WEBVTT 00:00:00.090 --> 00:00:02.490 The following content is provided under a Creative 00:00:02.490 --> 00:00:04.030 Commons license. 00:00:04.030 --> 00:00:06.330 Your support will help MIT OpenCourseWare 00:00:06.330 --> 00:00:10.690 continue to offer high-quality educational resources for free. 00:00:10.690 --> 00:00:13.320 To make a donation or view additional materials 00:00:13.320 --> 00:00:17.270 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:17.270 --> 00:00:18.200 at ocw.mit.edu. 00:00:32.082 --> 00:00:33.070 HERBERT GROSS: Hi. 00:00:33.070 --> 00:00:35.740 As I was getting myself prepared for the lecture, 00:00:35.740 --> 00:00:38.320 an old shaggy dog science story came 00:00:38.320 --> 00:00:41.380 to mind that's usually told as a tribute 00:00:41.380 --> 00:00:44.110 to the German scientist's thoroughness. 00:00:44.110 --> 00:00:47.110 This was the story of when all the scientists of the world 00:00:47.110 --> 00:00:49.840 got together and decided for an annual project. 00:00:49.840 --> 00:00:53.650 Each country would study exhaustively a different animal 00:00:53.650 --> 00:00:56.140 and report at the end of the year 00:00:56.140 --> 00:00:57.490 as to what they had found out. 00:00:57.490 --> 00:01:00.430 And at the end of the year, every country but Germany 00:01:00.430 --> 00:01:01.810 had been heard from. 00:01:01.810 --> 00:01:04.330 No one knew what had happened to the German scientists. 00:01:04.330 --> 00:01:09.250 And five years later, the German report came in, a huge epitome, 00:01:09.250 --> 00:01:13.860 and it was entitled, "Handbook of the Elephant, Volume 1." 00:01:13.860 --> 00:01:16.820 And the reason that this story comes to mind is twofold. 00:01:16.820 --> 00:01:19.750 First of all, the study of vector spaces 00:01:19.750 --> 00:01:25.390 has so many ramifications, that to study the subject thoroughly 00:01:25.390 --> 00:01:28.210 should be a full year's course by itself. 00:01:28.210 --> 00:01:32.170 And secondly, with respect to the particular topic of basis 00:01:32.170 --> 00:01:35.800 vectors and spanning vectors, and whether subelements 00:01:35.800 --> 00:01:38.050 are linearly independent, whereas we've 00:01:38.050 --> 00:01:41.950 invented a rather nice row reduced matrix technique 00:01:41.950 --> 00:01:47.330 for finding out what space is spanned by a set of vectors 00:01:47.330 --> 00:01:49.550 and what a basis for that space is, 00:01:49.550 --> 00:01:52.830 frequently we don't want that much information. 00:01:52.830 --> 00:01:55.760 And what I'm leading up to is a topic which you've all 00:01:55.760 --> 00:01:57.560 had in a previous context. 00:01:57.560 --> 00:01:59.000 It's called determinants. 00:01:59.000 --> 00:02:00.830 And today, what I would like to do 00:02:00.830 --> 00:02:06.470 is to study determinants within the framework of vector spaces. 00:02:06.470 --> 00:02:09.000 And the way it comes up is as follows. 00:02:09.000 --> 00:02:12.020 Suppose we have an n-dimensional vector space, v, 00:02:12.020 --> 00:02:14.730 and suppose that we pick a particular basis, 00:02:14.730 --> 00:02:18.270 u1 up to un, for v itself. 00:02:18.270 --> 00:02:20.010 Now, the idea is this. 00:02:20.010 --> 00:02:22.080 Knowing that the dimension of v is n, 00:02:22.080 --> 00:02:25.140 we immediately know that more than n vectors 00:02:25.140 --> 00:02:27.210 can't be a basis for v, because they 00:02:27.210 --> 00:02:29.370 will be linearly dependent rather than linearly 00:02:29.370 --> 00:02:31.080 independent. 00:02:31.080 --> 00:02:35.700 Fewer than n vectors of v cannot be a basis, 00:02:35.700 --> 00:02:39.180 because fewer than n vectors, since the dimension is n, 00:02:39.180 --> 00:02:43.950 cannot span v. Consequently, the only point of interest is what 00:02:43.950 --> 00:02:48.030 happens when we're given a set of n vectors and v. Namely, 00:02:48.030 --> 00:02:52.620 given the n vectors, alpha 1 up to alpha n and v, 00:02:52.620 --> 00:02:55.770 is this set a basis or isn't it? 00:02:55.770 --> 00:02:59.610 And rather than to use the row reduced matrix technique, here 00:02:59.610 --> 00:03:02.080 what we're saying is we don't want, 00:03:02.080 --> 00:03:05.190 for example, in many cases, to know what the betas look 00:03:05.190 --> 00:03:06.900 like that we talked about in our lecture 00:03:06.900 --> 00:03:08.670 on spanning vectors and the like. 00:03:08.670 --> 00:03:11.280 All we want to know is, are these vectors linearly 00:03:11.280 --> 00:03:13.560 independent or aren't they? 00:03:13.560 --> 00:03:16.440 And so what we do is we essentially 00:03:16.440 --> 00:03:18.247 invent a function machine. 00:03:18.247 --> 00:03:19.830 In other words, what we're going to do 00:03:19.830 --> 00:03:22.350 is to construct a function in which, 00:03:22.350 --> 00:03:26.010 given that the dimension of our space is n-dimensional, 00:03:26.010 --> 00:03:29.730 the input of our function machine 00:03:29.730 --> 00:03:33.420 will be any set of n vectors from that space. 00:03:33.420 --> 00:03:36.630 And the machine will be programmed to give 0 00:03:36.630 --> 00:03:40.030 as an output if the vectors are linearly dependent, 00:03:40.030 --> 00:03:43.730 in other words, if they are not a basis, and non-zero-- 00:03:43.730 --> 00:03:46.000 and I'll explain in more detail what non-zero-- 00:03:46.000 --> 00:03:49.650 why I picked non-zero rather than a specific non-zero 00:03:49.650 --> 00:03:51.900 number-- 00:03:51.900 --> 00:03:54.160 if the n vectors do form a basis. 00:03:54.160 --> 00:03:56.550 In other words, I am going to invent 00:03:56.550 --> 00:04:00.180 a function, D, capital D, to indicate the word determinant 00:04:00.180 --> 00:04:06.360 here, such that D of the n vectors, alpha 1 up to alpha n, 00:04:06.360 --> 00:04:10.770 will be 0 if and only if the set of n vectors 00:04:10.770 --> 00:04:12.990 are linearly dependent. 00:04:12.990 --> 00:04:14.880 Now, obviously, a function machine 00:04:14.880 --> 00:04:16.649 can do nothing by itself. 00:04:16.649 --> 00:04:17.839 We have to give-- 00:04:17.839 --> 00:04:21.160 and I hope this is well-known by now in our course-- 00:04:21.160 --> 00:04:24.390 we have to endow anything that we're 00:04:24.390 --> 00:04:27.190 working with with a particular structure 00:04:27.190 --> 00:04:31.140 so that is feels free to work logically for us. 00:04:31.140 --> 00:04:33.000 Well, among other things, what do we 00:04:33.000 --> 00:04:35.970 know for sure is a set of n vectors 00:04:35.970 --> 00:04:38.550 which are a basis for v? 00:04:38.550 --> 00:04:43.740 Since we were given that the specific basis that v is being 00:04:43.740 --> 00:04:48.296 referred to with respect to are u1 up tp un, 00:04:48.296 --> 00:04:53.550 then certainly, if u1 up to un is the input of the D machine, 00:04:53.550 --> 00:04:55.740 we want a non-zero output. 