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 00:00:32,002 --> 00:00:33,040 HERBERT GROSS: Hi. 9 00:00:33,040 --> 00:00:35,860 Several times in our course up until now, 10 00:00:35,860 --> 00:00:38,530 we have referred to linearity. 11 00:00:38,530 --> 00:00:40,960 We referred to it when we talked about systems 12 00:00:40,960 --> 00:00:42,580 of linear equations. 13 00:00:42,580 --> 00:00:45,220 We referred to it when we talked about systems 14 00:00:45,220 --> 00:00:47,290 of linear differential equations. 15 00:00:47,290 --> 00:00:50,680 We referred to it when we talked about Laplace transforms. 16 00:00:50,680 --> 00:00:53,290 We referred to it when we were mapping the xy 17 00:00:53,290 --> 00:00:57,130 plane into the uv plane and talking about differentials. 18 00:00:57,130 --> 00:00:59,830 And now, what we'd like to do in today's lesson 19 00:00:59,830 --> 00:01:03,580 is to show how the general concept of a vector space 20 00:01:03,580 --> 00:01:07,120 gives us a very nice vehicle to tie together 21 00:01:07,120 --> 00:01:10,435 all of these different applications of linearity. 22 00:01:10,435 --> 00:01:12,820 In fact, today's lesson is called 23 00:01:12,820 --> 00:01:14,830 "Linear Transformations." 24 00:01:14,830 --> 00:01:18,280 And by way of a very quick definition, 25 00:01:18,280 --> 00:01:21,970 one which I will not bother motivating because we've 26 00:01:21,970 --> 00:01:25,210 seen the definition on many, many different occasions. 27 00:01:25,210 --> 00:01:27,820 All we're doing now is formalizing it. 28 00:01:27,820 --> 00:01:32,980 Let's suppose I have two vector spaces, V and W, and a mapping 29 00:01:32,980 --> 00:01:36,610 f that carries V into W. Remember, 30 00:01:36,610 --> 00:01:38,860 these vector spaces are in particular sets. 31 00:01:38,860 --> 00:01:42,360 And we've talked about functions that map sets into sets. 32 00:01:42,360 --> 00:01:45,437 Now we're mapping a set into a set 33 00:01:45,437 --> 00:01:47,770 where the set has a structure, the structure of a vector 34 00:01:47,770 --> 00:01:48,670 space. 35 00:01:48,670 --> 00:01:52,510 And what we're saying is that a function that maps a vector 36 00:01:52,510 --> 00:01:58,480 space V into a vector space W is called a linear transformation 37 00:01:58,480 --> 00:02:03,070 if, given any linear combination of elements in the domain, 38 00:02:03,070 --> 00:02:07,630 say C1 alpha 1 plus C2 alpha 2, where C1 and C2 are 39 00:02:07,630 --> 00:02:11,350 real numbers, and alpha 1 and alpha 2 are vectors in V, 40 00:02:11,350 --> 00:02:17,500 if f of that is C1 f of alpha 1 plus C2 f of alpha 2-- 41 00:02:17,500 --> 00:02:20,710 the same linearity definition that we've had in the past. 42 00:02:20,710 --> 00:02:23,530 And again, if this is hard for you to visualize, 43 00:02:23,530 --> 00:02:25,420 think in terms of pictures. 44 00:02:25,420 --> 00:02:29,110 Think of V as, for example, being the xy plane. 45 00:02:29,110 --> 00:02:32,050 Think of W as denoting the uv plane. 46 00:02:32,050 --> 00:02:36,010 And we're mapping the xy plane, say, into the uv plane. 47 00:02:36,010 --> 00:02:39,520 Not all such mappings will be linear. 48 00:02:39,520 --> 00:02:42,370 We studied that when we talked about linear systems 49 00:02:42,370 --> 00:02:44,180 of equations. 50 00:02:44,180 --> 00:02:46,390 Now, what we would like to do now 51 00:02:46,390 --> 00:02:48,700 is to show how many of the properties 52 00:02:48,700 --> 00:02:51,490 of linearity that we've used in the past 53 00:02:51,490 --> 00:02:55,480 are immediate consequences of this particular definition. 54 00:02:55,480 --> 00:02:58,760 Well, let's take a look at some of the consequences. 55 00:02:58,760 --> 00:03:01,460 The first thing I claim is that if f 56 00:03:01,460 --> 00:03:06,550 is a linear transformation, it must map the 0 element of V 57 00:03:06,550 --> 00:03:11,860 into the 0 element of W. In other words, f acting on 0 maps 58 00:03:11,860 --> 00:03:13,000 it into 0. 59 00:03:13,000 --> 00:03:14,950 And the easiest way to prove that 60 00:03:14,950 --> 00:03:19,390 is to notice that since 0 is equal to 0 plus 0, 61 00:03:19,390 --> 00:03:23,260 by that property, f of 0 is certainly the same as f of 0 62 00:03:23,260 --> 00:03:24,820 plus 0. 63 00:03:24,820 --> 00:03:31,120 And now by linearity, f of 0 plus 0 is f of 0 plus f of 0. 64 00:03:31,120 --> 00:03:34,310 And now, comparing this with this, 65 00:03:34,310 --> 00:03:37,090 remember, these are vectors, the cancellation rule 66 00:03:37,090 --> 00:03:39,460 applies for vectors. 67 00:03:39,460 --> 00:03:42,430 We then obtain that f of 0 is 0. 68 00:03:42,430 --> 00:03:46,990 And by the way, this is a rather interesting subtlety 69 00:03:46,990 --> 00:03:47,890 that comes up. 70 00:03:47,890 --> 00:03:51,190 Remember when we talked about linear equations in the plane? 71 00:03:51,190 --> 00:03:54,640 In other words, f of x equals mx plus b. 72 00:03:54,640 --> 00:03:57,550 That is not the same as a linear transformation. 73 00:03:57,550 --> 00:03:59,740 In fact, let me just see if I can scribble something 74 00:03:59,740 --> 00:04:01,450 in over here. 75 00:04:01,450 --> 00:04:02,090 Look at this. 76 00:04:02,090 --> 00:04:07,000 Suppose I have that f of x is equal to mx plus b. 77 00:04:07,000 --> 00:04:09,080 See, that's the equation of a straight line. 78 00:04:09,080 --> 00:04:11,500 What is f of 0? 79 00:04:11,500 --> 00:04:16,540 If f of x is mx plus b, if I replace x by 0, I get what? 80 00:04:16,540 --> 00:04:19,588 0 times m, which is 0, plus b. 81 00:04:19,588 --> 00:04:24,020 And therefore, for f to be a linear transformation 82 00:04:24,020 --> 00:04:28,950 according to our definition, this must equal 0. 83 00:04:28,950 --> 00:04:33,750 And therefore, what this means is that b itself must equal 0. 84 00:04:33,750 --> 00:04:36,090 In other words, even though we think 85 00:04:36,090 --> 00:04:37,980 of a straight line as being linear, 86 00:04:37,980 --> 00:04:40,840 to represent a linear transformation, 87 00:04:40,840 --> 00:04:43,350 the straight line must pass through the origin. 88 00:04:43,350 --> 00:04:45,780 But I don't want to harp on that now. 89 00:04:45,780 --> 00:04:48,270 I want to let that go for the exercises 90 00:04:48,270 --> 00:04:51,910 because I'm afraid, again, if we spend too much time on details, 91 00:04:51,910 --> 00:04:54,120 you will fail to get the overall picture. 92 00:04:54,120 --> 00:04:57,030 But at any rate, one property of a linear transformation 93 00:04:57,030 --> 00:05:01,110 is that it maps the 0 vector into the 0 vector. 