47,280 --> 00:17:50,700 Namely, the fact that this is a 0 vector 349 00:17:50,700 --> 00:17:52,460 tells me that the space-- 350 00:17:52,460 --> 00:17:56,070 the space spanned by alpha 1, alpha 2, alpha 3, and alpha 4 351 00:17:56,070 --> 00:17:57,510 is certainly the same as the space 352 00:17:57,510 --> 00:18:00,570 spanned by these four vectors. 353 00:18:00,570 --> 00:18:03,060 But one of these vectors is the 0 vector, 354 00:18:03,060 --> 00:18:06,630 and the 0 vector spans no space, as we've already seen. 355 00:18:06,630 --> 00:18:09,600 In other words, the 0 vector is always redundant, 356 00:18:09,600 --> 00:18:12,150 because any scalar multiple of the 0 vector 357 00:18:12,150 --> 00:18:13,590 is still the 0 vector. 358 00:18:13,590 --> 00:18:17,190 So that doesn't change what any other linear combination would 359 00:18:17,190 --> 00:18:18,090 be. 360 00:18:18,090 --> 00:18:21,690 So I immediately know now that the space spanned by alpha 1, 361 00:18:21,690 --> 00:18:26,220 alpha 2, and alpha 3 can't be any more than three-dimensional 362 00:18:26,220 --> 00:18:26,740 space. 363 00:18:26,740 --> 00:18:30,180 In other words, it's also spanned by these three vectors. 364 00:18:30,180 --> 00:18:32,437 By the way, let me just continue to row 365 00:18:32,437 --> 00:18:34,270 reduce this for the sake of argument anyway. 366 00:18:34,270 --> 00:18:37,380 In other words, let me now replace the first row 367 00:18:37,380 --> 00:18:40,450 by the first minus twice the fourth. 368 00:18:40,450 --> 00:18:42,330 Or if I delete this, twice the third, 369 00:18:42,330 --> 00:18:43,500 but that's not important. 370 00:18:43,500 --> 00:18:46,860 If I do this, notice that the matrix I get now 371 00:18:46,860 --> 00:18:53,130 is 1, 0, 0, 8, 0, 1, 0, minus 1, a 0 row, and then 0, 0, 1, 372 00:18:53,130 --> 00:18:54,210 minus 2. 373 00:18:54,210 --> 00:18:58,320 Let me give the three non-zero vectors here special names. 374 00:18:58,320 --> 00:19:01,770 Let me call this vector beta 1, this vector, beta 2, 375 00:19:01,770 --> 00:19:03,220 this vector, beta 3. 376 00:19:03,220 --> 00:19:04,577 And what does that mean? 377 00:19:04,577 --> 00:19:05,160 It means what? 378 00:19:05,160 --> 00:19:07,830 That beta 1 is u1 plus 8, u4. 379 00:19:07,830 --> 00:19:10,020 Beta 2 is u2 minus u4. 380 00:19:10,020 --> 00:19:13,290 Beta 3 is u3 minus 2u, 4. 381 00:19:13,290 --> 00:19:16,680 In other words, this is beta 1, beta 2, and beta 3. 382 00:19:16,680 --> 00:19:19,650 What do we know about beta 1, beta 2, and beta 3? 383 00:19:19,650 --> 00:19:22,830 Because they were obtained from the matrix of the alphas 384 00:19:22,830 --> 00:19:26,550 by row reduction, they span the same space as alpha 1, 385 00:19:26,550 --> 00:19:28,810 alpha 2, alpha 3, alpha 4. 386 00:19:28,810 --> 00:19:30,780 In other words, the first thing we know 387 00:19:30,780 --> 00:19:33,570 is that the space spanned by alpha 1, alpha 2, alpha 3, 388 00:19:33,570 --> 00:19:36,720 and alpha 4 is not four-dimensional space, 389 00:19:36,720 --> 00:19:38,610 but rather, it's the space spanned 390 00:19:38,610 --> 00:19:42,547 by beta 1, beta 2, and beta 3. 391 00:19:42,547 --> 00:19:44,130 By the way, that immediately tells me, 392 00:19:44,130 --> 00:19:47,880 since beta 1, beta 2, and beta 3 span my space, w, 393 00:19:47,880 --> 00:19:50,640 that I'm looking for, that the dimension of that space 394 00:19:50,640 --> 00:19:54,690 can be no greater than 3, because 3 vectors span it. 395 00:19:54,690 --> 00:19:56,910 Now, at this stage of the game, the question 396 00:19:56,910 --> 00:19:59,790 might come up, what's so great about the betas? 397 00:19:59,790 --> 00:20:03,150 Why do we have the betas rather than the alphas? 398 00:20:03,150 --> 00:20:05,020 Do the betas help us at all? 399 00:20:05,020 --> 00:20:08,160 And my claim is, making this metaphysical for a moment, 400 00:20:08,160 --> 00:20:10,680 if you just look at what the betas look like over here, just 401 00:20:10,680 --> 00:20:12,055 look at what the betas look like, 402 00:20:12,055 --> 00:20:14,490 forget about the u4 component, notice 403 00:20:14,490 --> 00:20:17,640 that the betas without the u4 component 404 00:20:17,640 --> 00:20:19,940 look like the identity matrix. 405 00:20:19,940 --> 00:20:23,830 See, 1, 0, 0, 0, 1, 0, 0, 0, 1. 