00:04:55.740 --> 00:04:59.340 For the sake of just normalizing things, 00:04:59.340 --> 00:05:03.160 let's program the machine so that D of u1 up to un 00:05:03.160 --> 00:05:05.750 will be 1. 00:05:05.750 --> 00:05:08.290 Now, notice, by the way, this only 00:05:08.290 --> 00:05:10.480 tells us one special basis. 00:05:10.480 --> 00:05:13.450 There are many bases that I could have chosen for v, 00:05:13.450 --> 00:05:15.760 and certainly condition one by itself 00:05:15.760 --> 00:05:19.210 isn't going to tell me any-- has no information programmed 00:05:19.210 --> 00:05:21.550 in it to tell me anything other than what 00:05:21.550 --> 00:05:24.220 it does to u1 up to un. 00:05:24.220 --> 00:05:28.360 As a second input to my D machine for programming it, 00:05:28.360 --> 00:05:31.450 I certainly know that given the n vectors, 00:05:31.450 --> 00:05:34.600 if any two of the vectors happen to be equal, then 00:05:34.600 --> 00:05:37.210 those vectors are linearly dependent. 00:05:37.210 --> 00:05:40.330 Consequently, I instruct my D machine, 00:05:40.330 --> 00:05:45.030 that if the input is a set of vectors alpha 1 up to alpha n, 00:05:45.030 --> 00:05:47.830 and at least two of the alphas are equal-- 00:05:47.830 --> 00:05:49.750 and the way we say that mathematically 00:05:49.750 --> 00:05:55.060 is that alpha i equals alpha j for a sum i unequal to j-- 00:05:55.060 --> 00:05:57.520 that if two of the vectors in the set are equal, 00:05:57.520 --> 00:06:00.580 we tell the machine to grind out 0 as an output. 00:06:00.580 --> 00:06:02.470 And by the way, at this stage we should 00:06:02.470 --> 00:06:05.590 be very careful to recognize that this is not 00:06:05.590 --> 00:06:07.720 the only way in which a set of vectors 00:06:07.720 --> 00:06:09.170 can be linearly dependent. 00:06:09.170 --> 00:06:11.230 There are many ways in which a set of vectors 00:06:11.230 --> 00:06:14.290 can be linearly dependent, even if no two of the vectors 00:06:14.290 --> 00:06:15.560 are equal. 00:06:15.560 --> 00:06:18.490 So consequently, given a linearly dependent set 00:06:18.490 --> 00:06:21.230 of vectors in which no two are equal, 00:06:21.230 --> 00:06:25.720 notice that condition two here will not work or do 00:06:25.720 --> 00:06:26.660 anything for me. 00:06:26.660 --> 00:06:30.160 So all I have done by conditions one and two 00:06:30.160 --> 00:06:33.700 is I have endowed my D machine with two standards. 00:06:33.700 --> 00:06:37.150 One particular case in which the output of the D machine 00:06:37.150 --> 00:06:40.780 will be the number 1, and a particular situation 00:06:40.780 --> 00:06:43.990 in which the output of the D machine will be 0. 00:06:43.990 --> 00:06:45.850 But I have not solved the problem 00:06:45.850 --> 00:06:49.630 that I wanted so far, namely, to make sure 00:06:49.630 --> 00:06:53.740 that D grinds out 0 if and only if the set of n vectors 00:06:53.740 --> 00:06:56.320 are linearly dependent. 00:06:56.320 --> 00:07:01.020 Let me now endow D with a third property. 00:07:01.020 --> 00:07:04.360 It is a property that will suggest linearity to you, 00:07:04.360 --> 00:07:07.540 and it's one which is amazingly powerful. 00:07:07.540 --> 00:07:10.420 And what I mean by amazingly powerful hopefully 00:07:10.420 --> 00:07:12.580 will become clear in a few moments. 00:07:12.580 --> 00:07:14.920 But the third property is this. 00:07:14.920 --> 00:07:16.840 And rather than work with n-tuples 00:07:16.840 --> 00:07:19.060 here, to make things easier to read, 00:07:19.060 --> 00:07:21.350 let me just take a 3-tuple over here. 00:07:21.350 --> 00:07:24.040 Suppose I have a three-dimensional vector space, 00:07:24.040 --> 00:07:26.530 and I pick three vectors in that space. 00:07:26.530 --> 00:07:28.240 Suppose one of those vectors-- the way 00:07:28.240 --> 00:07:29.770 I've written that here, it happens 00:07:29.770 --> 00:07:31.090 to be the second vector. 00:07:31.090 --> 00:07:34.240 Suppose one of those vectors is itself written 00:07:34.240 --> 00:07:37.180 as a sum of two other vectors in the space. 00:07:37.180 --> 00:07:41.500 Then I tell the D machine to compute this as follows. 00:07:41.500 --> 00:07:44.800 Linearize this, in other words, compute this 00:07:44.800 --> 00:07:49.810 as if it were the sum of two separate determinants, one 00:07:49.810 --> 00:07:53.080 of which had the alpha 2 missing, and the other of which 00:07:53.080 --> 00:07:54.670 had the beta 2 missing. 00:07:54.670 --> 00:07:56.650 In other words, what I do is is I 00:07:56.650 --> 00:07:58.660 endow the machine with the property 00:07:58.660 --> 00:08:04.210 that D of alpha 1 comma, alpha 2 plus beta 2 comma, alpha 3 00:08:04.210 --> 00:08:09.850 will be D of alpha 1, alpha 2, alpha 3 plus D 00:08:09.850 --> 00:08:13.410 of alpha 1, beta 2, alpha 3. 00:08:16.420 --> 00:08:18.875 And thirdly-- well, thirdly. 00:08:18.875 --> 00:08:19.750 I don't mean thirdly. 00:08:19.750 --> 00:08:21.790 I mean the other lineal property. 00:08:21.790 --> 00:08:27.880 If one of the input vectors is multiplied by a scalar, 00:08:27.880 --> 00:08:29.690 I can factor the scalar out. 00:08:29.690 --> 00:08:33.190 In other words, if the three vectors that are being tested 00:08:33.190 --> 00:08:36.790 in my three-dimensional space have the form alpha 1, 00:08:36.790 --> 00:08:42.549 c alpha 2, and alpha 3, D of alpha 1 comma, c alpha 2 comma, 00:08:42.549 --> 00:08:44.340 alpha 3 will be-- 00:08:44.340 --> 00:08:45.970 see, factor of the c out. 00:08:45.970 --> 00:08:50.200 c times D of alpha 1, alpha 2, alpha 3. 00:08:50.200 --> 00:08:53.950 And it's extremely important to notice that the c does not 00:08:53.950 --> 00:08:58.360 have to be a scalar multiple of all three vectors in the input. 00:08:58.360 --> 00:09:00.610 Notice, that the fact that the c was 00:09:00.610 --> 00:09:05.230 multiplying one of the vectors is enough to factor the c out. 00:09:05.230 --> 00:09:07.660 Without belaboring this point, all I wanted to say 00:09:07.660 --> 00:09:10.950 was, that if the c were multiplying each of the alphas, 00:09:10.950 --> 00:09:14.230 in other words, if this were D of c alpha 1 comma, 00:09:14.230 --> 00:09:19.900 c alpha 2 comma, c alpha 3, this would equal c cubed D alpha 1, 00:09:19.900 --> 00:09:21.880 alpha 2, alpha 3, because we would 00:09:21.880 --> 00:09:24.490 fact about the c each time-- 00:09:24.490 --> 00:09:26.860 one for each vector. 00:09:26.860 --> 00:09:30.270 But let me illustrate this in terms of a 2 by 2-- 00:09:30.270 --> 00:09:32.770 a two-dimensional example for you. 00:09:32.770 --> 00:09:35.170 Suppose I'm dealing in two-dimensional space, 00:09:35.170 --> 00:09:39.170 and I'm dealing with respect to a particular basis, u1 and u2. 00:09:39.170 --> 00:09:42.460 And let's suppose that relative to that basis, u1 and u2, 00:09:42.460 --> 00:09:45.040 alpha 1 is the vector 3 comma, 1, 00:09:45.040 --> 00:09:49.570 beta 1 is the vector 6 comma, 7, and beta 2 00:09:49.570 --> 00:09:53.930 is the vector 4 common, 5. 