94 00:05:01,110 --> 00:05:04,780 Now, of course, other vectors may be mapped into the 0 vector 95 00:05:04,780 --> 00:05:05,280 as well. 96 00:05:05,280 --> 00:05:07,950 I mean, we'll talk about that in more detail in the form 97 00:05:07,950 --> 00:05:09,630 of a review, in fact, later on. 98 00:05:09,630 --> 00:05:12,060 But for the time being, let me show you 99 00:05:12,060 --> 00:05:14,610 the next property of linear transformations 100 00:05:14,610 --> 00:05:15,940 that are very important. 101 00:05:15,940 --> 00:05:19,410 And that is that if V1 and V2 are 102 00:05:19,410 --> 00:05:23,520 mapped into 0 by the linear transformation f, 103 00:05:23,520 --> 00:05:26,640 then any linear combination of V1 and V2 104 00:05:26,640 --> 00:05:28,710 is also mapped into 0. 105 00:05:28,710 --> 00:05:31,320 And by the way, you may think when 106 00:05:31,320 --> 00:05:33,210 I say this-- this may ring a bell 107 00:05:33,210 --> 00:05:34,980 and remind you of what we were doing 108 00:05:34,980 --> 00:05:39,120 when we talked about homogeneous linear differential equations. 109 00:05:39,120 --> 00:05:41,040 And the interesting point is that this 110 00:05:41,040 --> 00:05:44,160 is exactly the same as what we were doing there, 111 00:05:44,160 --> 00:05:48,540 only now, we're not restricted to be dealing only 112 00:05:48,540 --> 00:05:50,500 with linear differential equations. 113 00:05:50,500 --> 00:05:53,580 This is now any linear transformation from one vector 114 00:05:53,580 --> 00:05:54,750 space into another. 115 00:05:54,750 --> 00:05:56,220 And the proof goes exactly the way 116 00:05:56,220 --> 00:05:58,800 it did in the linear differential equation case. 117 00:05:58,800 --> 00:06:03,060 Namely, given f of C1V1 plus C2V2, 118 00:06:03,060 --> 00:06:06,010 we certainly know that by linearity, that 119 00:06:06,010 --> 00:06:09,270 C1 f of V1 plus C2 f of V2. 120 00:06:09,270 --> 00:06:13,320 But given that f of V1 is 0, and f of V2 is 0, 121 00:06:13,320 --> 00:06:16,530 certainly, this then becomes 0, which 122 00:06:16,530 --> 00:06:18,450 is what we wanted to show. 123 00:06:18,450 --> 00:06:22,090 In other words, if you have a linear transformation, 124 00:06:22,090 --> 00:06:25,260 the set of elements that are mapped into the 0 element 125 00:06:25,260 --> 00:06:26,700 are more than a set. 126 00:06:26,700 --> 00:06:31,305 They are themselves a vector space, a subspace of V, 127 00:06:31,305 --> 00:06:34,080 a subspace of the domain of f. 128 00:06:34,080 --> 00:06:35,310 Because why? 129 00:06:35,310 --> 00:06:42,480 Any linear combinations of elements mapped into 0 by f is 130 00:06:42,480 --> 00:06:44,400 itself-- any linear combination is itself-- 131 00:06:44,400 --> 00:06:45,840 mapped into f. 132 00:06:45,840 --> 00:06:48,360 In other words, let's define n to be 133 00:06:48,360 --> 00:06:54,300 the set of all elements eta in V, such that f of eta is 0. 134 00:06:54,300 --> 00:06:58,790 And that space is called the null space of f. 135 00:06:58,790 --> 00:07:02,940 It's a subspace of the domain of fV. 136 00:07:02,940 --> 00:07:06,360 And all we're saying is by our previous remark, 137 00:07:06,360 --> 00:07:10,850 if two elements belong to n, their sum belongs to n, 138 00:07:10,850 --> 00:07:14,490 and any scalar multiple of an element in n also belongs to n. 139 00:07:14,490 --> 00:07:20,280 So n is indeed a subspace of V. It's a very important subspace. 140 00:07:20,280 --> 00:07:22,050 It's called the null space. 141 00:07:22,050 --> 00:07:24,160 And one of things about the null space-- 142 00:07:24,160 --> 00:07:26,730 and by the way, many people get confused by this name. 143 00:07:26,730 --> 00:07:30,660 They say the null space, they think of "null" as being empty, 144 00:07:30,660 --> 00:07:31,360 nothing. 145 00:07:31,360 --> 00:07:33,540 And they say that null space has nothing in it. 146 00:07:33,540 --> 00:07:34,170 No. 147 00:07:34,170 --> 00:07:36,090 Think of it in terms of the mapping. 148 00:07:36,090 --> 00:07:39,540 The null space is part of a domain. 149 00:07:39,540 --> 00:07:42,210 And what it is-- it's that part of the domain 150 00:07:42,210 --> 00:07:46,500 that's mapped into the 0 element of the image. 151 00:07:46,500 --> 00:07:47,340 You see? 152 00:07:47,340 --> 00:07:48,330 It is not nothing. 153 00:07:48,330 --> 00:07:51,780 It's the set of all elements that are mapped into 0. 154 00:07:51,780 --> 00:07:54,450 And what we're saying is that that subspace 155 00:07:54,450 --> 00:07:56,610 is enough to determine the entire mapping. 156 00:07:56,610 --> 00:07:59,160 Let me show what I mean by that abstractly. 157 00:07:59,160 --> 00:08:02,520 I claim that n determines the image of f. 158 00:08:02,520 --> 00:08:06,990 Namely, I claim that if I know that f of alpha equals f 159 00:08:06,990 --> 00:08:10,050 of beta, and f is a linear transformation, 160 00:08:10,050 --> 00:08:15,000 alpha and beta cannot be very random elements of V. In fact, 161 00:08:15,000 --> 00:08:17,880 what I claim is that the difference between alpha 162 00:08:17,880 --> 00:08:20,970 and beta must be an element of the null space n. 163 00:08:20,970 --> 00:08:23,190 Or another way of saying that, I claim 164 00:08:23,190 --> 00:08:27,120 that alpha has the form beta plus eta, where eta is 165 00:08:27,120 --> 00:08:29,480 an element of the null space n. 166 00:08:29,480 --> 00:08:32,400 And again, this may ring a familiar bell for you 167 00:08:32,400 --> 00:08:34,500 when you think in terms of linear differential 168 00:08:34,500 --> 00:08:37,260 equations and the like, where we essentially 169 00:08:37,260 --> 00:08:42,240 saw that any two solutions of a differential equation, 170 00:08:42,240 --> 00:08:44,610 that their difference was a solution 171 00:08:44,610 --> 00:08:48,330 of the homogeneous linear differential equation. 172 00:08:48,330 --> 00:08:50,580 But again, I don't want to dwell on that now. 173 00:08:50,580 --> 00:08:55,090 I want to talk on the most abstract version of this. 174 00:08:55,090 --> 00:08:56,652 How do I know that this is true? 175 00:08:56,652 --> 00:08:57,360 Well, look at it. 176 00:08:57,360 --> 00:09:01,500 What does it mean to say that f of alpha equals f of beta? 177 00:09:01,500 --> 00:09:05,100 It means that f of alpha minus f of beta is 0. 178 00:09:05,100 --> 00:09:08,790 But by linearity, f of alpha minus f of beta 179 00:09:08,790 --> 00:09:11,490 is the same as f of alpha minus beta. 180 00:09:11,490 --> 00:09:15,030 Therefore, the fact that f of alpha minus f of beta is 0 181 00:09:15,030 --> 00:09:18,030 says that f of alpha minus beta is 0. 182 00:09:18,030 --> 00:09:20,490 By definition of the null space n, 183 00:09:20,490 --> 00:09:23,790 the fact that alpha minus beta is mapped into 0 184 00:09:23,790 --> 00:09:26,480 means that alpha minus beta belongs to n. 185 00:09:26,480 --> 00:09:28,350 And that's the same as saying what? 186 00:09:28,350 --> 00:09:31,230 That alpha minus beta is some element in n. 187 00:09:31,230 --> 00:09:35,880 Or alpha is beta plus some element in n-- 188 00:09:35,880 --> 00:09:37,410 same thing as we asserted. 