406 00:20:23,830 --> 00:20:28,110 My claim is that the betas are, indeed, a very special basis. 407 00:20:28,110 --> 00:20:29,990 And let me show you what I mean by that. 408 00:20:29,990 --> 00:20:33,430 What does it mean for a vector to be in the space spanned 409 00:20:33,430 --> 00:20:35,830 by beta 1, beta 2, and beta 3? 410 00:20:35,830 --> 00:20:39,280 Well it means that that vector must be a linear combination 411 00:20:39,280 --> 00:20:41,590 of beta 1, beta 2, and beta 3. 412 00:20:41,590 --> 00:20:43,930 Let me write down that linear combination in the form 413 00:20:43,930 --> 00:20:49,820 x1 beta 1 plus x2 beta 2 plus x3 beta 3, where x1, x2, 414 00:20:49,820 --> 00:20:53,230 x3 represent arbitrary constants here. 415 00:20:53,230 --> 00:20:55,630 By the way, don't be upset that I've used xs now 416 00:20:55,630 --> 00:20:58,180 for the first time instead of cs. 417 00:20:58,180 --> 00:20:59,890 I've essentially done the same thing 418 00:20:59,890 --> 00:21:03,080 that we did when we were dealing with arrows in the plane. 419 00:21:03,080 --> 00:21:05,980 When we were talking about a typical arrow, 420 00:21:05,980 --> 00:21:09,700 we refer to it as being xi plus yj 421 00:21:09,700 --> 00:21:12,220 rather than refer to it as being ai plus 422 00:21:12,220 --> 00:21:16,660 bj to indicate, somehow how or other, that we had arbitrarily 423 00:21:16,660 --> 00:21:19,360 chosen the constants, the coefficients. 424 00:21:19,360 --> 00:21:20,920 It's sort of like in algebra. 425 00:21:20,920 --> 00:21:24,520 There's no law that says, when you have a constant, that you 426 00:21:24,520 --> 00:21:27,580 couldn't call the constant x, or if you have a variable, 427 00:21:27,580 --> 00:21:29,440 you couldn't call the variable c. 428 00:21:29,440 --> 00:21:31,570 But somehow or other, when we see a c, 429 00:21:31,570 --> 00:21:33,700 we assume it's a constant in algebra. 430 00:21:33,700 --> 00:21:35,830 When we see an x, we assume it's a variable. 431 00:21:35,830 --> 00:21:38,380 The xs are just thrown in here to suggest 432 00:21:38,380 --> 00:21:41,620 that I'm looking at all linear combinations of beta 1, beta 433 00:21:41,620 --> 00:21:43,000 2, and beta 3. 434 00:21:43,000 --> 00:21:46,630 And now, watch how beautifully the form of the betas 435 00:21:46,630 --> 00:21:48,070 comes to help us now. 436 00:21:48,070 --> 00:21:52,060 Let's look explicitly to see what x1 beta 1 plus x2 beta 2 437 00:21:52,060 --> 00:21:54,340 plus x3 beta 3 really is. 438 00:21:54,340 --> 00:21:57,460 Let's come back to what our betas were. 439 00:21:57,460 --> 00:22:01,630 To multiply beta 1 by x1 means that we multiply each component 440 00:22:01,630 --> 00:22:02,860 by x1. 441 00:22:02,860 --> 00:22:06,310 So this will be what? x1, 0, 0, 8x1. 442 00:22:06,310 --> 00:22:09,810 We multiply each component of beta 2 by x2. 443 00:22:09,810 --> 00:22:13,000 That's 0, x2, 0, minus x2. 444 00:22:13,000 --> 00:22:16,450 We multiply each component of beta 3 by x3. 445 00:22:16,450 --> 00:22:19,810 So I'll get 0, 0, x3 minus 2x3. 446 00:22:19,810 --> 00:22:22,930 Notice, again, that these components as 4-tuples 447 00:22:22,930 --> 00:22:24,400 are relative to the us. 448 00:22:24,400 --> 00:22:30,310 In other words, x1 beta 1 plus x2 beta 2 plus x3 beta 3 449 00:22:30,310 --> 00:22:33,730 are the three 4-tuples given here, 450 00:22:33,730 --> 00:22:35,920 where it's understood that the columns values 451 00:22:35,920 --> 00:22:38,800 are u1, u2, u3, and u4. 452 00:22:38,800 --> 00:22:41,230 Now, I want to add these up. 453 00:22:41,230 --> 00:22:44,470 If I add them up, I add component by component. 454 00:22:44,470 --> 00:22:47,440 Here's the beauty of that row reduced matrix form 455 00:22:47,440 --> 00:22:48,790 that yielded the betas. 456 00:22:48,790 --> 00:22:50,530 Notice, that when I add these up, 457 00:22:50,530 --> 00:22:53,260 x1 appears only in the first vector. 458 00:22:53,260 --> 00:22:56,950 There are 0s every place else, x2 in the second, 459 00:22:56,950 --> 00:22:58,600 x3 only in the third. 460 00:22:58,600 --> 00:23:00,760 Notice, that when I add these up, I get what? 461 00:23:00,760 --> 00:23:07,510 x1, x2, x3, 8x1 minus x2 minus 2x3. 