00:09:53.930 --> 00:09:57.460 Let's recall the fact that, one way or another, 00:09:57.460 --> 00:10:00.190 we already know how to expand 2 by 2 determinants, 00:10:00.190 --> 00:10:02.350 even though we may not know rigorously 00:10:02.350 --> 00:10:04.000 why the rules were chosen. 00:10:04.000 --> 00:10:07.510 The idea is, what would alpha 1 plus beta 1 be? 00:10:07.510 --> 00:10:11.950 It would be the 2-tuple whose first entry was 3 plus 6 00:10:11.950 --> 00:10:15.190 and whose second entry was 1 plus 7. 00:10:15.190 --> 00:10:18.985 In other words, the first row of the determinant 00:10:18.985 --> 00:10:22.170 that I'm going to be talking about has as its entries 00:10:22.170 --> 00:10:24.940 3 plus 6 and 1 plus 7. 00:10:24.940 --> 00:10:28.960 Beta 2, which will make up the second row with my determinant, 00:10:28.960 --> 00:10:31.580 has as its components 4 and 5. 00:10:31.580 --> 00:10:34.840 So the second row of my determinant is 4, 5. 00:10:34.840 --> 00:10:37.720 And using the usual rule for multiplying determinants, 00:10:37.720 --> 00:10:43.300 this is 5 times 9 minus 4 times 8, which is 13. 00:10:43.300 --> 00:10:45.790 And by the way, just to refresh your memories here, 00:10:45.790 --> 00:10:49.840 what if I computed in terms of my D language and the alpha 1, 00:10:49.840 --> 00:10:52.090 alpha 2, and beta 2 is used here? 00:10:52.090 --> 00:11:00.370 This is D of alpha 1 plus beta 1 comma, beta 2. 00:11:00.370 --> 00:11:05.140 On the other hand, what would D of alpha 1, beta 2 be? 00:11:05.140 --> 00:11:07.960 D of alpha 1 beta 2 would be the 2 00:11:07.960 --> 00:11:11.800 by 2 determinant, whose first row was 3, 1, 00:11:11.800 --> 00:11:14.970 and whose second row was 4, 5. 00:11:14.970 --> 00:11:20.320 D of alpha 1, beta 2 would be the determinant 00:11:20.320 --> 00:11:23.670 whose first row-- 00:11:23.670 --> 00:11:24.230 I'm sorry. 00:11:24.230 --> 00:11:27.960 D of beta 1 comma, beta 2, would be the determinant 00:11:27.960 --> 00:11:32.590 whose first row was 6, 7 and whose second row was 4, 5. 00:11:32.590 --> 00:11:36.240 In other words, these matrices here, what is this again? 00:11:36.240 --> 00:11:42.510 This is D of alpha 1, beta 2. 00:11:42.510 --> 00:11:50.300 And this one here is D of beta 1, beta 2. 00:11:50.300 --> 00:11:51.830 And notice, that this determinant 00:11:51.830 --> 00:11:54.440 by the traditional way of expanding is 11. 00:11:54.440 --> 00:11:57.380 This determinant is 30 minus 28, which is 2. 00:11:57.380 --> 00:11:59.300 11 plus 2 is 13. 00:11:59.300 --> 00:12:02.300 And we see that, at least in this case, D of alpha 1 00:12:02.300 --> 00:12:05.380 plus beta 1 comma, beta 2, is the same as D 00:12:05.380 --> 00:12:10.070 of alpha 1 comma, beta 2 plus D of beta 1 comma, beta 2. 00:12:10.070 --> 00:12:13.640 So at least the first part of property three 00:12:13.640 --> 00:12:16.460 is obeyed in this particular example. 00:12:16.460 --> 00:12:18.710 And to show what the second property is, 00:12:18.710 --> 00:12:20.540 let's suppose alpha 1 now-- 00:12:20.540 --> 00:12:21.590 I pick a new alpha 1. 00:12:21.590 --> 00:12:23.570 We'll call that 2 comma, 6. 00:12:23.570 --> 00:12:26.300 Suppose alpha 2 is 3 comma, 4. 00:12:26.300 --> 00:12:30.650 Then the determinant of alpha 1 and alpha 2 would be what? 00:12:30.650 --> 00:12:33.583 2, 6, 3, 4. 00:12:33.583 --> 00:12:34.250 And that's what? 00:12:34.250 --> 00:12:37.130 8 minus 18, which is minus 10. 00:12:37.130 --> 00:12:40.580 On the other hand, notice that a common-- 00:12:40.580 --> 00:12:42.680 that this vector here could have been 00:12:42.680 --> 00:12:45.890 written as twice 1 comma, 3. 00:12:45.890 --> 00:12:47.900 I can, therefore, factor out the 2. 00:12:47.900 --> 00:12:49.670 See, notice this very carefully. 00:12:49.670 --> 00:12:54.950 Notice that the alphas, which are written in a row here-- 00:12:54.950 --> 00:12:56.450 one row this way-- 00:12:56.450 --> 00:13:01.080 each alpha forms a row when I use the matrix interpretation. 00:13:01.080 --> 00:13:06.930 So what I'm saying is, that 2 comma, 6 is one vector. 00:13:06.930 --> 00:13:10.070 It's a 2-tuple with respect to the basis u1 and u2. 00:13:10.070 --> 00:13:13.490 All I'm saying is factor out the 2 from the first row, 00:13:13.490 --> 00:13:14.750 from the first vector. 00:13:14.750 --> 00:13:19.670 That leaves me with 2 outside, 1, 3, 3, 4 inside. 00:13:19.670 --> 00:13:21.880 And notice, that 4 times-- 00:13:21.880 --> 00:13:26.220 4 minus 9 is minus 5 times 2 is also minus 10. 00:13:26.220 --> 00:13:29.090 Notice that these, indeed, are equal, and notice 00:13:29.090 --> 00:13:33.140 that I could factor out the 2 simply by virtue of the fact 00:13:33.140 --> 00:13:35.600 that it was a common factor in one row. 00:13:35.600 --> 00:13:39.020 It did not have to be a common factor in both rows. 00:13:39.020 --> 00:13:41.810 In fact, if it had been a common factor in both rows, 00:13:41.810 --> 00:13:44.540 I would have had to factor out a 2 twice, 00:13:44.540 --> 00:13:47.240 in other words, a 4 over here. 00:13:47.240 --> 00:13:49.740 Now, I can go on with things like this, 00:13:49.740 --> 00:13:52.790 but, again, I want to stress the overview. 00:13:52.790 --> 00:13:56.480 And the key point is that these three simple properties, which 00:13:56.480 --> 00:13:59.210 I've just called simply one, two, and three, those three 00:13:59.210 --> 00:14:03.740 properties that I've programmed the D machine with 00:14:03.740 --> 00:14:06.650 are enough to completely determine 00:14:06.650 --> 00:14:09.980 D. In other words, if I wanted to be dramatic over here, 00:14:09.980 --> 00:14:14.660 the key point is that one, two, and three completely determine 00:14:14.660 --> 00:14:17.240 D. I'll put an exclamation point down there. 00:14:17.240 --> 00:14:20.480 Because what I claim is, that once these three properties are 00:14:20.480 --> 00:14:23.690 obeyed, there is no possible way for D 00:14:23.690 --> 00:14:28.450 to behave other than in a very unique well-defined way. 00:14:28.450 --> 00:14:31.580 That D now has a perfectly well-defined structure. 00:14:31.580 --> 00:14:34.130 Let me give you some examples of that structure. 00:14:34.130 --> 00:14:35.990 I'll just prove a couple of theorems. 00:14:35.990 --> 00:14:38.120 I'll prove them in two-dimensional space, 00:14:38.120 --> 00:14:40.580 leaving it for the exercises to work 00:14:40.580 --> 00:14:44.120 on higher dimensional space, but this doesn't get too cluttered. 00:14:44.120 --> 00:14:46.760 First of all, what I claim is that the D machine 00:14:46.760 --> 00:14:49.190 is so finicky, that if you interchange 00:14:49.190 --> 00:14:53.780 the order in which the two vectors are given, 00:14:53.780 --> 00:14:58.700 you change the sign of the output. 00:14:58.700 --> 00:15:01.820 In other words, D of alpha 1, alpha 2, in that order, 00:15:01.820 --> 00:15:05.360 is the negative of D of alpha 2, alpha 1. 