189 00:09:37,410 --> 00:09:39,540 By the way, another interesting way 190 00:09:39,540 --> 00:09:42,570 of seeing one to oneness of a linear transformation 191 00:09:42,570 --> 00:09:45,330 is that since the only way that two elements can have 192 00:09:45,330 --> 00:09:47,970 the same image is for one of them 193 00:09:47,970 --> 00:09:51,510 to equal the other plus something that belongs to n, 194 00:09:51,510 --> 00:09:54,750 the only way we can be sure that alpha equals 195 00:09:54,750 --> 00:09:57,420 beta is if this element must be 0. 196 00:09:57,420 --> 00:10:00,060 You see, alpha equals beta plus 0. 197 00:10:00,060 --> 00:10:01,990 We'll say that alpha equals beta. 198 00:10:01,990 --> 00:10:06,660 So in particular, if a linear transformation is one to one, 199 00:10:06,660 --> 00:10:10,080 the null space must consist of 0 alone. 200 00:10:10,080 --> 00:10:13,620 And conversely, if the null space consists of 0 alone, 201 00:10:13,620 --> 00:10:16,240 the linear transformation is 1 to 1. 202 00:10:16,240 --> 00:10:18,060 Notice, by the way, the null space 203 00:10:18,060 --> 00:10:21,420 can never be empty because by our first property, 204 00:10:21,420 --> 00:10:25,320 we show that the 0 vector is always mapped into 0. 205 00:10:25,320 --> 00:10:28,680 So at least the 0 vector belongs to the null space. 206 00:10:28,680 --> 00:10:31,050 If other vectors belong to the null space, 207 00:10:31,050 --> 00:10:33,600 the linear transformation will not be 1 to 1. 208 00:10:33,600 --> 00:10:36,900 If 0 is the only vector that belongs to the null space of f, 209 00:10:36,900 --> 00:10:40,260 then the linear transformation will be 1 to 1. 210 00:10:40,260 --> 00:10:42,630 Now, I think the time has come to illustrate 211 00:10:42,630 --> 00:10:46,000 some of these remarks in terms of concrete examples. 212 00:10:46,000 --> 00:10:48,720 For my first example, let's recall 213 00:10:48,720 --> 00:10:54,000 that the mapping that maps functions into their derivative 214 00:10:54,000 --> 00:10:58,470 is a well-defined linear transformation on the set 215 00:10:58,470 --> 00:10:59,670 of differentiable functions. 216 00:10:59,670 --> 00:11:01,270 Remember, the derivative of a sum 217 00:11:01,270 --> 00:11:02,648 is the sum of the derivatives. 218 00:11:02,648 --> 00:11:04,440 The derivative of a scalar times a function 219 00:11:04,440 --> 00:11:06,820 is a scalar times the derivative of the function. 220 00:11:06,820 --> 00:11:08,970 In other words, then, the mapping D, 221 00:11:08,970 --> 00:11:14,740 which maps a function into its derivative, is a vector space. 222 00:11:14,740 --> 00:11:17,310 It's a linear transformation from one vector 223 00:11:17,310 --> 00:11:18,880 space to another. 224 00:11:18,880 --> 00:11:21,690 In particular, what does the null space look like? 225 00:11:21,690 --> 00:11:23,820 The null space of D must be what? 226 00:11:23,820 --> 00:11:27,930 All functions f, such that D maps f into 0. 227 00:11:27,930 --> 00:11:32,310 The only way that D of f can be 0 is if f prime is 0. 228 00:11:32,310 --> 00:11:35,760 But f prime being identically 0 is the same as saying 229 00:11:35,760 --> 00:11:37,360 f equals a constant. 230 00:11:37,360 --> 00:11:39,570 Therefore, the null space in this case 231 00:11:39,570 --> 00:11:41,880 is the set of all constants. 232 00:11:41,880 --> 00:11:44,160 Notice, by the way, the null space 233 00:11:44,160 --> 00:11:46,140 does not consist of 0 alone. 234 00:11:46,140 --> 00:11:48,750 It consists of infinitely many numbers, namely 235 00:11:48,750 --> 00:11:50,040 all the constants. 236 00:11:50,040 --> 00:11:52,620 By the way, that shouldn't be too surprising 237 00:11:52,620 --> 00:11:56,970 because after all, the derivative is not a one to one 238 00:11:56,970 --> 00:11:59,520 mapping of differentiable functions 239 00:11:59,520 --> 00:12:00,930 into the set of functions. 240 00:12:00,930 --> 00:12:04,830 Namely, two different functions can have the same derivative. 241 00:12:04,830 --> 00:12:06,780 In fact, what is the property? 242 00:12:06,780 --> 00:12:12,480 For D of f to equal D of g, it's necessary and sufficient 243 00:12:12,480 --> 00:12:15,590 that f and g differ by a constant. 244 00:12:15,590 --> 00:12:20,070 And notice that that's the same as saying what we said before, 245 00:12:20,070 --> 00:12:23,700 that for two elements to have the same image 246 00:12:23,700 --> 00:12:25,560 under a linear transformation, they 247 00:12:25,560 --> 00:12:29,220 must differ by, at most, an element in the null space. 248 00:12:29,220 --> 00:12:32,850 That's exactly what happens in this particular example. 249 00:12:32,850 --> 00:12:35,130 As a second example, let's go back 250 00:12:35,130 --> 00:12:38,970 to our study of linear differential equations. 251 00:12:38,970 --> 00:12:43,590 We wanted to find the solution of the equation L of y 252 00:12:43,590 --> 00:12:44,970 equals f of x. 253 00:12:44,970 --> 00:12:49,080 Notice that L is a linear transformation mapping 254 00:12:49,080 --> 00:12:50,970 functions into functions. 255 00:12:50,970 --> 00:12:52,470 The interesting point was, you may 256 00:12:52,470 --> 00:12:56,550 recall, that we emphasized the homogeneous equation, 257 00:12:56,550 --> 00:12:59,310 where the right-hand side was replaced by 0. 258 00:12:59,310 --> 00:13:02,300 Notice that the null space of the mapping L-- 259 00:13:02,300 --> 00:13:03,690 see, the null space-- 260 00:13:03,690 --> 00:13:04,500 namely what? 261 00:13:04,500 --> 00:13:09,113 All functions y that map into 0 with respect to L. 262 00:13:09,113 --> 00:13:09,780 That means what? 263 00:13:09,780 --> 00:13:11,310 L of y is 0. 264 00:13:11,310 --> 00:13:12,840 Well, the null space is just what? 265 00:13:12,840 --> 00:13:15,300 The solution set of L of y equals 0. 266 00:13:15,300 --> 00:13:17,790 That's just the homogeneous equation. 267 00:13:17,790 --> 00:13:19,830 And to show you what we mean by this, 268 00:13:19,830 --> 00:13:22,440 notice what the general theory was for linear differential 269 00:13:22,440 --> 00:13:23,310 equations. 270 00:13:23,310 --> 00:13:26,850 Namely, the general solution was the homogeneous solution 271 00:13:26,850 --> 00:13:29,580 plus a particular solution, noticing 272 00:13:29,580 --> 00:13:31,470 that for the homogeneous solution, 273 00:13:31,470 --> 00:13:33,340 it was characterized by the fact that it 274 00:13:33,340 --> 00:13:34,920 belonged to the null space. 275 00:13:34,920 --> 00:13:37,620 Namely, L of y sub h was 0. 