462 00:23:07,510 --> 00:23:09,580 In other words, any linear combination 463 00:23:09,580 --> 00:23:13,480 of the betas in terms of the us has a very interesting 464 00:23:13,480 --> 00:23:14,740 representation. 465 00:23:14,740 --> 00:23:19,680 Namely, x1 beta 1 plus x2 beta 2 plus x3 beta 3 466 00:23:19,680 --> 00:23:25,390 is simply x1, u1 plus x2, u2 plus x3, u3. 467 00:23:25,390 --> 00:23:30,310 Then plus 8x1 minus x2 minus 2x3, x4. 468 00:23:30,310 --> 00:23:32,740 By the way, as an aside, notice that I can now 469 00:23:32,740 --> 00:23:36,400 very quickly show you that the betas are linearly independent. 470 00:23:36,400 --> 00:23:38,920 Namely, for linear independence, notice 471 00:23:38,920 --> 00:23:41,770 that I want to show that the only way this can be 0 472 00:23:41,770 --> 00:23:44,080 is if all the xs are 0. 473 00:23:44,080 --> 00:23:46,570 But notice that since this combination comes out 474 00:23:46,570 --> 00:23:49,300 to be this, the only way this can be 0 475 00:23:49,300 --> 00:23:51,040 is to have 0s every place. 476 00:23:51,040 --> 00:23:56,140 That says 0 must equal x1, 0 equals x2, 0 equals x3. 477 00:23:56,140 --> 00:23:59,080 In other words, writing this out in more detail, x1 beta 478 00:23:59,080 --> 00:24:03,160 1 plus x2 beta 2 plus x3 beta 3 equals 0 479 00:24:03,160 --> 00:24:07,960 means this, which in turn means that x1, x2, and x3 must all 480 00:24:07,960 --> 00:24:08,770 be 0. 481 00:24:08,770 --> 00:24:10,900 That proves, in particular, that the betas 482 00:24:10,900 --> 00:24:12,490 are linearly independent. 483 00:24:12,490 --> 00:24:15,550 And therefore, my space, w, not only 484 00:24:15,550 --> 00:24:18,370 is spanned by beta 1, beta 2, and beta 3, 485 00:24:18,370 --> 00:24:19,850 but they are a basis. 486 00:24:19,850 --> 00:24:21,790 That's why I write the square brackets here. 487 00:24:21,790 --> 00:24:26,500 And so my dimensions of w is actually 3. 488 00:24:26,500 --> 00:24:30,010 And by the way, let me show you, again, what this thing means. 489 00:24:30,010 --> 00:24:32,590 It means that, you see, if something isn't w, 490 00:24:32,590 --> 00:24:36,160 If it's written as a 4-tuple, all I have to do 491 00:24:36,160 --> 00:24:38,890 is know what the first three components are, 492 00:24:38,890 --> 00:24:41,590 because the fourth one is expressed 493 00:24:41,590 --> 00:24:43,030 in terms of the other three. 494 00:24:43,030 --> 00:24:47,960 For example, the u1 cannot belong to my space, w. 495 00:24:47,960 --> 00:24:48,610 Why? 496 00:24:48,610 --> 00:24:55,170 Because u1 has x1 equal to 1, x2 equal to 0, x3 equal to 0. 497 00:24:55,170 --> 00:24:57,010 And this tells us what? 498 00:24:57,010 --> 00:25:01,840 That when x1 is 1, and x2 and x3 are 0, x4 must be 8. 499 00:25:01,840 --> 00:25:05,860 In other words, the only member of w that begins with 1, 500 00:25:05,860 --> 00:25:11,253 0, 0, is 1, 0, 0, 8, and that's exactly what beta 1 is. 501 00:25:11,253 --> 00:25:12,670 In other words, I leave it for you 502 00:25:12,670 --> 00:25:18,340 the check that u1, u2, and u3 do not belong to us space, w. 503 00:25:18,340 --> 00:25:19,560 Well, so much for that. 504 00:25:19,560 --> 00:25:22,620 Let me review now very quickly what we've shown. 505 00:25:22,620 --> 00:25:25,480 These betas are a rather remarkable basis. 506 00:25:25,480 --> 00:25:28,980 In other words, I claim that beta 1, beta 2, beta 3 507 00:25:28,980 --> 00:25:31,410 is a natural basis for w, you see, 508 00:25:31,410 --> 00:25:33,030 much nicer than the alphas. 509 00:25:33,030 --> 00:25:34,110 Why is that? 510 00:25:34,110 --> 00:25:37,050 Well, let me pick any 4-tuple in w. 511 00:25:37,050 --> 00:25:40,530 Let me call it x1, x2, x3, x4. 512 00:25:40,530 --> 00:25:45,180 My claim is that once I know that that 4-tuple is in w, 513 00:25:45,180 --> 00:25:48,420 I can write down the linear combination of the betas 514 00:25:48,420 --> 00:25:50,880 that this vector is by inspection. 515 00:25:50,880 --> 00:25:54,510 Namely, what I just showed was that that 4-tuple is what? 516 00:25:54,510 --> 00:26:00,270 It's x1 beta 1 plus x2 beta 2 plus x3 beta 3. 517 00:26:00,270 --> 00:26:02,280 Or equivalently, if you want to see it 518 00:26:02,280 --> 00:26:07,140 in terms of the 4-tuple of notation in terms of the us, 519 00:26:07,140 --> 00:26:10,020 the first three components can be chosen at random, 520 00:26:10,020 --> 00:26:13,210 and then the fourth component is given by this. 521 00:26:13,210 --> 00:26:15,570 And by way of a quick example, notice 522 00:26:15,570 --> 00:26:18,570 that as a 4-tuple alpha 1 was what? 523 00:26:18,570 --> 00:26:21,100 It was 1, 1, 2, 3. 