00:15:05.360 --> 00:15:07.910 And the proof, again, follows very nicely 00:15:07.910 --> 00:15:10.040 in terms of our game structure. 00:15:10.040 --> 00:15:13.190 The gimmick to begin with is that we must be clever enough 00:15:13.190 --> 00:15:16.610 to decide that we'll compute the determinant of alpha 1 00:15:16.610 --> 00:15:20.210 plus alpha 2 comma, alpha 1 plus alpha 2. 00:15:20.210 --> 00:15:22.380 On the one hand, by our second property, 00:15:22.380 --> 00:15:26.390 since two of the vectors making up the set being tested 00:15:26.390 --> 00:15:29.960 are equal, it means that that determinant must be 0. 00:15:29.960 --> 00:15:31.820 On the other hand, by property three, 00:15:31.820 --> 00:15:35.780 by linearity, splitting this up as two terms, 00:15:35.780 --> 00:15:40.730 notice that this is D of alpha 1 comma, alpha 1 plus alpha 2 00:15:40.730 --> 00:15:45.420 plus D of alpha 2 comma, alpha 1 plus alpha 2. 00:15:45.420 --> 00:15:47.390 In other words, this result here. 00:15:47.390 --> 00:15:51.470 Now, in turn, noticing that these form a sum, 00:15:51.470 --> 00:15:54.080 I can rewrite this one as what? 00:15:54.080 --> 00:16:00.620 D of alpha 1 comma, alpha 1 plus D of alpha 1 comma, alpha 2. 00:16:00.620 --> 00:16:04.310 This one is D of alpha 2 comma, alpha 1 00:16:04.310 --> 00:16:08.090 plus D of alpha 2 comma, alpha 2. 00:16:08.090 --> 00:16:10.850 I've just written this whole thing out on this next line. 00:16:10.850 --> 00:16:16.500 By property two, D of alpha 1 comma, alpha 1 must be 0. 00:16:16.500 --> 00:16:19.130 D of alpha 2 comma, alpha 2 must be 0, 00:16:19.130 --> 00:16:24.380 because after all, you see, two vectors are equal in this. 00:16:24.380 --> 00:16:25.850 So what do I have left? 00:16:25.850 --> 00:16:30.950 Comparing this with this, I have the D of alpha 1 comma, alpha 2 00:16:30.950 --> 00:16:34.520 plus D of alpha 2 comma, alpha 1 is 0. 00:16:34.520 --> 00:16:37.130 And that means that the number-- and keep that in mind, 00:16:37.130 --> 00:16:38.510 the determinant is a number. 00:16:38.510 --> 00:16:40.790 It maps the n vectors into a number, which 00:16:40.790 --> 00:16:42.380 is either 0 or non-zero. 00:16:42.380 --> 00:16:45.260 But the determinant of alpha 1 comma, alpha 2, 00:16:45.260 --> 00:16:48.890 therefore, must be the negative of the term of alpha 2, 00:16:48.890 --> 00:16:52.010 alpha 1, because their sum is 0. 00:16:52.010 --> 00:16:54.170 The second theorem-- and this is the one 00:16:54.170 --> 00:16:58.280 that relates determinants to matrices, a very fantastic 00:16:58.280 --> 00:17:01.850 result, one that has tremendous practical application that 00:17:01.850 --> 00:17:04.319 will also talk about later in the lecture. 00:17:04.319 --> 00:17:08.359 And that is, that if I take any one of my input vectors 00:17:08.359 --> 00:17:13.490 and replace it by itself, plus a scalar multiple of another, 00:17:13.490 --> 00:17:15.740 I do not change the determinant. 00:17:15.740 --> 00:17:19.910 In other words, for example, if I replace alpha 1 by alpha 1 00:17:19.910 --> 00:17:24.550 plus a scalar multiple of alpha 2, and I leave alpha 2 alone, 00:17:24.550 --> 00:17:27.530 the determinant of alpha 1 and alpha 2 00:17:27.530 --> 00:17:30.170 will be the same as the determinant of alpha 1 00:17:30.170 --> 00:17:33.140 plus c alpha 2 comma, alpha 2. 00:17:33.140 --> 00:17:35.570 And again, the proof is very easy, 00:17:35.570 --> 00:17:38.870 namely, look at the expression D of alpha 1 00:17:38.870 --> 00:17:41.390 plus c alpha 2 comma, alpha 2. 00:17:41.390 --> 00:17:44.120 Use the linearity property that this 00:17:44.120 --> 00:17:51.050 is D of alpha 1 comma, alpha 2 plus D of c alpha 2 comma, 00:17:51.050 --> 00:17:51.770 alpha 2. 00:17:54.390 --> 00:17:57.530 Then use the second part of the linearity property 00:17:57.530 --> 00:18:00.780 that the constant factor, c, can be taken outside here. 00:18:00.780 --> 00:18:03.410 So this becomes D of alpha 1, alpha 2 00:18:03.410 --> 00:18:07.520 plus c times D of alpha 2 comma, alpha 2. 00:18:07.520 --> 00:18:11.780 And notice, that since alpha 2 is repeated here, 00:18:11.780 --> 00:18:13.310 this determinant is 0. 00:18:13.310 --> 00:18:14.930 And consequently, what we've proven 00:18:14.930 --> 00:18:18.302 is that this is equal to this. 00:18:18.302 --> 00:18:20.010 Now, I've just proven those two theorems, 00:18:20.010 --> 00:18:21.620 because that's about all I need. 00:18:21.620 --> 00:18:23.780 Let me show you now, that if I had never 00:18:23.780 --> 00:18:27.140 seen that shortcut method of expanding a 2 00:18:27.140 --> 00:18:29.780 by 2 determinant, as learned in high school, 00:18:29.780 --> 00:18:32.630 how these three properties uniquely 00:18:32.630 --> 00:18:34.265 determine what D has to mean. 00:18:34.265 --> 00:18:36.890 In other words, why I said that these three properties uniquely 00:18:36.890 --> 00:18:38.630 determine D. 00:18:38.630 --> 00:18:42.020 As an example, let me take a two-dimensional vector space 00:18:42.020 --> 00:18:42.950 again. 00:18:42.950 --> 00:18:47.150 Let me pick u1 and u2 as a specific basis, 00:18:47.150 --> 00:18:48.260 and that's important. 00:18:48.260 --> 00:18:51.200 I've picked a specific basis, u1, u2. 00:18:51.200 --> 00:18:53.000 Now, I pick any two vectors. 00:18:53.000 --> 00:18:56.990 Since v is a two-dimensional space, I pick any two vectors-- 00:18:56.990 --> 00:18:59.210 alpha 1 and alpha 2. 00:18:59.210 --> 00:19:01.790 And because of this notation, this 00:19:01.790 --> 00:19:05.360 means that alpha 1 and alpha 2 are both linear combinations 00:19:05.360 --> 00:19:06.830 of u1 and u2. 00:19:06.830 --> 00:19:10.130 Say alpha 1 is this, and alpha 2 is this. 00:19:10.130 --> 00:19:13.520 Therefore, what is D of alpha 1 comma, alpha 2? 00:19:13.520 --> 00:19:20.460 Well, by direct substitution, D of alpha 1, alpha 2, just 00:19:20.460 --> 00:19:23.430 replace alpha 1 by what it's equal to here, alpha 2 00:19:23.430 --> 00:19:25.020 by what it's equal to here. 00:19:25.020 --> 00:19:29.290 And we get that D of alpha 1, alpha 2 is this expression. 00:19:29.290 --> 00:19:31.800 Now, we use the linearity property. 00:19:31.800 --> 00:19:34.230 We treat this as one number for the time being 00:19:34.230 --> 00:19:36.090 and split this up as a sum. 00:19:36.090 --> 00:19:38.040 In other words, it's going to be D 00:19:38.040 --> 00:19:44.940 of this term comma, this whole term plus D of this term 00:19:44.940 --> 00:19:47.020 comma, this whole term. 00:19:47.020 --> 00:19:51.150 In other words, if I do that, this breaks down to this. 00:19:51.150 --> 00:19:54.300 Now, I notice that each of these two 00:19:54.300 --> 00:19:59.760 is a 2-tuple in which the second entry is a sum of two terms. 