276 00:13:37,620 --> 00:13:41,220 The particular solution was just anything in the space, 277 00:13:41,220 --> 00:13:45,450 namely L of yp equals f of x, same as we have over here. 278 00:13:45,450 --> 00:13:48,570 And so that structure for linear differential equations 279 00:13:48,570 --> 00:13:53,040 was not an exception, but rather one concrete illustration 280 00:13:53,040 --> 00:13:55,260 of what a really a transformation means 281 00:13:55,260 --> 00:13:57,210 in an abstract vector space. 282 00:13:57,210 --> 00:14:01,140 As a third and final and more visual example, 283 00:14:01,140 --> 00:14:05,310 let's go back to mapping the xy plane into the uv plane. 284 00:14:05,310 --> 00:14:08,700 Let's suppose I take the mapping u equals x plus y, 285 00:14:08,700 --> 00:14:11,820 v equals 2x plus 2y. 286 00:14:11,820 --> 00:14:13,770 And notice, by the way, that's just 287 00:14:13,770 --> 00:14:17,070 another way of saying v equals twice u. 288 00:14:17,070 --> 00:14:19,770 The induced function is the one that 289 00:14:19,770 --> 00:14:23,400 carries x comma y in the xy plane 290 00:14:23,400 --> 00:14:28,600 into u comma v in the uv plane, so my induced function f bar 291 00:14:28,600 --> 00:14:35,050 is the one that maps x comma y into x plus y comma 2x plus 2y. 292 00:14:35,050 --> 00:14:37,390 w 293 00:14:37,390 --> 00:14:41,110 And you can look at x comma y as being the vector xi 294 00:14:41,110 --> 00:14:43,570 plus yj in the xy plane. 295 00:14:43,570 --> 00:14:46,360 Or you can look at it as being the point x comma 296 00:14:46,360 --> 00:14:47,775 y in the xy plane. 297 00:14:47,775 --> 00:14:49,150 I don't care which interpretation 298 00:14:49,150 --> 00:14:50,570 you want to use there. 299 00:14:50,570 --> 00:14:52,300 But the point is, what we now would 300 00:14:52,300 --> 00:14:54,310 like to know in terms of the null space 301 00:14:54,310 --> 00:14:58,450 is what points x comma y are mapped into 0, 302 00:14:58,450 --> 00:15:03,070 0 in the uv plane under the mapping f bar? 303 00:15:03,070 --> 00:15:05,530 Or in terms of vectors, what vectors 304 00:15:05,530 --> 00:15:10,990 xi plus yj are mapped into the 0 vector under the mapping f bar? 305 00:15:10,990 --> 00:15:16,420 Well, notice that for this to be 0, we must have what? 306 00:15:16,420 --> 00:15:18,280 That both components are 0. 307 00:15:18,280 --> 00:15:22,750 That means that x plus y must be 0 and that 2x plus 2y is 0. 308 00:15:22,750 --> 00:15:25,610 But of course, that's the same as this condition. 309 00:15:25,610 --> 00:15:30,940 In other words, f bar of x comma y is 0, meaning what? 310 00:15:30,940 --> 00:15:37,300 It's mapped into the origin if and only if x plus y is 0. 311 00:15:37,300 --> 00:15:41,800 In other words, the null space of f bar 312 00:15:41,800 --> 00:15:45,400 is the set of all points in the plane x comma y 313 00:15:45,400 --> 00:15:47,530 for which x plus y equals 0. 314 00:15:47,530 --> 00:15:49,810 Or again, if you want to state this vectorially, 315 00:15:49,810 --> 00:15:53,080 it's all vectors xi plus yj in the xy 316 00:15:53,080 --> 00:15:56,770 plane, for which x plus y is 0. 317 00:15:56,770 --> 00:15:59,140 And what this means pictorially is this. 318 00:15:59,140 --> 00:16:02,170 Notice that the set n is nothing more than what we 319 00:16:02,170 --> 00:16:04,470 call the line y equals minus x. 320 00:16:04,470 --> 00:16:08,430 In the set of all points x comma y, such as x plus y equals 0, 321 00:16:08,430 --> 00:16:10,780 is the same as the set of all points x comma y, 322 00:16:10,780 --> 00:16:12,700 such that y equals minus x. 323 00:16:12,700 --> 00:16:15,670 y equals minus x, or x plus y equals 0, 324 00:16:15,670 --> 00:16:17,680 is a straight line through the origin. 325 00:16:17,680 --> 00:16:19,420 Notice again that key point. 326 00:16:19,420 --> 00:16:22,240 f of 0 must be 0 for a linear transformation. 327 00:16:22,240 --> 00:16:24,370 If that line didn't pass through the origin, 328 00:16:24,370 --> 00:16:26,500 the mapping wouldn't be linear. 329 00:16:26,500 --> 00:16:29,260 Now, notice, by the way, if I come back here, 330 00:16:29,260 --> 00:16:32,830 the mapping defined by u and v maps 331 00:16:32,830 --> 00:16:37,130 the xy plane into the single line v equals 2u. 332 00:16:37,130 --> 00:16:39,730 In other words, the xy plane is mapped 333 00:16:39,730 --> 00:16:42,700 into the line v equals 2u. 334 00:16:42,700 --> 00:16:45,940 And the entire line x plus y equals 0 335 00:16:45,940 --> 00:16:48,290 is mapped into the origin over here. 336 00:16:48,290 --> 00:16:52,180 In other words, the null space here is an entire line. 337 00:16:52,180 --> 00:16:53,080 What line is it? 338 00:16:53,080 --> 00:16:55,270 It's the line x plus y equals 0. 339 00:16:55,270 --> 00:16:57,280 Why is it called the null space? 340 00:16:57,280 --> 00:17:00,280 It's called the null space not because it has no points, 341 00:17:00,280 --> 00:17:03,610 but rather because its image consists of the single point 342 00:17:03,610 --> 00:17:06,430 0 in the uv plane. 343 00:17:06,430 --> 00:17:11,170 And by the way, notice again the parallelism here and that 344 00:17:11,170 --> 00:17:13,420 which would happen when we studied linear differential 345 00:17:13,420 --> 00:17:14,420 equations. 346 00:17:14,420 --> 00:17:16,030 Let's suppose for the sake of argument 347 00:17:16,030 --> 00:17:19,790 that I want to find all points x comma y that map into the point 348 00:17:19,790 --> 00:17:23,589 1 comma 2 in the uv plane. 349 00:17:23,589 --> 00:17:26,230 Essentially, what I want is a particular solution here. 350 00:17:26,230 --> 00:17:29,890 I would like to find one point that maps into 1 comma 2. 351 00:17:29,890 --> 00:17:33,580 And my claim is that knowing the null space, once I know one 352 00:17:33,580 --> 00:17:35,530 such point, I know them all. 353 00:17:35,530 --> 00:17:40,690 Well, one point that maps into 1 comma 2 trivially is 0 comma 1. 354 00:17:40,690 --> 00:17:46,090 Namely, 0 plus 1 is 1, and twice 0 plus twice 1 is 2. 355 00:17:46,090 --> 00:17:51,550 So 0, 1 is mapped into 1 comma 2 by the mapping f bar. 356 00:17:51,550 --> 00:17:55,480 Now the question is, what other points map into 1 comma 2? 357 00:17:55,480 --> 00:17:59,530 And the answer is it's every point on the straight line 358 00:17:59,530 --> 00:18:03,550 which passes through 0 comma 1 and is parallel to the line 359 00:18:03,550 --> 00:18:04,920 x plus y equals 0. 360 00:18:04,920 --> 00:18:07,570 In other words, the line x plus y 361 00:18:07,570 --> 00:18:12,490 equals 1 is the line which maps into the point 1 comma 2 362 00:18:12,490 --> 00:18:14,290 under the mapping f bar. 363 00:18:14,290 --> 00:18:18,280 And the easiest way to see this, I would say, is vectorially. 