524 00:26:21,100 --> 00:26:26,390 In other words, it was u1 plus u2 plus 2u3 plus 3u4. 525 00:26:26,390 --> 00:26:29,520 Notice that because the first three coefficients were 526 00:26:29,520 --> 00:26:33,360 1, 1, 2 relative to the betas, the coefficients 527 00:26:33,360 --> 00:26:35,320 are also 1, 1, 2. 528 00:26:35,320 --> 00:26:38,880 In other words, I can write alpha 1 as the 3-tuple 1, 529 00:26:38,880 --> 00:26:43,470 1, 2, provided I tacitly understand here 530 00:26:43,470 --> 00:26:45,840 that I'm making reference to the special betas 531 00:26:45,840 --> 00:26:47,430 as being my basis. 532 00:26:47,430 --> 00:26:49,890 You see, in other words, referring to the alpha 1, 533 00:26:49,890 --> 00:26:52,860 alpha 2, alpha 3, alpha 4 that we're talking about, 534 00:26:52,860 --> 00:26:55,350 just drop the fourth-- 535 00:26:55,350 --> 00:27:01,750 in terms of the 4-tuples, drop the use of four coefficient 536 00:27:01,750 --> 00:27:03,760 and just write down the remaining 3-tuple. 537 00:27:03,760 --> 00:27:06,100 In other words, alpha 2 is 2 beta 1 538 00:27:06,100 --> 00:27:08,440 plus 3 beta 2 plus 4 beta 3. 539 00:27:08,440 --> 00:27:12,520 Alpha 3 is 3 beta 1 plus 7 beta 2 plus 6 beta 3. 540 00:27:12,520 --> 00:27:17,120 Alpha 4 is 4 beta 1 plus 5 beta 2 plus 9 beta 3. 541 00:27:17,120 --> 00:27:19,510 You see, the betas are a very nice basis, 542 00:27:19,510 --> 00:27:21,310 because they're that special basis 543 00:27:21,310 --> 00:27:23,170 that once I know what the vector looks 544 00:27:23,170 --> 00:27:26,140 like as a 4-tuple with respect to the us, 545 00:27:26,140 --> 00:27:29,050 I can immediately write down what it looks like with respect 546 00:27:29,050 --> 00:27:29,980 to the betas. 547 00:27:29,980 --> 00:27:31,960 Let me give you another example. 548 00:27:31,960 --> 00:27:34,377 Suppose I just pick at random-- 549 00:27:34,377 --> 00:27:35,710 well, it's not picked at random. 550 00:27:35,710 --> 00:27:36,970 I picked one that works here. 551 00:27:36,970 --> 00:27:41,215 But let's suppose I just take the 4-tuple 2, 5, 3, 5, 552 00:27:41,215 --> 00:27:43,000 and I want to know whether that's in w. 553 00:27:43,000 --> 00:27:44,860 And by the way, again, let me remind you 554 00:27:44,860 --> 00:27:48,110 just one more time that 4-tuple stands for what? 555 00:27:48,110 --> 00:27:52,360 2u1 plus 5u2 plus 3u3 plus 5u4. 556 00:27:52,360 --> 00:27:54,490 Does that 4-tuple belong to w? 557 00:27:54,490 --> 00:27:57,820 The answer is, it belongs to w if and only 558 00:27:57,820 --> 00:28:00,120 if it's equal to what? 559 00:28:00,120 --> 00:28:04,690 2 beta 1 plus 5 beta 2 plus 3 beta 3. 560 00:28:04,690 --> 00:28:08,320 That's the beauty of that special basis of the betas. 561 00:28:08,320 --> 00:28:12,400 Well, 2 beta 1 is simply 2, 0, 0, 16. 562 00:28:12,400 --> 00:28:17,140 5 beta 2 is 0, 5, 0, minus 5, and 3 beta 3 563 00:28:17,140 --> 00:28:19,090 is 0, 0, 3, minus 6. 564 00:28:19,090 --> 00:28:22,900 Recalling, of course, that beta 1 was 1, 0, 0, 8. 565 00:28:22,900 --> 00:28:26,320 Beta 2 was 0, 1, 0, minus 1, and that beta 3 566 00:28:26,320 --> 00:28:28,400 was 0, 0, 1, minus 2. 567 00:28:28,400 --> 00:28:30,400 So again, this is all with respect-- 568 00:28:30,400 --> 00:28:32,710 as 4-tuples with respect to the us. 569 00:28:32,710 --> 00:28:34,630 If I add these up, I get what? 570 00:28:34,630 --> 00:28:39,130 2, 5, 3, 5, which is exactly the vector I'm dealing with. 571 00:28:39,130 --> 00:28:41,680 So 2, 5, 3, 5 belongs to w. 572 00:28:41,680 --> 00:28:45,210 In fact, another way of looking at this is once the 2, 5, 573 00:28:45,210 --> 00:28:49,300 and the 3 are given as the first three components, the only way 574 00:28:49,300 --> 00:28:51,730 that the 4-tuple can belong to w is 575 00:28:51,730 --> 00:28:53,927 if that fourth component is 5. 576 00:28:53,927 --> 00:28:55,510 In other words, again, looking at this 577 00:28:55,510 --> 00:28:57,670 from one more point of view, remember over 578 00:28:57,670 --> 00:29:01,090 here we showed that the fourth component in terms of the us 579 00:29:01,090 --> 00:29:04,600 had to be eight times the first minus the second minus twice 580 00:29:04,600 --> 00:29:06,040 the third. 