00:19:59.760 --> 00:20:02.730 So I now split this up into what? 00:20:02.730 --> 00:20:09.456 D of this comma, this plus d of this comma, this. 00:20:09.456 --> 00:20:15.180 This term becomes D of this comma, this term plus D 00:20:15.180 --> 00:20:17.940 of this comma, this term. 00:20:17.940 --> 00:20:20.130 And I hope it doesn't sound boorish 00:20:20.130 --> 00:20:23.370 on my part to keep saying "D of this comma, this." 00:20:23.370 --> 00:20:25.800 I prefer to say that for you so that you 00:20:25.800 --> 00:20:27.600 can watch what I'm doing, and then 00:20:27.600 --> 00:20:33.000 have you just be able afterwards to read what these things mean. 00:20:33.000 --> 00:20:35.250 Now, when I'm down to here, notice 00:20:35.250 --> 00:20:37.740 that by my second part of linearity 00:20:37.740 --> 00:20:41.520 I can factor out the a11 from here. 00:20:41.520 --> 00:20:44.340 I can factor out the a21, because it's 00:20:44.340 --> 00:20:46.060 a common factor of this term. 00:20:46.060 --> 00:20:48.510 In other words, I can write this term 00:20:48.510 --> 00:20:53.220 as a11, a21, D of u1 comma, u2. 00:20:53.220 --> 00:20:55.540 And again, sparing you the details, 00:20:55.540 --> 00:21:00.870 notice I can factor out an a11, a22 from this term. 00:21:00.870 --> 00:21:06.210 I can factor out an a12, a21 from this term, an a12, 00:21:06.210 --> 00:21:08.460 a22 from this term. 00:21:08.460 --> 00:21:12.870 And that this now can be written in this particular way. 00:21:12.870 --> 00:21:16.560 Now, what properties do I know that D is endowed with? 00:21:16.560 --> 00:21:18.480 I know, first of all, that whenever 00:21:18.480 --> 00:21:22.050 D operates on a set of vectors where at least two of them 00:21:22.050 --> 00:21:25.360 are equal, D must give 0 as an output. 00:21:25.360 --> 00:21:28.620 So D of u1 comma, u1 is 0 again. 00:21:28.620 --> 00:21:32.640 u1 comma, u2 is the particular basis with respect 00:21:32.640 --> 00:21:35.400 to which I define v, you see. 00:21:35.400 --> 00:21:41.280 Therefore, by property one, D of u1 comma, u2 must be 1. 00:21:41.280 --> 00:21:46.860 D of u2 comma, u1 is just the permuted order of u1 and u2. 00:21:46.860 --> 00:21:52.800 Therefore, by our first theorem, that must be minus D of u1, u2. 00:21:52.800 --> 00:21:54.930 Therefore, it must be minus 1. 00:21:54.930 --> 00:21:58.260 And finally, D of u2 comma, u2 is 0. 00:21:58.260 --> 00:22:00.210 And if I now collect everything that I 00:22:00.210 --> 00:22:02.200 have left here, what do I have? 00:22:02.200 --> 00:22:08.020 I have the D of alpha 1 comma, alpha 2 is equal to what? 00:22:08.020 --> 00:22:15.460 It's equal to a11, a22, see, times 1, minus a12, a21. 00:22:15.460 --> 00:22:17.700 In other words, this is the determinant 00:22:17.700 --> 00:22:20.310 of alpha 1 comma, alpha 2. 00:22:20.310 --> 00:22:24.210 And let me, again, put an exclamation point here. 00:22:24.210 --> 00:22:26.400 Because if you now go back to the high school 00:22:26.400 --> 00:22:29.010 way of computing this determinant-- 00:22:29.010 --> 00:22:30.510 remember how we did it? 00:22:30.510 --> 00:22:32.060 What would we have obtained? 00:22:32.060 --> 00:22:34.050 And, again, let me just come back to this board for a second 00:22:34.050 --> 00:22:34.580 here. 00:22:34.580 --> 00:22:36.835 Remember how you used the matrix of coefficients? 00:22:36.835 --> 00:22:38.460 So take the determinant, would be what? 00:22:38.460 --> 00:22:44.690 a11, a22 minus a12, a21. 00:22:44.690 --> 00:22:47.820 So, again, notice two things that happen here. 00:22:47.820 --> 00:22:49.860 First of all, I get the same answer 00:22:49.860 --> 00:22:51.570 as I would have got the traditional way. 00:22:51.570 --> 00:22:53.750 And second, just as a minor aside 00:22:53.750 --> 00:22:56.170 that I'll emphasize more later in the lecture 00:22:56.170 --> 00:22:58.650 and also in the exercises, notice 00:22:58.650 --> 00:23:01.980 that you can begin to suspect that the actual value 00:23:01.980 --> 00:23:04.440 of the determinant of alpha 1 and alpha 2 00:23:04.440 --> 00:23:07.410 should depend on what basis was chosen. 00:23:07.410 --> 00:23:09.870 Because relative to a different basis, 00:23:09.870 --> 00:23:12.570 notice that the coefficients might very well 00:23:12.570 --> 00:23:15.300 be different for alpha 1 and alpha 2. 00:23:15.300 --> 00:23:18.090 But I don't want to mention that right now. 00:23:18.090 --> 00:23:20.190 I think, as I say, what the amazing result 00:23:20.190 --> 00:23:22.590 is, that what I meant by saying that these three 00:23:22.590 --> 00:23:26.250 properties completely determine D, is the fact 00:23:26.250 --> 00:23:27.900 that with these three properties it 00:23:27.900 --> 00:23:32.410 turns out that the high school definition was ironclad. 00:23:32.410 --> 00:23:35.350 Meaning, there was no other possible way 00:23:35.350 --> 00:23:37.660 to define what the determinant should 00:23:37.660 --> 00:23:41.800 be if you wanted these three properties to be obeyed. 00:23:41.800 --> 00:23:43.690 By the way, I have some quick checks. 00:23:43.690 --> 00:23:45.400 Notice that relative to the basis 00:23:45.400 --> 00:23:48.850 u1, u2, u1 can be written this way, 00:23:48.850 --> 00:23:50.770 u2 can be written this way. 00:23:50.770 --> 00:23:53.890 We know that D of u1, u2 should be 1. 00:23:53.890 --> 00:23:57.310 And using the old fashioned way for checking this, 00:23:57.310 --> 00:23:58.690 what would the determinant be? 00:23:58.690 --> 00:24:03.880 It would be 1 times 1 minus 0 times 0, which is 1. 00:24:03.880 --> 00:24:08.170 Secondly, we also know that property two should yield 0 00:24:08.170 --> 00:24:12.220 as a determinant if these two vectors that were making up 00:24:12.220 --> 00:24:13.750 the input were equal. 00:24:13.750 --> 00:24:17.230 Let's call the vectors au1, bu2. 00:24:17.230 --> 00:24:19.930 So the 2 by 2 determinant I would get in this case 00:24:19.930 --> 00:24:21.580 is ab, ab. 00:24:21.580 --> 00:24:23.890 And if I expand that determinant, it's what? 00:24:23.890 --> 00:24:28.970 a times b minus a times b, which is 0. 00:24:28.970 --> 00:24:33.450 And now, let me come back to that idea of what 00:24:33.450 --> 00:24:35.460 motivated the whole block of material 00:24:35.460 --> 00:24:39.490 here, the trouble with writing vectors as n-tuples. 00:24:39.490 --> 00:24:42.870 That there really is something that requires great care. 00:24:42.870 --> 00:24:46.690 In advanced applications, one must always be wary of this. 00:24:46.690 --> 00:24:49.410 I am not going to give you the advanced applications here. 00:24:49.410 --> 00:24:52.500 All I want you to do is to become prepared 00:24:52.500 --> 00:24:54.480 against the pitfalls. 00:24:54.480 --> 00:24:57.120 And the thing is, that in dealing with determinants, 00:24:57.120 --> 00:25:00.900 as I said before, you must be very, very careful about what 00:25:00.900 --> 00:25:04.380 basis you're referring to for a given vector space. 00:25:04.380 --> 00:25:07.500 Obviously, the vectors do not depend on the basis, 00:25:07.500 --> 00:25:09.420 but their representation does. 00:25:09.420 --> 00:25:11.190 Let me give you a for instance. 