364 00:18:18,280 --> 00:18:20,380 Pick any point on this line. 365 00:18:20,380 --> 00:18:22,990 Look at it as a vector. 366 00:18:22,990 --> 00:18:26,470 That vector is the sum of these two vectors. 367 00:18:29,320 --> 00:18:32,410 This vector, being a member of the null space, 368 00:18:32,410 --> 00:18:34,070 is mapped into-- 369 00:18:34,070 --> 00:18:34,570 I'm sorry. 370 00:18:34,570 --> 00:18:36,710 I don't want to do it this way. 371 00:18:36,710 --> 00:18:38,220 I'd rather do it-- excuse me. 372 00:18:38,220 --> 00:18:39,880 I'd rather do it this way. 373 00:18:39,880 --> 00:18:43,180 I want to utilize the fact that I know that this vector is 374 00:18:43,180 --> 00:18:45,520 mapped into this vector. 375 00:18:45,520 --> 00:18:49,210 So I'll break down this vector into components this way. 376 00:18:49,210 --> 00:18:53,590 I'll take one component like this and one like this. 377 00:18:53,590 --> 00:18:57,910 You see, this component here is indeed the same as this vector 378 00:18:57,910 --> 00:18:59,930 because these two lines are parallel. 379 00:18:59,930 --> 00:19:04,150 So therefore, f of this is f of this plus f of this. 380 00:19:04,150 --> 00:19:08,080 By linearity, that's f of this plus f of this. 381 00:19:08,080 --> 00:19:12,200 This being a vector in the null space maps into 0, 382 00:19:12,200 --> 00:19:15,640 so the image is just whatever f of this happens to be. 383 00:19:15,640 --> 00:19:17,230 You see the idea again? 384 00:19:17,230 --> 00:19:19,450 Call this vector, if you will, alpha. 385 00:19:19,450 --> 00:19:21,310 Call this vector here beta. 386 00:19:21,310 --> 00:19:24,740 f of alpha plus beta is f of alpha, 387 00:19:24,740 --> 00:19:27,430 which is 0, because alpha is in the null space, 388 00:19:27,430 --> 00:19:31,030 plus f of beta, which is just 1 comma 2. 389 00:19:31,030 --> 00:19:33,970 Again, I think I've made too much of a mountain out 390 00:19:33,970 --> 00:19:36,850 of a molehill, not because this isn't important, 391 00:19:36,850 --> 00:19:40,240 but because you will recall back in our block four 392 00:19:40,240 --> 00:19:44,260 treatment of linear equations, linear algebraic equations, 393 00:19:44,260 --> 00:19:45,880 we discussed things like this. 394 00:19:45,880 --> 00:19:47,890 And I don't want to get too deeply involved 395 00:19:47,890 --> 00:19:49,390 in reviewing the details. 396 00:19:49,390 --> 00:19:51,460 What I am interested in seeing is 397 00:19:51,460 --> 00:19:55,390 that you see the structure of what a null space really is. 398 00:19:55,390 --> 00:19:57,400 And by the way, let's continue on 399 00:19:57,400 --> 00:20:00,310 with what the properties of a linear transformation 400 00:20:00,310 --> 00:20:01,445 happen to be. 401 00:20:01,445 --> 00:20:03,070 Let's leave this go for the time being. 402 00:20:03,070 --> 00:20:05,210 That's enough in terms of examples. 403 00:20:05,210 --> 00:20:07,470 Let's go on with the properties. 404 00:20:07,470 --> 00:20:11,740 Our fourth property is that if V is a vector space, 405 00:20:11,740 --> 00:20:14,940 and we pick a particular basis, u1 up to un, 406 00:20:14,940 --> 00:20:17,710 then V is neatly determined. 407 00:20:17,710 --> 00:20:21,400 I say "neatly determined" by its effect on the basis vectors 408 00:20:21,400 --> 00:20:22,810 u1 up to un. 409 00:20:22,810 --> 00:20:25,850 What I mean by neatly determined is the following. 410 00:20:25,850 --> 00:20:27,400 Let's do a specific example. 411 00:20:27,400 --> 00:20:29,950 Suppose for the sake of argument that n is 2. 412 00:20:29,950 --> 00:20:34,720 Let's look at V, which has as its basis u1 and u2. 413 00:20:34,720 --> 00:20:39,730 Pick any vector little v and v. Relative to u1 and u2, 414 00:20:39,730 --> 00:20:44,450 v has a unique representation x1u1 plus x2u2. 415 00:20:44,450 --> 00:20:47,230 f of v by the linear property of f-- f 416 00:20:47,230 --> 00:20:51,160 being a linear transformation-- is simply what? x1 f 417 00:20:51,160 --> 00:20:54,130 of u1 plus x2 f of u2. 418 00:20:54,130 --> 00:20:58,930 And notice that once we know what f does to u1 and u2, 419 00:20:58,930 --> 00:21:02,680 we now know what it does to every vector in the space. 420 00:21:02,680 --> 00:21:06,760 Namely, you see, to find what f does to V, 421 00:21:06,760 --> 00:21:10,120 we simply look at the same vector as V, 422 00:21:10,120 --> 00:21:15,520 only with u1 and u2 replaced by f of u1 and f of v2. 423 00:21:15,520 --> 00:21:17,560 And by the way, because we can do this, 424 00:21:17,560 --> 00:21:20,380 it implies that linear transformations 425 00:21:20,380 --> 00:21:23,650 may be represented also very conveniently in terms 426 00:21:23,650 --> 00:21:25,150 of matrix notation. 427 00:21:25,150 --> 00:21:28,990 And I think that both of these last two remarks, 4 and 5, 428 00:21:28,990 --> 00:21:32,680 can be best illustrated in terms of an example. 429 00:21:32,680 --> 00:21:35,925 Let me first give you a fairly abstract example 430 00:21:35,925 --> 00:21:37,840 so that you see the terminology. 431 00:21:37,840 --> 00:21:39,610 And then I'll make this more concrete. 432 00:21:39,610 --> 00:21:43,435 And the concrete example will be our finale for today's lesson. 433 00:21:43,435 --> 00:21:48,370 But first of all, by means of an example-- 434 00:21:48,370 --> 00:21:49,360 example number 4. 435 00:21:49,360 --> 00:21:53,710 Again, let V be a two-dimensional space 436 00:21:53,710 --> 00:21:55,570 with u1 and u2 as a basis. 437 00:21:55,570 --> 00:21:57,700 Let's assume now for the sake of argument 438 00:21:57,700 --> 00:22:03,390 that the linear transformation f maps V into V itself-- 439 00:22:03,390 --> 00:22:05,810 in other words, that V and W are equal in this case, 440 00:22:05,810 --> 00:22:07,090 just for the sake of argument. 441 00:22:07,090 --> 00:22:08,320 Now, what did I just say? 442 00:22:08,320 --> 00:22:11,260 I said before that the transformation 443 00:22:11,260 --> 00:22:13,780 is uniquely determined once we know what it does 444 00:22:13,780 --> 00:22:15,910 to each element of a basis. 445 00:22:15,910 --> 00:22:18,670 What does f do to u2 and u2? 446 00:22:18,670 --> 00:22:21,730 Well, since f is mapping V into V, 447 00:22:21,730 --> 00:22:24,880 and since this implies that every vector in V 448 00:22:24,880 --> 00:22:27,910 is going to be with respect to a coordinate system u1, 449 00:22:27,910 --> 00:22:31,840 u2, what I do know is that whatever f of u1 and f of u2 450 00:22:31,840 --> 00:22:34,480 are, they are linear combinations of the basis 451 00:22:34,480 --> 00:22:36,970 vectors of V u1 and u2. 