581 00:29:06,040 --> 00:29:09,790 And given 2, 5, 3, the only way that can happen is what? 582 00:29:09,790 --> 00:29:13,540 8 times 2 minus 5 minus twice three 583 00:29:13,540 --> 00:29:16,630 and that happens to be five. 584 00:29:16,630 --> 00:29:18,460 Now we come to the home stretch. 585 00:29:18,460 --> 00:29:22,810 And that is, knowing now what the basis looks 586 00:29:22,810 --> 00:29:25,120 like in terms of the betas, we come back 587 00:29:25,120 --> 00:29:26,560 to a more crucial question. 588 00:29:26,560 --> 00:29:29,110 And that is, look it, we were given alpha 1, alpha 589 00:29:29,110 --> 00:29:30,910 2, alpha 3, alpha 4. 590 00:29:30,910 --> 00:29:33,070 Those were the vectors that we were given. 591 00:29:33,070 --> 00:29:35,170 Maybe for the problem that we're dealing with, 592 00:29:35,170 --> 00:29:36,700 it's important to have everything 593 00:29:36,700 --> 00:29:38,200 relative to the alphas. 594 00:29:38,200 --> 00:29:43,030 So now what we're saying is, knowing what 2, 5, 3, 5 looks 595 00:29:43,030 --> 00:29:45,430 like as a linear combination of the betas, 596 00:29:45,430 --> 00:29:48,970 how can we conveniently express that as a linear combination 597 00:29:48,970 --> 00:29:50,020 of the alphas? 598 00:29:50,020 --> 00:29:53,440 And again, the answer comes from row reduced matrices. 599 00:29:53,440 --> 00:29:55,720 In other words, all we're going to say is this. 600 00:29:55,720 --> 00:29:58,060 Look it, we know what this vector looks 601 00:29:58,060 --> 00:29:59,590 like in terms of the betas. 602 00:29:59,590 --> 00:30:03,820 It's 2 beta 1 plus 5 beta 2 plus 3 beta 3. 603 00:30:03,820 --> 00:30:08,110 Suppose we could now express the betas in terms of the alphas. 604 00:30:08,110 --> 00:30:10,150 See, sort of like before with the matrices. 605 00:30:10,150 --> 00:30:12,430 We sort of want to talk about inverting-- 606 00:30:12,430 --> 00:30:15,040 interchanging the roles of the vectors. 607 00:30:15,040 --> 00:30:16,130 Here's what I'm saying. 608 00:30:16,130 --> 00:30:19,240 Look it, we already know from over here 609 00:30:19,240 --> 00:30:23,560 that alpha 1 is beta 1 plus beta 2 plus 2 beta 3, et cetera. 610 00:30:23,560 --> 00:30:25,270 Let me do the following thing. 611 00:30:25,270 --> 00:30:27,820 Let me write down an augmented matrix. 612 00:30:27,820 --> 00:30:29,350 The first three columns will stand 613 00:30:29,350 --> 00:30:33,340 for beta 1, beta 2, and beta 3, and the last four columns 614 00:30:33,340 --> 00:30:37,130 will stand for alpha 1, alpha 2, alpha 3, and alpha 4. 615 00:30:37,130 --> 00:30:40,430 Notice that I can write alpha 1 by two different names. 616 00:30:40,430 --> 00:30:44,460 It's 1, 1, 2 relative to the betas and 1, 0, 0, 617 00:30:44,460 --> 00:30:46,360 0 relative to the alphas. 618 00:30:46,360 --> 00:30:49,300 And similarly, alpha 2, alpha 3, alpha 4 619 00:30:49,300 --> 00:30:52,370 can be represented as I've written these things over here. 620 00:30:52,370 --> 00:30:53,440 Now, what have I done? 621 00:30:53,440 --> 00:30:57,350 I have written each vector in two different ways-- 622 00:30:57,350 --> 00:31:01,000 one relative to the betas and one relative to the alphas. 623 00:31:01,000 --> 00:31:03,530 If I now row reduce this matrix-- 624 00:31:03,530 --> 00:31:06,040 notice that I'm dealing with vectors-- 625 00:31:06,040 --> 00:31:09,070 whatever vector I get by one name 626 00:31:09,070 --> 00:31:12,610 must be the same vector with respect to the other name. 627 00:31:12,610 --> 00:31:16,300 In other words, if I start row reducing now in the usual way, 628 00:31:16,300 --> 00:31:18,940 this matrix becomes this matrix. 629 00:31:18,940 --> 00:31:21,010 And again, to show what we're saying is, 630 00:31:21,010 --> 00:31:23,340 this is just another way of saying what? 631 00:31:23,340 --> 00:31:27,070 That, for example, this is beta 2, 632 00:31:27,070 --> 00:31:30,010 and this says that the vector beta 2 is the same as alpha 2 633 00:31:30,010 --> 00:31:32,140 minus 2 alpha 1. 634 00:31:32,140 --> 00:31:33,295 This happens to be-- 635 00:31:33,295 --> 00:31:34,420 oh, let's look at this one. 636 00:31:34,420 --> 00:31:37,290 This is beta 2 plus beta 3. 