00:25:11.190 --> 00:25:17.250 Using v and u1 and u2 as an example number 1, 00:25:17.250 --> 00:25:21.220 suppose we let alpha 1 be the 2-tuple, 3 comma, 4. 00:25:21.220 --> 00:25:23.650 In other words, 3u1 plus 4u2. 00:25:23.650 --> 00:25:26.370 And let alpha 2 be 2 comma, 5. 00:25:26.370 --> 00:25:28.950 Then by what we've just proven, the determinant 00:25:28.950 --> 00:25:31.920 of alpha 1 comma, alpha 2 must be 00:25:31.920 --> 00:25:34.110 the determinant of what matrix? 00:25:34.110 --> 00:25:36.810 The one whose first row is 3 comma, 4, and whose 00:25:36.810 --> 00:25:38.940 second row is 2 comma, 5. 00:25:38.940 --> 00:25:41.730 If I compute that determinant-- and by the way, notice, 00:25:41.730 --> 00:25:43.290 now that I know that the shortcut 00:25:43.290 --> 00:25:45.582 way has to be the right answer, I 00:25:45.582 --> 00:25:47.040 don't do this the long way anymore. 00:25:47.040 --> 00:25:51.630 I just say, OK, it's 3 times 5 minus 2 times 4 15 minus 8, 00:25:51.630 --> 00:25:52.590 which is 7. 00:25:52.590 --> 00:25:56.670 So the determinant of alpha 1, alpha 2 is 7. 00:25:56.670 --> 00:25:57.930 So far so good. 00:25:57.930 --> 00:26:00.450 But let me point out the following thing. 00:26:00.450 --> 00:26:04.320 Notice, that alpha 1 and alpha 2 are also linearly independent. 00:26:04.320 --> 00:26:09.750 Namely, 3 common, 4 is not a scalar multiple of 2 comma, 5. 00:26:09.750 --> 00:26:12.480 Consequently, since alpha 1 and alpha 2 our linearly 00:26:12.480 --> 00:26:16.260 independent, and since v is a two-dimensional space, 00:26:16.260 --> 00:26:20.430 it means that alpha 1 and alpha 2 are themselves a basis for v. 00:26:20.430 --> 00:26:24.570 In other words, v is equal to the space spanned by alpha 1 00:26:24.570 --> 00:26:26.400 and alpha 2 as a basis. 00:26:26.400 --> 00:26:28.590 Notice, that relative to this new basis, 00:26:28.590 --> 00:26:34.140 alpha 1 comma, alpha 2, alpha 1 is 1 alpha 1 plus 0 alpha 2, 00:26:34.140 --> 00:26:37.740 alpha 2 is 0 alpha 1 plus alpha 2. 00:26:37.740 --> 00:26:40.560 And therefore, using the traditional method 00:26:40.560 --> 00:26:42.780 of computing the determinant, notice 00:26:42.780 --> 00:26:45.900 that the determinant of alpha 1 comma, alpha 2 00:26:45.900 --> 00:26:48.990 relative to the basis alpha 1 and alpha 2 00:26:48.990 --> 00:26:50.120 would be what determinant? 00:26:50.120 --> 00:26:51.953 It would be to the determinant of the matrix 00:26:51.953 --> 00:26:55.980 whose first row was 1, 0, and whose second role was 0, 1. 00:26:55.980 --> 00:26:57.750 And that is 1. 00:26:57.750 --> 00:26:59.190 This looks like a contradiction. 00:26:59.190 --> 00:27:01.830 You see, on the one hand, we have the D of alpha 1 00:27:01.830 --> 00:27:03.420 and alpha 2 is 1. 00:27:03.420 --> 00:27:09.210 But down here, we just saw that D of alpha 1 and alpha 2 was 7. 00:27:09.210 --> 00:27:11.870 Which is correct? 00:27:11.870 --> 00:27:14.420 Well, the answer is, they're both correct. 00:27:14.420 --> 00:27:17.630 That the first observation is, that if the determinant 00:27:17.630 --> 00:27:21.800 of alpha 1 and alpha 2 is not 0, the value 00:27:21.800 --> 00:27:24.090 depends on the particular basis. 00:27:24.090 --> 00:27:26.870 In other words, if the determinant is 0, 00:27:26.870 --> 00:27:29.030 it turns out that no matter what basis 00:27:29.030 --> 00:27:32.150 you use to represent the determinant of alpha 1 00:27:32.150 --> 00:27:36.005 and alpha 2, you will get 0 once the terminator is 00:27:36.005 --> 00:27:37.340 0 in one basis. 00:27:37.340 --> 00:27:40.160 And if the determinant is not 0 with respect 00:27:40.160 --> 00:27:43.880 to a particular basis, it will be non-zero with respect 00:27:43.880 --> 00:27:45.200 to all bases. 00:27:45.200 --> 00:27:49.620 But what non-zero number it will be does depend on the basis. 00:27:49.620 --> 00:27:51.950 And that was why way back at the beginning 00:27:51.950 --> 00:27:57.080 we told the D machine simply to give a non-zero output 00:27:57.080 --> 00:28:01.670 if the input was a set of n linearly-independent vectors. 00:28:01.670 --> 00:28:05.960 In summary, when one says that the determinant of alpha 1 00:28:05.960 --> 00:28:09.350 and alpha 2 is 7, it's tacitly assumed 00:28:09.350 --> 00:28:14.330 that D is being referred to with respect to the particular basis 00:28:14.330 --> 00:28:15.770 u1, u2. 00:28:15.770 --> 00:28:19.010 On the other hand, when one says that D of alpha 1 and alpha 2 00:28:19.010 --> 00:28:22.130 is 1, it's tacitly assumed that the basis 00:28:22.130 --> 00:28:26.810 that we're using to express v is alpha 1 and alpha 2. 00:28:26.810 --> 00:28:30.440 Again, I'll drill that more in the exercises. 00:28:30.440 --> 00:28:32.810 Let me go on now to generalize what 00:28:32.810 --> 00:28:35.510 happens in n-dimensional space, keeping in mind 00:28:35.510 --> 00:28:37.370 the fact that when we generalize what 00:28:37.370 --> 00:28:41.210 happens in n-dimensional space, the actual proofs become 00:28:41.210 --> 00:28:44.690 messier, but the theory remains very much the same. 00:28:44.690 --> 00:28:49.040 I prefer to leave the messy details to the exercises, 00:28:49.040 --> 00:28:51.080 either optional or required, depending 00:28:51.080 --> 00:28:53.490 on how hard the exercises may be. 00:28:53.490 --> 00:28:55.790 But let me summarize what the results are 00:28:55.790 --> 00:28:58.010 for any n-dimensional space, keeping 00:28:58.010 --> 00:29:00.440 in mind we've proven the results rigorously 00:29:00.440 --> 00:29:02.780 for the case n equals 2. 00:29:02.780 --> 00:29:04.490 The generalization is this. 00:29:04.490 --> 00:29:06.470 If I have an n-dimensional vector space, 00:29:06.470 --> 00:29:11.000 v, with respect to a particular basis, u1 up to un, 00:29:11.000 --> 00:29:16.010 and if alpha 1 up to alpha n are n vectors chosen from v so 00:29:16.010 --> 00:29:19.490 that they are linear combinations of the us, 00:29:19.490 --> 00:29:22.040 say, in the traditional way that we've written this 00:29:22.040 --> 00:29:24.860 all the time, it turns out the following. 00:29:24.860 --> 00:29:28.820 That the determinant of alpha 1 up to alpha n-- 00:29:28.820 --> 00:29:31.940 first of all, it's conventional and convenient to write 00:29:31.940 --> 00:29:34.550 the determinant as if it were a matrix, only 00:29:34.550 --> 00:29:37.850 replacing the square brackets by sort of absolute value 00:29:37.850 --> 00:29:39.560 signs-- vertical lines. 