452 00:22:36,970 --> 00:22:39,700 So let's say for the sake of argument that f of u1 453 00:22:39,700 --> 00:22:42,100 is this linear combination. f of u2 454 00:22:42,100 --> 00:22:43,870 is this linear combination, where we're 455 00:22:43,870 --> 00:22:48,220 using the usual double subscript notation that seems to indicate 456 00:22:48,220 --> 00:22:50,560 a matrix treatment coming up. 457 00:22:50,560 --> 00:22:53,140 Well, let me show you now what happens next. 458 00:22:53,140 --> 00:22:56,440 Let's suppose we again pick our arbitrary vector 459 00:22:56,440 --> 00:22:58,030 and fix it-- some arbitrary vector 460 00:22:58,030 --> 00:23:02,920 V in our space V. Let's suppose that relative to u1 and u2, 461 00:23:02,920 --> 00:23:06,830 that vector v is x1u1 plus x2u2. 462 00:23:06,830 --> 00:23:10,090 Well, we just saw that f of v must therefore 463 00:23:10,090 --> 00:23:14,290 be x1 f of u1 plus x2 f of u2. 464 00:23:14,290 --> 00:23:17,290 But now, we know what f of u1 and f of u2 look 465 00:23:17,290 --> 00:23:21,730 like explicitly as linear combinations of u1 and u2. 466 00:23:21,730 --> 00:23:25,720 And therefore, simply by substituting these expressions 467 00:23:25,720 --> 00:23:29,040 for f of u1 and f of u2 in here, leaving again 468 00:23:29,040 --> 00:23:32,460 the details for you to verify, I find 469 00:23:32,460 --> 00:23:35,880 that relative to a 2-tuple notation, where 470 00:23:35,880 --> 00:23:40,820 we're using u1 and u2 as our basis vectors, that f of v 471 00:23:40,820 --> 00:23:48,690 is x1a11 plus x2a21 comma x1a12 plus x2a22. 472 00:23:48,690 --> 00:23:52,520 Now, this, at first glance, may not 473 00:23:52,520 --> 00:23:54,620 seem to suggest a matrix to you. 474 00:23:54,620 --> 00:23:56,300 But at second glance, hopefully it 475 00:23:56,300 --> 00:24:02,930 will, namely, if I look at the matrix a11, a21, a12, a22. 476 00:24:02,930 --> 00:24:06,110 And notice, by the way, now, a very important subtlety 477 00:24:06,110 --> 00:24:07,040 that's come up here. 478 00:24:07,040 --> 00:24:08,690 In the past, the matrix has always 479 00:24:08,690 --> 00:24:15,830 been written a11, a12, a21, a22 to match the coefficients 480 00:24:15,830 --> 00:24:19,280 as they occur in these expressions, here. 481 00:24:19,280 --> 00:24:21,740 But notice, I know what the answer has to be. 482 00:24:21,740 --> 00:24:23,120 It has to be this. 483 00:24:23,120 --> 00:24:26,060 All I'm saying is if you look at this particular matrix, 484 00:24:26,060 --> 00:24:26,720 this is what? 485 00:24:26,720 --> 00:24:30,230 A11, x1 plus a21, x2. 486 00:24:30,230 --> 00:24:33,470 And if you want the matrix to represent the right 2-tuple, 487 00:24:33,470 --> 00:24:35,210 it has to be written this way. 488 00:24:35,210 --> 00:24:38,480 By the way, this is a two-row matrix. 489 00:24:38,480 --> 00:24:40,700 To make the thing come out exactly right, 490 00:24:40,700 --> 00:24:42,640 I simply take the transpose of this. 491 00:24:42,640 --> 00:24:46,400 In other words, notice that this, without the transpose, 492 00:24:46,400 --> 00:24:49,390 comes out to be two rows and one column, 493 00:24:49,390 --> 00:24:52,100 where this is the first row, this is the second row. 494 00:24:52,100 --> 00:24:55,550 To make this look exactly right, I put the transpose in here. 495 00:24:55,550 --> 00:24:57,950 By the way, remember that we have already 496 00:24:57,950 --> 00:25:04,850 seen that to take the transpose of the product of two matrices 497 00:25:04,850 --> 00:25:08,390 A, B, it's B transpose A transpose. 498 00:25:08,390 --> 00:25:10,040 So the transpose would be what? 499 00:25:10,040 --> 00:25:13,970 You transpose this, which is x1, x2, and multiply that 500 00:25:13,970 --> 00:25:20,540 by the transpose of this, which is a11, a12, a21, a22. 501 00:25:20,540 --> 00:25:23,330 In other words, if you do want this matrix 502 00:25:23,330 --> 00:25:26,720 to look exactly like the matrix of coefficients, what we're 503 00:25:26,720 --> 00:25:31,010 saying is you are going to have to write the vector x1, x2 504 00:25:31,010 --> 00:25:34,250 on the left-hand side rather than on the right-hand side. 505 00:25:34,250 --> 00:25:37,850 And by the way, that's why in many textbooks that 506 00:25:37,850 --> 00:25:39,860 deal with matrix algebra and the like, 507 00:25:39,860 --> 00:25:42,500 they always put the variable on the left-hand side. 508 00:25:42,500 --> 00:25:45,080 It's because if you put the variable on the left-hand side, 509 00:25:45,080 --> 00:25:47,030 you can identify the matrix as it 510 00:25:47,030 --> 00:25:51,680 stands with the matrix of coefficients verbatim, here. 511 00:25:51,680 --> 00:25:53,570 If you don't do that, if you want 512 00:25:53,570 --> 00:25:56,750 to write the x1, x2 on the right, 513 00:25:56,750 --> 00:25:59,780 then you have to remember that the matrix that you're 514 00:25:59,780 --> 00:26:04,070 multiplying it by has its first column, not its first row, 515 00:26:04,070 --> 00:26:06,950 representing f of u1, and its second column 516 00:26:06,950 --> 00:26:08,780 representing f of u2. 517 00:26:08,780 --> 00:26:12,260 And now, let's illustrate that by means of a still more 518 00:26:12,260 --> 00:26:14,610 concrete example. 519 00:26:14,610 --> 00:26:18,590 Let's again let f be a linear transformation 520 00:26:18,590 --> 00:26:24,680 that maps V into V. Again, let's assume that V is given 521 00:26:24,680 --> 00:26:27,590 in terms of a basis u1 and u2. 522 00:26:27,590 --> 00:26:31,220 But let's become more concrete now and replace the as 523 00:26:31,220 --> 00:26:32,660 by specific numbers. 524 00:26:32,660 --> 00:26:35,330 Let's assume, for example, that we know that f of u1 525 00:26:35,330 --> 00:26:37,550 is 3u1 plus 4u2. 526 00:26:37,550 --> 00:26:40,940 f of u2 is 5u1 plus 7u2. 527 00:26:40,940 --> 00:26:43,400 What we're saying now is that by linearity, we 528 00:26:43,400 --> 00:26:48,620 know exactly what f does to every vector in V, 529 00:26:48,620 --> 00:26:50,720 simply in terms of what it does here. 530 00:26:50,720 --> 00:26:52,230 Now, let's see how that works. 531 00:26:52,230 --> 00:26:56,130 Let me pick an arbitrary vector in capital V. Let's say, 532 00:26:56,130 --> 00:26:59,120 for example, it's 2u1 plus u2. 533 00:26:59,120 --> 00:27:02,660 According to my theory of the previous example, what must 534 00:27:02,660 --> 00:27:04,790 f of V equal? 535 00:27:04,790 --> 00:27:10,280 Well, what I do is I take the matrix whose first column has 536 00:27:10,280 --> 00:27:13,340 3 and 4 as entries and whose second column 537 00:27:13,340 --> 00:27:15,830 has 5 and 7 as entries. 538 00:27:15,830 --> 00:27:20,270 And I multiply that by the column matrix 2, 1. 539 00:27:20,270 --> 00:27:23,420 And to make this come out to be a row vector, 540 00:27:23,420 --> 00:27:25,760 I take the transpose of this. 541 00:27:25,760 --> 00:27:28,930 And that, indeed, does turn out to be what? 542 00:27:28,930 --> 00:27:31,015 3 times 2, 5 times 1. 