637 00:31:37,290 --> 00:31:38,530 And that's the same as what? 638 00:31:38,530 --> 00:31:41,650 Minus 4 alpha 1 plus alpha 4. 639 00:31:41,650 --> 00:31:43,390 And you see, what I'm going to try to do 640 00:31:43,390 --> 00:31:46,720 is row reduce this part here and hopefully be 641 00:31:46,720 --> 00:31:49,360 able to express the betas in terms of the alphas. 642 00:31:49,360 --> 00:31:53,860 So what I now do is continue my row reduction technique. 643 00:31:53,860 --> 00:31:58,150 And you say, if I do that, my next step gives me this, 644 00:31:58,150 --> 00:32:00,640 and this is a rather remarkable thing. 645 00:32:00,640 --> 00:32:04,270 Because you'll now notice that reading my first three 646 00:32:04,270 --> 00:32:08,380 components here, this tells me that the third vector, 647 00:32:08,380 --> 00:32:11,860 at least in terms of the betas, is the 0 vector. 648 00:32:11,860 --> 00:32:13,360 By the way, what's another name-- 649 00:32:13,360 --> 00:32:14,777 see, what I'm saying this is what? 650 00:32:14,777 --> 00:32:18,480 0 beta 1 plus 0 beta 2 plus 0 beta 3. 651 00:32:18,480 --> 00:32:20,230 What's another name for that vector? 652 00:32:20,230 --> 00:32:24,500 It's 5 alpha 1 minus 4 alpha 2 plus alpha 3. 653 00:32:24,500 --> 00:32:26,980 In other words, as an aside, 5 alpha 654 00:32:26,980 --> 00:32:31,000 1 minus 4 alpha 2 plus alpha 3 is the 0 vector. 655 00:32:31,000 --> 00:32:37,450 Therefore, by transposing, alpha 3 is 4 alpha 2 minus 5 alpha 1. 656 00:32:37,450 --> 00:32:39,670 And now, indeed, we have not only 657 00:32:39,670 --> 00:32:42,880 seen that the alphas were linearly dependent, 658 00:32:42,880 --> 00:32:44,440 we have found the redundancy. 659 00:32:44,440 --> 00:32:47,770 That in the order given, alpha 3 is a linear combination 660 00:32:47,770 --> 00:32:49,360 of alpha 1 and alpha 2. 661 00:32:49,360 --> 00:32:52,120 By the way, just as a computational check, 662 00:32:52,120 --> 00:32:56,140 notice by our definitions of alpha 1 alpha 2 that 4 alpha 2 663 00:32:56,140 --> 00:32:58,510 is 8, 12, 16, 20. 664 00:32:58,510 --> 00:33:03,670 Minus 5 alpha 1 is minus 5, minus 5, minus 10, minus 15. 665 00:33:03,670 --> 00:33:08,350 If I add these two, I get 3, 7, 6, 5, 666 00:33:08,350 --> 00:33:10,540 which, as you will recall from looking 667 00:33:10,540 --> 00:33:14,050 at our definition of alpha 3, is, indeed, alpha 3. 668 00:33:14,050 --> 00:33:16,090 At any rate, what I'd not like to point out 669 00:33:16,090 --> 00:33:18,010 is that what this piece of information 670 00:33:18,010 --> 00:33:22,920 tells me is that this is the redundant piece of information. 671 00:33:22,920 --> 00:33:27,310 That what we really know is that the betas do not need alpha 3 672 00:33:27,310 --> 00:33:28,120 to express them. 673 00:33:28,120 --> 00:33:28,900 Why? 674 00:33:28,900 --> 00:33:30,370 Because the betas could originally 675 00:33:30,370 --> 00:33:32,500 be expressed in terms of alpha 1, alpha 2, 676 00:33:32,500 --> 00:33:34,000 alpha 3, and alpha 4. 677 00:33:34,000 --> 00:33:35,910 So you don't lose track of that. 678 00:33:35,910 --> 00:33:38,080 The space spanned by the betas is 679 00:33:38,080 --> 00:33:40,600 the same as the space spanned by the alphas. 680 00:33:40,600 --> 00:33:42,880 So, in particular, each of the betas 681 00:33:42,880 --> 00:33:45,670 must be a linear combination of the alphas by definition 682 00:33:45,670 --> 00:33:48,160 of spanning space, just as each of the alphas 683 00:33:48,160 --> 00:33:50,410 must be a linear combination of the betas. 684 00:33:50,410 --> 00:33:52,240 So I know that the base can be expressed 685 00:33:52,240 --> 00:33:53,620 in terms of the alphas. 686 00:33:53,620 --> 00:33:56,620 But now, I've seen that alpha 3 is redundant, 687 00:33:56,620 --> 00:33:59,260 because alpha 3 can be expressed in terms of alpha 1 688 00:33:59,260 --> 00:34:00,370 and alpha 2. 689 00:34:00,370 --> 00:34:02,350 So what I now have is what? 690 00:34:02,350 --> 00:34:05,050 That beta 1, beta 2, and beta 3 can 691 00:34:05,050 --> 00:34:08,389 be expressed in terms of alpha 1, alpha 2, and alpha 4. 692 00:34:08,389 --> 00:34:11,260 See, notice that with this row deleted, 693 00:34:11,260 --> 00:34:14,650 the alpha 3 column has all 0s in it. 694 00:34:14,650 --> 00:34:19,210 And by the way, if I now finally complete my row reduction, 695 00:34:19,210 --> 00:34:25,000 you see, I wind up with this matrix, this 3 by 7 matrix. 