00:29:39.560 --> 00:29:44.650 What we do is, notice that alpha 1 is written as an n-tuple, 00:29:44.650 --> 00:29:47.060 a11 up to a1n. 00:29:47.060 --> 00:29:49.460 In other words, the first rule of the determinant 00:29:49.460 --> 00:29:54.470 represents alpha 1 as an n-tuple vector relative to the basis 00:29:54.470 --> 00:29:55.820 u1 up to un. 00:29:55.820 --> 00:29:59.450 And a similar thing holds for alpha 2 through alpha n. 00:29:59.450 --> 00:30:01.880 And you may remember, I told you what the recipe 00:30:01.880 --> 00:30:04.442 was when we were dealing with-- 00:30:04.442 --> 00:30:05.900 I told you what the recipe was when 00:30:05.900 --> 00:30:08.300 we were dealing with cross products and the like. 00:30:08.300 --> 00:30:10.130 And the idea is simply this. 00:30:10.130 --> 00:30:12.650 What you do is is you start in the upper left-hand corner 00:30:12.650 --> 00:30:14.250 writing a plus sign. 00:30:14.250 --> 00:30:18.110 Then you alternate going along rows and columns 00:30:18.110 --> 00:30:21.170 in any order you want writing plus, minus, plus, minus, 00:30:21.170 --> 00:30:23.690 et cetera, plus, minus, plus, minus, et cetera. 00:30:23.690 --> 00:30:26.150 Pick any row or column that you want. 00:30:26.150 --> 00:30:29.090 And then you go down that row or column factoring 00:30:29.090 --> 00:30:32.960 out a particular term, the plus sign telling you to take out 00:30:32.960 --> 00:30:37.280 the term as it is, the minus sign telling you to take out 00:30:37.280 --> 00:30:40.910 the term and change its sign, and you multiply that term 00:30:40.910 --> 00:30:44.030 by the n minus 1 by n minus 1 matrix that's 00:30:44.030 --> 00:30:47.390 left when you strike out the row and column in which that term 00:30:47.390 --> 00:30:48.210 occurs. 00:30:48.210 --> 00:30:50.360 Now, if that still sounds like a tongue-twister, 00:30:50.360 --> 00:30:54.050 let's do this in terms of a specific semi-abstract, 00:30:54.050 --> 00:30:56.960 semi-concrete, four-dimensional case. 00:30:56.960 --> 00:31:00.380 Namely, let's suppose v is a four-dimensional vector space 00:31:00.380 --> 00:31:04.520 relative to a particular basis, u1, u2, u3, and u4. 00:31:04.520 --> 00:31:09.670 Suppose alpha 1, alpha 2, alpha 3, and alpha 4 00:31:09.670 --> 00:31:14.360 are these four specific vectors of v. And again-- 00:31:14.360 --> 00:31:16.190 I can't keep emphasizing this too much, 00:31:16.190 --> 00:31:18.800 even though I hope some of you are bored because you know it 00:31:18.800 --> 00:31:19.610 so well-- 00:31:19.610 --> 00:31:21.860 it's always understood when I write this 00:31:21.860 --> 00:31:25.820 that the components are relative to u1, u2, u3, u4. 00:31:25.820 --> 00:31:31.340 For example, this is 3u1 plus 5u2 plus 7u3 plus 2u4. 00:31:31.340 --> 00:31:34.670 Now, the traditional way of expanding this determinant 00:31:34.670 --> 00:31:38.640 is you write down the rows as the n-tuples. 00:31:38.640 --> 00:31:41.610 See, you write down-- this is your first row, second row, 00:31:41.610 --> 00:31:43.140 third row, fourth row. 00:31:43.140 --> 00:31:46.910 So you notice that the determinant 00:31:46.910 --> 00:31:52.520 is an n by an array of numbers that you have four vectors. 00:31:52.520 --> 00:31:55.432 Each vector is a 4-tuple, and this is 00:31:55.432 --> 00:31:56.640 how you get this determinant. 00:31:56.640 --> 00:31:57.780 Now, what do we do next? 00:31:57.780 --> 00:31:59.360 Let's say, for the sake of argument, 00:31:59.360 --> 00:32:03.560 I elect to expand this determinant along the top row. 00:32:03.560 --> 00:32:05.970 I first take the 1 out. 00:32:05.970 --> 00:32:08.540 Because it has a plus sign, that comes out as 1. 00:32:08.540 --> 00:32:11.040 I multiply that by the 3 by 3 determinant 00:32:11.040 --> 00:32:14.790 that's left when I strike out the row and column in which 1 00:32:14.790 --> 00:32:16.320 appears. 00:32:16.320 --> 00:32:19.650 I take out 2, but make it a minus, because of the signature 00:32:19.650 --> 00:32:23.160 here, and multiply that by the 3 by 3 determinant that's 00:32:23.160 --> 00:32:26.730 left when I strike out the row and column in which 2 appears. 00:32:26.730 --> 00:32:30.330 I factor out 1 as it is, and multiply that by the 3 00:32:30.330 --> 00:32:32.280 by 3 determinant that's left when 00:32:32.280 --> 00:32:36.670 I strike out the row and column in which this 1 appears. 00:32:36.670 --> 00:32:38.475 And finally, I take out this minus 1 00:32:38.475 --> 00:32:41.800 and change its sign, because of this minus code over here. 00:32:41.800 --> 00:32:44.580 In other words, this comes out as a plus 1 multiplied by the 3 00:32:44.580 --> 00:32:47.160 by 3 determinant that's left when I strike out 00:32:47.160 --> 00:32:50.130 the row and column in which minus 1 appears in. 00:32:50.130 --> 00:32:52.650 And to make a long story short, all we're saying 00:32:52.650 --> 00:32:55.800 is that this 4 by 4 determinant can 00:32:55.800 --> 00:33:00.330 be written as a sum of four 3 by 3 determinants, 00:33:00.330 --> 00:33:04.770 namely, this, which is precisely what I was 00:33:04.770 --> 00:33:06.960 saying the long way up here. 00:33:06.960 --> 00:33:10.350 Now, what we say is, each of these 3 by 3s 00:33:10.350 --> 00:33:12.960 can be written as three 2 by 2s. 00:33:12.960 --> 00:33:14.670 Namely, I can factor out-- 00:33:14.670 --> 00:33:16.530 say again, I'm going across the top row. 00:33:16.530 --> 00:33:18.150 I can pick any row or any column, 00:33:18.150 --> 00:33:20.370 but let me stick to the top row, like we did in i, j, 00:33:20.370 --> 00:33:21.900 and k, for the time being. 00:33:21.900 --> 00:33:22.950 But this becomes what? 00:33:22.950 --> 00:33:28.410 5 times this 2 by 2 matrix, minus 3 times 00:33:28.410 --> 00:33:34.330 this 2 by 2 matrix, plus 1 times this 2 by 2 matrix. 00:33:34.330 --> 00:33:36.960 In other words, this expression here. 00:33:36.960 --> 00:33:39.190 I can now do that for each of these. 00:33:39.190 --> 00:33:43.680 So each of the 3 by 3s gives me three 2 by 2s. 00:33:43.680 --> 00:33:46.710 So altogether, I would have 12 2 by 2s 00:33:46.710 --> 00:33:49.320 and go through this whole mess to compute this thing. 00:33:49.320 --> 00:33:52.290 And this can be very, very difficult, sure. 00:33:52.290 --> 00:33:54.270 This was only a four-dimensional case. 00:33:54.270 --> 00:33:56.610 Imagine trying to apply this technique 00:33:56.610 --> 00:33:59.340 for a 10-dimensional vector space, 00:33:59.340 --> 00:34:00.810 for the sake of argument. 00:34:00.810 --> 00:34:06.460 It turns out that that recipe that we called 00:34:06.460 --> 00:34:11.270 theorem two gives us a tie-in between row 00:34:11.270 --> 00:34:15.780 reduced matrices and a quick way to compute determinants. 00:34:15.780 --> 00:34:17.578 For example, let's suppose we still 00:34:17.578 --> 00:34:19.620 wanted to compute the same determinant that we've 00:34:19.620 --> 00:34:21.370 written down over here. 00:34:21.370 --> 00:34:24.556 Let's call that-- let's write it again over here. 00:34:24.556 --> 00:34:26.909 What we already know is, if I were 00:34:26.909 --> 00:34:30.989 to replace the second vector by the second minus twice 00:34:30.989 --> 00:34:33.840 the first, the determinant does not change. 