543 00:27:31,015 --> 00:27:32,180 That adds up to 11. 544 00:27:32,180 --> 00:27:33,290 4 times 2 is 8. 545 00:27:33,290 --> 00:27:34,550 7 times 1 is 7. 546 00:27:34,550 --> 00:27:35,780 That adds up to 15. 547 00:27:35,780 --> 00:27:39,320 That's the vector 11 comma 15, relative to the basis 548 00:27:39,320 --> 00:27:40,670 u1 and u2. 549 00:27:40,670 --> 00:27:43,340 Again, if you don't like the idea 550 00:27:43,340 --> 00:27:46,790 of having to transpose this, if you 551 00:27:46,790 --> 00:27:50,540 want to be able to write the matrix 3, 4, 5, 7 directly, 552 00:27:50,540 --> 00:27:53,690 the way you do it is write that matrix directly. 553 00:27:53,690 --> 00:27:58,640 But now, write the vector v as a row vector on the left side 554 00:27:58,640 --> 00:28:00,100 of this matrix. 555 00:28:00,100 --> 00:28:00,600 You see? 556 00:28:00,600 --> 00:28:02,308 Again, I'll get the same answer, won't I? 557 00:28:02,308 --> 00:28:03,770 2 times 3 is 6. 558 00:28:03,770 --> 00:28:05,150 1 times 5 is 5. 559 00:28:05,150 --> 00:28:07,550 6 plus 5 is 11. 560 00:28:07,550 --> 00:28:09,110 2 times 4 is 8. 561 00:28:09,110 --> 00:28:10,520 1 times 7 is 7. 562 00:28:10,520 --> 00:28:11,500 I get 15. 563 00:28:11,500 --> 00:28:14,690 Now, whichever way I do this, notice that f of v 564 00:28:14,690 --> 00:28:17,450 is 11u1 plus 15u2. 565 00:28:17,450 --> 00:28:19,640 And notice also, there was nothing sacred 566 00:28:19,640 --> 00:28:23,540 about this choice of v. I just picked a concrete example 567 00:28:23,540 --> 00:28:28,130 to show you how I can determine what the image of any vector 568 00:28:28,130 --> 00:28:30,830 v with respect to the linear transformation f 569 00:28:30,830 --> 00:28:34,370 looks like once I know the matrix of coefficients 570 00:28:34,370 --> 00:28:37,700 of f relative to the basis u1 and u2. 571 00:28:37,700 --> 00:28:40,280 By the way, in every lecture so far 572 00:28:40,280 --> 00:28:44,120 I've been coming back to this rather difficult point 573 00:28:44,120 --> 00:28:47,810 that everything seems to depend on what basis we choose. 574 00:28:47,810 --> 00:28:49,670 And the interesting point is that whereas 575 00:28:49,670 --> 00:28:52,930 a linear transformation is defined independently 576 00:28:52,930 --> 00:28:56,590 of any coordinate system, the matrix that represents 577 00:28:56,590 --> 00:29:00,370 that linear transformation does indeed depend 578 00:29:00,370 --> 00:29:01,780 on the coordinate system. 579 00:29:01,780 --> 00:29:03,730 For example, let me just say that. 580 00:29:03,730 --> 00:29:06,970 The matrix of f does depend on the basis that you've chosen. 581 00:29:06,970 --> 00:29:09,070 And by means of an example, let's 582 00:29:09,070 --> 00:29:15,710 let f u1, u2, and v be the same as in the previous exercise. 583 00:29:15,710 --> 00:29:16,900 Remember what that is, now. 584 00:29:16,900 --> 00:29:20,140 Now, I'm going to pick a new basis for my vector space 585 00:29:20,140 --> 00:29:23,160 v. In fact, let's call capital V the same thing, too. 586 00:29:23,160 --> 00:29:27,080 Let all this be the same as in exercise 5, example 5. 587 00:29:27,080 --> 00:29:29,740 Now, I'm going to pick a new basis, alpha 1 and alpha 2. 588 00:29:29,740 --> 00:29:34,300 Relative to u1 and u2, alpha 1 is 2u1 plus 3u2. 589 00:29:34,300 --> 00:29:36,520 Alpha 2 is u1 plus u2. 590 00:29:36,520 --> 00:29:38,560 Since these are not constant multiples of one 591 00:29:38,560 --> 00:29:41,560 another, alpha 1 and alpha 2 are linearly independent. 592 00:29:41,560 --> 00:29:43,390 Because they're linearly independent, 593 00:29:43,390 --> 00:29:46,210 they form a basis because they are what? 594 00:29:46,210 --> 00:29:47,740 The dimension of V is 2. 595 00:29:47,740 --> 00:29:50,800 And so any set of two linearly independent vectors 596 00:29:50,800 --> 00:29:53,890 in a two-dimensional space will be a basis for that space. 597 00:29:53,890 --> 00:29:57,190 So I now have a new basis for V, alpha 1 alpha 2. 598 00:29:57,190 --> 00:30:00,130 By the way, in the same way that I can express the alphas 599 00:30:00,130 --> 00:30:04,300 in terms of the us, I can invert this by any method that I wish 600 00:30:04,300 --> 00:30:06,160 and show that the us can be expressed 601 00:30:06,160 --> 00:30:08,590 in terms of the alphas in particularly 602 00:30:08,590 --> 00:30:10,450 in this particular manner. 603 00:30:10,450 --> 00:30:12,880 What this means now, of course, is that V is still 604 00:30:12,880 --> 00:30:14,410 the same vector space. 605 00:30:14,410 --> 00:30:18,640 But I can view it as being with respect to the basis alpha 1, 606 00:30:18,640 --> 00:30:19,630 alpha 2. 607 00:30:19,630 --> 00:30:23,350 But it's the same vector space that I had back in example 5, 608 00:30:23,350 --> 00:30:27,520 when I was viewing this with respect to the basis u1 and u2. 609 00:30:27,520 --> 00:30:29,740 Now the thing that comes up now is can I 610 00:30:29,740 --> 00:30:32,680 convert everything from example 5 611 00:30:32,680 --> 00:30:35,080 into the language of the alphas? 612 00:30:35,080 --> 00:30:40,480 For example, we knew in example 5 that v was 2u1 plus u2. 613 00:30:40,480 --> 00:30:43,480 But now, knowing what u1 and u2 look like in terms 614 00:30:43,480 --> 00:30:46,270 of the alphas, I simply make this substitution, 615 00:30:46,270 --> 00:30:48,910 replacing u1 and u2 by what they look 616 00:30:48,910 --> 00:30:50,440 like in terms of the alphas. 617 00:30:50,440 --> 00:30:54,460 And I conclude that in terms of the alphas, v is minus alpha 1 618 00:30:54,460 --> 00:30:56,150 plus 4 alpha 2. 619 00:30:56,150 --> 00:31:00,370 In other words, notice that v is the 2-tuple minus 1 620 00:31:00,370 --> 00:31:04,290 comma 4 relative to the basis alpha 1, alpha 2. 621 00:31:04,290 --> 00:31:08,920 But if the 2-tuple 2 comma 1 relative to u1 and u2 is 622 00:31:08,920 --> 00:31:09,760 the basis-- 623 00:31:09,760 --> 00:31:13,450 but that we already knew from the previous lecture 624 00:31:13,450 --> 00:31:16,540 that this was the case, that the 2-tuple depended 625 00:31:16,540 --> 00:31:18,537 on the basis that was chosen. 626 00:31:18,537 --> 00:31:20,620 But now, I'd like to show you that the matrix also 627 00:31:20,620 --> 00:31:22,330 depends on the basis chosen. 628 00:31:22,330 --> 00:31:25,180 Well, how do we write f of v relative 629 00:31:25,180 --> 00:31:26,950 to alpha 1 and alpha 2? 630 00:31:26,950 --> 00:31:30,280 According to my general theory, all I have to know 631 00:31:30,280 --> 00:31:34,210 is what alpha 1 and alpha 2 get mapped into. 632 00:31:34,210 --> 00:31:35,280 And then I'm home free. 633 00:31:35,280 --> 00:31:37,530 In other words, I would like to know what f of alpha 1 634 00:31:37,530 --> 00:31:39,430 and f of alpha 2 look like, but now 635 00:31:39,430 --> 00:31:42,640 as linear combinations of alpha 1 and alpha 2. 