696 00:34:25,000 --> 00:34:27,320 And this tells me, in particular, what? 697 00:34:27,320 --> 00:34:30,880 How to express beta 1, beta 2, and beta 3 in terms 698 00:34:30,880 --> 00:34:33,460 of the alphas, in particular, beta 1 699 00:34:33,460 --> 00:34:37,060 is the 4-tuple 7, 1, 0, minus 2. 700 00:34:37,060 --> 00:34:40,870 And notice now, I don't want to say this 701 00:34:40,870 --> 00:34:43,630 as a 4-tuple in the following sense. 702 00:34:43,630 --> 00:34:48,400 Notice, if I now were to write 7, 1, 0, minus 2, 703 00:34:48,400 --> 00:34:49,600 that would be redundant. 704 00:34:49,600 --> 00:34:53,560 Because now I'm talking relative to the alphas, not the us. 705 00:34:53,560 --> 00:34:54,760 The important thing is what? 706 00:34:54,760 --> 00:34:59,470 That beta 1 is 7 alpha 1 plus alpha 2 minus 2 alpha 4. 707 00:34:59,470 --> 00:35:03,253 Beta 2 is minus 2 alpha 1 plus alpha 2-- 708 00:35:03,253 --> 00:35:04,420 well, just that's all it is. 709 00:35:04,420 --> 00:35:08,810 And beta 3 is minus 2 alpha 1 minus alpha 2 plus alpha 4. 710 00:35:08,810 --> 00:35:11,110 This tells me that the betas are a basis, 711 00:35:11,110 --> 00:35:14,260 the alphas weren't the basis, but alpha 1, alpha 2, 712 00:35:14,260 --> 00:35:16,240 and alpha 4 as a basis. 713 00:35:16,240 --> 00:35:18,410 Getting back to our original problem, 714 00:35:18,410 --> 00:35:21,670 now that we're expressed beta 1, beta 2, and beta 3 715 00:35:21,670 --> 00:35:24,460 in terms of alpha 1, alpha 2, and alpha 4, 716 00:35:24,460 --> 00:35:30,010 knowing that 2, 5, 3, 5 is 2 beta 1 plus 5 beta 2 plus 3 717 00:35:30,010 --> 00:35:32,920 beta 3, we can now replace beta 1 718 00:35:32,920 --> 00:35:36,630 by what it's equal to in terms of the alphas, beta 2 719 00:35:36,630 --> 00:35:40,930 by what it's equal to in terms of the alphas, beta 3-- 720 00:35:40,930 --> 00:35:45,130 well, beta 2 is equal to minus 2 alpha 1 plus alpha 2. 721 00:35:45,130 --> 00:35:49,300 Beta 3 is minus 2 alpha 1 minus alpha 2 plus alpha 4. 722 00:35:49,300 --> 00:35:51,310 Replacing the betas by what they're 723 00:35:51,310 --> 00:35:55,330 equal to in terms of the alphas, I wind up with this expression. 724 00:35:55,330 --> 00:35:58,900 And I find that 2, 5, 3, 5 is the following 725 00:35:58,900 --> 00:36:02,080 linear combination of alpha 1, alpha 2, and alpha 4. 726 00:36:02,080 --> 00:36:06,790 Namely, minus 2 alpha 1 plus 4 alpha 2 minus alpha 4. 727 00:36:06,790 --> 00:36:09,730 And I leave it for you to check that, if you actually take 728 00:36:09,730 --> 00:36:12,670 alpha 1, alpha 2, and alpha 4 as given 729 00:36:12,670 --> 00:36:14,600 at the beginning of this example, 730 00:36:14,600 --> 00:36:18,100 you will indeed find that this linear combination, minus 2 731 00:36:18,100 --> 00:36:21,370 alpha 1 plus 4 alpha 2 minus alpha 4, 732 00:36:21,370 --> 00:36:26,250 is indeed the vector 2, 5, 3, 5 that we're talking about. 733 00:36:26,250 --> 00:36:28,520 Well, as I mentioned to you, I want 734 00:36:28,520 --> 00:36:30,850 you to get an overview of what we're talking about. 735 00:36:30,850 --> 00:36:33,190 There are going to be plenty of exercises on this. 736 00:36:33,190 --> 00:36:35,260 Let me just summarize, or at least 737 00:36:35,260 --> 00:36:38,050 give a partial summary, of what we have really 738 00:36:38,050 --> 00:36:41,080 done computationally in this example. 739 00:36:41,080 --> 00:36:43,780 We started with four-dimensional space, v, 740 00:36:43,780 --> 00:36:47,920 relative to a basis u1, u2, u3, u4. 741 00:36:47,920 --> 00:36:49,420 We picked four vectors-- 742 00:36:49,420 --> 00:36:52,930 alpha 1, alpha 2, alpha 3, alpha 4-- in that space, 743 00:36:52,930 --> 00:36:56,530 and asked to see what space was spanned by those four vectors. 744 00:36:56,530 --> 00:36:59,715 And we showed that there were three very special vectors-- 745 00:36:59,715 --> 00:37:01,340 they had a very nice form-- 746 00:37:01,340 --> 00:37:05,990 called beta 1, beta 2, beta 3, such that v had-- 747 00:37:05,990 --> 00:37:09,640 the w had beta 1, beta 2, and beta 3 as a basis 748 00:37:09,640 --> 00:37:11,870 so that the dimension w was 3. 