636 00:31:42,640 --> 00:31:44,330 All right, how do I do this? 637 00:31:44,330 --> 00:31:48,970 How do I express f of alpha 1 and f of alpha 2 638 00:31:48,970 --> 00:31:51,790 as linear combinations of alpha 1 and alpha 2? 639 00:31:51,790 --> 00:31:54,760 Well, from the previous example, the easiest thing to do 640 00:31:54,760 --> 00:31:57,610 is to express f of alpha 1 and f of alpha 2 641 00:31:57,610 --> 00:31:59,470 in terms of u1 and u2. 642 00:31:59,470 --> 00:32:02,440 Namely, f of alpha 1 is what? 643 00:32:02,440 --> 00:32:06,940 The matrix relative to u1 and u2 is 3, 5, 4, 7. 644 00:32:06,940 --> 00:32:12,430 Alpha 1 written in u1 and u2 components is 2u1 plus 3u2. 645 00:32:12,430 --> 00:32:17,400 So f of alpha 1 is the 2-tuple 21 comma 29, 646 00:32:17,400 --> 00:32:20,180 where this is relative to u1 and u2. 647 00:32:20,180 --> 00:32:24,880 In other words, f of alpha 1 is 21u1 plus 29u2. 648 00:32:24,880 --> 00:32:28,450 We know what alpha 1 and alpha 2 look like in terms of alpha 1 649 00:32:28,450 --> 00:32:29,500 and alpha 2. 650 00:32:29,500 --> 00:32:31,210 We just were talking about that. 651 00:32:31,210 --> 00:32:35,620 Leaving the details to you, replacing u1 and u2 652 00:32:35,620 --> 00:32:38,830 by what they're equal to in terms of alpha 1 alpha 2, 653 00:32:38,830 --> 00:32:43,360 we see that f of alpha 1 is 8 alpha 1 plus 5 alpha 2. 654 00:32:43,360 --> 00:32:47,740 We have now, you see, expressed f of alpha 1 655 00:32:47,740 --> 00:32:52,060 as a 2-tuple relative to alpha 1 and alpha 2. 656 00:32:52,060 --> 00:32:58,590 In a similar way, f of alpha 2 is given by this, which, again, 657 00:32:58,590 --> 00:33:00,510 replacing u1 and u2 by what they're 658 00:33:00,510 --> 00:33:04,080 equal to in terms of the alphas, turns out to be this. 659 00:33:04,080 --> 00:33:10,470 So I now know the matrix of f in terms of what 660 00:33:10,470 --> 00:33:12,060 it does to alpha 1 and alpha 2. 661 00:33:12,060 --> 00:33:13,830 In fact, what will that matrix do? 662 00:33:13,830 --> 00:33:16,620 Writing it in terms of the transposed idea here, 663 00:33:16,620 --> 00:33:20,190 the first column of my matrix will be 8, 5. 664 00:33:20,190 --> 00:33:22,950 The second column will be 3, 2. 665 00:33:22,950 --> 00:33:27,510 So the matrix now of f is 8, 3, 5, 2. 666 00:33:27,510 --> 00:33:32,250 I multiply that by the vector v, which as we just saw over here, 667 00:33:32,250 --> 00:33:36,320 is minus 1 comma 4. 668 00:33:36,320 --> 00:33:38,750 If I carry out this operation, I have what? 669 00:33:38,750 --> 00:33:43,580 Minus 8 plus 12, which is 4, minus 5 plus 8, which is 3. 670 00:33:43,580 --> 00:33:46,850 So f of v is now the 2-tuple 4 comma 671 00:33:46,850 --> 00:33:50,640 3 relative to the basis alpha 1, alpha 2. 672 00:33:50,640 --> 00:33:54,920 In other words, f of v is 4 alpha 1 plus 3 alpha 2. 673 00:33:54,920 --> 00:33:58,900 And remembering that alpha 1 was 2u1 plus 3u2, 674 00:33:58,900 --> 00:34:03,320 and that alpha 2 was u1 plus u2, I can now as a check 675 00:34:03,320 --> 00:34:06,100 convert this into us. 676 00:34:06,100 --> 00:34:07,548 And I then find what? 677 00:34:07,548 --> 00:34:08,090 What is this? 678 00:34:08,090 --> 00:34:09,739 8 plus 3 is 11. 679 00:34:09,739 --> 00:34:12,020 12 plus 3 is 15. 680 00:34:12,020 --> 00:34:16,639 f of v is, indeed, 11u1 plus 15u2, 681 00:34:16,639 --> 00:34:21,560 which does check with our result of the example-- 682 00:34:21,560 --> 00:34:22,880 what number was it? 683 00:34:22,880 --> 00:34:27,110 Example 5, the previous example, where we did the same problem 684 00:34:27,110 --> 00:34:28,400 in terms of the us. 685 00:34:28,400 --> 00:34:31,580 And what I want you to see is this. 686 00:34:31,580 --> 00:34:33,739 Obviously, v is a fixed vector. 687 00:34:33,739 --> 00:34:36,570 f is a fixed linear transformation. 688 00:34:36,570 --> 00:34:39,260 So f of v must be the same vector, 689 00:34:39,260 --> 00:34:40,900 no matter what the basis is. 690 00:34:40,900 --> 00:34:45,679 But in terms of one basis, f of v is the 2-tuple 4 comma 3. 691 00:34:45,679 --> 00:34:50,060 In terms of another basis, it's the 2-tuple 11 comma 15. 692 00:34:50,060 --> 00:34:52,909 In terms of one basis, f is represented 693 00:34:52,909 --> 00:34:55,760 by the matrix 8, 3, 5, 2. 694 00:34:55,760 --> 00:34:58,640 In terms of another basis, f is represented 695 00:34:58,640 --> 00:35:01,190 by the matrix 3, 5, 4, 7. 696 00:35:01,190 --> 00:35:04,310 So what you see, indeed, is nothing more than this-- 697 00:35:04,310 --> 00:35:06,080 that there are two difficult problems 698 00:35:06,080 --> 00:35:07,700 that we have to deal with. 699 00:35:07,700 --> 00:35:09,740 One problem, which is relatively easy, 700 00:35:09,740 --> 00:35:11,930 once you get the basic ideas down, 701 00:35:11,930 --> 00:35:14,330 is the fact that linear transformations have nothing 702 00:35:14,330 --> 00:35:15,900 to do with the basis. 703 00:35:15,900 --> 00:35:16,400 You see? 704 00:35:16,400 --> 00:35:19,640 It's defined for a vector space. 705 00:35:19,640 --> 00:35:22,050 But the difficult point is that in many cases 706 00:35:22,050 --> 00:35:24,080 we are dealing with a specific basis. 707 00:35:24,080 --> 00:35:26,360 In the middle of a problem, we change from one basis 708 00:35:26,360 --> 00:35:27,290 to another. 709 00:35:27,290 --> 00:35:29,420 Consequently, the basis that represents 710 00:35:29,420 --> 00:35:33,350 the linear transformation will, in general, change, 711 00:35:33,350 --> 00:35:35,960 as we go from one representation to another. 712 00:35:35,960 --> 00:35:39,050 And this is where a lot of computational difficulty 713 00:35:39,050 --> 00:35:40,340 seems to creep in. 714 00:35:40,340 --> 00:35:43,820 Again, I will hammer this home during the exercises. 715 00:35:43,820 --> 00:35:46,700 But for now, what I hope you have gotten out of this lecture 716 00:35:46,700 --> 00:35:50,840 is the overall view of what a linear transformation means 717 00:35:50,840 --> 00:35:53,570 and how, in terms of vector spaces, 718 00:35:53,570 --> 00:35:58,700 the concept of linearity finds a very nice unified home 719 00:35:58,700 --> 00:35:59,540 to live in. 720 00:35:59,540 --> 00:36:04,770 At any rate then, until next time, goodbye. 721 00:36:04,770 --> 00:36:07,170 Funding for the publication of this video 722 00:36:07,170 --> 00:36:12,030 was provided by the Gabriella and Paul Rosenbaum Foundation. 723 00:36:12,030 --> 00:36:16,200 Help OCW continue to provide free and open access to MIT 724 00:36:16,200 --> 00:36:21,635 courses by making a donation at ocw.mit.edu/donate.