749 00:37:11,870 --> 00:37:14,900 We also showed in this particular problem 750 00:37:14,900 --> 00:37:17,060 that not only was the dimension three, 751 00:37:17,060 --> 00:37:22,790 but once you picked the u1, u2, and u3 components arbitrarily, 752 00:37:22,790 --> 00:37:26,090 the u4 component was determined by these 753 00:37:26,090 --> 00:37:27,450 in this particular way. 754 00:37:27,450 --> 00:37:29,330 In other words, another way of saying this 755 00:37:29,330 --> 00:37:35,360 is that w consisted of all linear combinations 756 00:37:35,360 --> 00:37:42,160 of the betas, where these are the same xs that appear here. 757 00:37:42,160 --> 00:37:44,470 And as a case in point, we showed 758 00:37:44,470 --> 00:37:49,180 that a particular vector, 2, 5, 3, 5, belonged to w. 759 00:37:49,180 --> 00:37:52,570 That when we wrote that vector relative to the betas, 760 00:37:52,570 --> 00:37:54,880 that vector was 2, 5, 3. 761 00:37:54,880 --> 00:37:57,620 When we wrote that same vector relative to the alphas, 762 00:37:57,620 --> 00:37:59,960 it was minus 2, 4, minus 1. 763 00:37:59,960 --> 00:38:04,510 Notice, by the way, a very important thing here. 764 00:38:04,510 --> 00:38:08,860 You see, every single time that we talked about two vectors are 765 00:38:08,860 --> 00:38:12,670 equal if and only if they were equal component by component, 766 00:38:12,670 --> 00:38:15,700 when we wrote them as n-tuples, it was assumed 767 00:38:15,700 --> 00:38:17,650 that the basis never changed. 768 00:38:17,650 --> 00:38:21,820 Look it, forget about the betas and the alphas here. 769 00:38:21,820 --> 00:38:25,690 Look at the 3-tuple 2, 5, 3 and the 3-tuple minus 2, 770 00:38:25,690 --> 00:38:27,330 4, minus 1. 771 00:38:27,330 --> 00:38:29,620 Certainly, they are different 3-tuples, 772 00:38:29,620 --> 00:38:32,410 but they happen to name the same vector, 773 00:38:32,410 --> 00:38:36,730 because one is a 3-tuple coded relative to the betas 774 00:38:36,730 --> 00:38:40,630 and the other is a 3-tuple coded relative to the alphas. 775 00:38:40,630 --> 00:38:43,930 In other words, once we pick a particular basis, 776 00:38:43,930 --> 00:38:46,360 we must remember that our coding system 777 00:38:46,360 --> 00:38:48,130 is relative to that basis. 778 00:38:48,130 --> 00:38:52,240 If, for some reason, it becomes convenient to change the basis, 779 00:38:52,240 --> 00:38:54,400 then we have to be very, very careful 780 00:38:54,400 --> 00:38:57,620 and remember that, therefore, our coding system has changed. 781 00:38:57,620 --> 00:39:01,080 And that's why, in most advanced applications using vector 782 00:39:01,080 --> 00:39:03,430 spaces, a great degree of difficulty 783 00:39:03,430 --> 00:39:05,230 comes in, because what we've done 784 00:39:05,230 --> 00:39:07,300 is we've changed basis vectors. 785 00:39:07,300 --> 00:39:11,560 And people who are so rigid that they keep visualizing that you 786 00:39:11,560 --> 00:39:14,470 are using the same n-tuples over and over again 787 00:39:14,470 --> 00:39:17,530 get confused as to how two different n-tuples can 788 00:39:17,530 --> 00:39:18,880 name the same vector. 789 00:39:18,880 --> 00:39:21,460 Well, at any rate, I think that's more than enough 790 00:39:21,460 --> 00:39:22,780 for one session. 791 00:39:22,780 --> 00:39:25,180 Do the exercises very carefully. 792 00:39:25,180 --> 00:39:25,930 Try your best. 793 00:39:25,930 --> 00:39:28,300 As I say, if you have any trouble after the exercises, 794 00:39:28,300 --> 00:39:29,950 review the film again. 795 00:39:29,950 --> 00:39:31,490 Listen to the lecture again. 796 00:39:31,490 --> 00:39:34,030 And what we will do next time is pick up 797 00:39:34,030 --> 00:39:36,460 some consequences of vector spaces 798 00:39:36,460 --> 00:39:38,800 that tie-in with other aspects of the course 799 00:39:38,800 --> 00:39:40,000 that we've already studied. 800 00:39:40,000 --> 00:39:42,180 But at any rate, until next time, goodbye. 801 00:39:44,690 --> 00:39:47,090 Funding for the publication of this video 802 00:39:47,090 --> 00:39:51,950 was provided by the Gabriella and Paul Rosenbaum Foundation. 803 00:39:51,950 --> 00:39:56,120 Help OCW continue to provide free and open access to MIT 804 00:39:56,120 --> 00:40:01,555 courses by making a donation at ocw.mit.edu/donate.