00:34:33.840 --> 00:34:35.100 That's what theorem two said. 00:34:35.100 --> 00:34:37.350 If you replace one vector by itself 00:34:37.350 --> 00:34:38.790 plus a scalar multiple of another, 00:34:38.790 --> 00:34:40.590 you don't change the determinant. 00:34:40.590 --> 00:34:43.380 Similarly, if I were to replace this vector 00:34:43.380 --> 00:34:46.260 by three times the first vector, I 00:34:46.260 --> 00:34:48.239 would still have the same determinant. 00:34:48.239 --> 00:34:50.730 And finally, if I were to replace this vector 00:34:50.730 --> 00:34:52.920 by three times the first vector, I 00:34:52.920 --> 00:34:55.825 would still have the same determinant. 00:34:55.825 --> 00:34:58.200 I don't know if you've noticed what I've been driving at. 00:34:58.200 --> 00:35:01.380 But wasn't what I was saying here the same computational 00:35:01.380 --> 00:35:05.050 steps that one goes through in row reducing a matrix? 00:35:05.050 --> 00:35:08.050 In other words, let me row reduce this matrix 00:35:08.050 --> 00:35:12.720 so I get 0s every place, say, in my first column, 00:35:12.720 --> 00:35:14.130 except in the upper column. 00:35:14.130 --> 00:35:16.440 By the way, notice over here I always 00:35:16.440 --> 00:35:20.730 have the habit of putting a 1 in the upper left-hand corner. 00:35:20.730 --> 00:35:24.250 That's simply to facilitate the arithmetic. 00:35:24.250 --> 00:35:26.460 I hope by this stage of the game you realize, 00:35:26.460 --> 00:35:29.730 if this weren't a 1, I can always divide through by it, 00:35:29.730 --> 00:35:32.388 or what have you, and do the arithmetic another way. 00:35:32.388 --> 00:35:33.930 I just do this thing for convenience. 00:35:33.930 --> 00:35:37.680 But the important point is that this determinant is now 00:35:37.680 --> 00:35:39.780 equal to this determinant. 00:35:39.780 --> 00:35:41.070 And be very careful. 00:35:41.070 --> 00:35:43.430 Notice, if these had been matrices-- in other words, 00:35:43.430 --> 00:35:46.440 if I had put square brackets here and had made these 00:35:46.440 --> 00:35:47.710 matrices-- 00:35:47.710 --> 00:35:53.160 we would call the two matrices row equivalent but different 00:35:53.160 --> 00:35:54.020 matrices. 00:35:54.020 --> 00:35:56.340 Notice, the determinant is a number. 00:35:56.340 --> 00:35:59.130 And what we're saying is that the number named 00:35:59.130 --> 00:36:03.330 by this determinant is equal to the number named 00:36:03.330 --> 00:36:06.300 by this determinant-- 00:36:06.300 --> 00:36:08.190 equal. 00:36:08.190 --> 00:36:10.110 Here's the key point. 00:36:10.110 --> 00:36:13.980 I now notice that I have three 0s in the first column. 00:36:13.980 --> 00:36:17.610 I, therefore, elect to expand this determinant 00:36:17.610 --> 00:36:19.440 along the first column. 00:36:19.440 --> 00:36:20.510 Why do I do this? 00:36:20.510 --> 00:36:21.510 Well, let's take a look. 00:36:21.510 --> 00:36:24.870 First of all, I get a 1 multiplied by this 3 00:36:24.870 --> 00:36:27.630 by 3 determinant that's left when I strike out the row 00:36:27.630 --> 00:36:29.790 and column in which 1 appears. 00:36:29.790 --> 00:36:33.900 The beauty is that when I form the other three determinants, 3 00:36:33.900 --> 00:36:37.870 by 3 determinants, they're all going to be multiplied by 0. 00:36:37.870 --> 00:36:39.870 For example, over here, I factor out-- well 00:36:39.870 --> 00:36:41.910 minus 0, which is still 0. 00:36:41.910 --> 00:36:43.710 What do I multiply that by? 00:36:43.710 --> 00:36:46.240 I multiply that by the 3 by 3 determinant 00:36:46.240 --> 00:36:48.090 that's left when I strike out the row 00:36:48.090 --> 00:36:50.670 and column that this 0 appears in, namely, 00:36:50.670 --> 00:36:53.610 the 3 by 3 determinant, 2, 1, minus 1, 00:36:53.610 --> 00:36:56.490 minus 1, 4, 5, 2, 3, 4. 00:36:56.490 --> 00:36:58.980 But since that's being multiplied by 0, 00:36:58.980 --> 00:37:00.510 the product will be 0. 00:37:00.510 --> 00:37:04.620 To make a long story short, to evaluate this determinant, 00:37:04.620 --> 00:37:05.930 it's simply what? 00:37:05.930 --> 00:37:09.920 Plus 1 times this 3 by 3 determinant. 00:37:09.920 --> 00:37:13.190 I put the 1 in parentheses here to indicate that I really 00:37:13.190 --> 00:37:14.570 factored that 1 out. 00:37:14.570 --> 00:37:18.470 But really what we're saying is that this 4 by 4 determinant 00:37:18.470 --> 00:37:21.860 has the same value as this 3 by 3 determinant. 00:37:21.860 --> 00:37:24.860 And now, I row reduce this 3 by 3 determinant. 00:37:24.860 --> 00:37:29.000 I replace the second row, in other words, the second vector, 00:37:29.000 --> 00:37:32.630 by the second plus the first, the third by the third minus 00:37:32.630 --> 00:37:33.800 twice the first. 00:37:33.800 --> 00:37:37.010 I now wind up with this 3 by 3 determinant. 00:37:37.010 --> 00:37:39.590 Since I don't change the value of the determinant 00:37:39.590 --> 00:37:41.540 when I replace one vector by itself 00:37:41.540 --> 00:37:43.430 plus a scalar multiple of another, 00:37:43.430 --> 00:37:46.550 then notice very quickly that what happens over here 00:37:46.550 --> 00:37:49.100 is that these two determinants are equal. 00:37:49.100 --> 00:37:54.830 Again, I elect to expand along the first column factoring out 00:37:54.830 --> 00:37:59.780 the plus 1 being left with the determinant whose entries are 00:37:59.780 --> 00:38:01.730 5, 8, 1, 1. 00:38:01.730 --> 00:38:04.610 And the other two terms contribute nothing, 00:38:04.610 --> 00:38:07.690 because the coefficient is 0. 00:38:07.690 --> 00:38:09.770 But I already know how to expand-- 00:38:09.770 --> 00:38:13.422 I've proven that-- how to expand the 2 by 2 matrix, determinant. 00:38:13.422 --> 00:38:14.630 That's just going to be what? 00:38:14.630 --> 00:38:17.630 5 minus 8 or minus 3. 00:38:17.630 --> 00:38:20.240 And that's how this shortcut row reduction works-- 00:38:20.240 --> 00:38:24.410 much, much more elegantly than the brute force technique. 00:38:24.410 --> 00:38:26.510 And by the way, this is the technique 00:38:26.510 --> 00:38:28.940 used in most, if not all, computers 00:38:28.940 --> 00:38:33.050 in determining the determinant of a set of n vectors 00:38:33.050 --> 00:38:35.240 in an n-dimensional vector space. 00:38:35.240 --> 00:38:38.040 Well, as I say, this was meant as an overview. 00:38:38.040 --> 00:38:40.880 I hope you now see the overall picture of what determinants 00:38:40.880 --> 00:38:41.870 are all about. 00:38:41.870 --> 00:38:44.660 And in the exercises, I will try to give you 00:38:44.660 --> 00:38:48.560 some more sophisticated and elaborate details. 00:38:48.560 --> 00:38:51.410 Next time what we will do is talk 00:38:51.410 --> 00:38:54.710 about an application of how determinants 00:38:54.710 --> 00:38:58.477 are used in certain aspects of vector spaces. 00:38:58.477 --> 00:39:00.560 In particular, we're going to talk about something 00:39:00.560 --> 00:39:04.220 called eigenvalues or eigenvectors, but more 00:39:04.220 --> 00:39:05.780 about that next time. 00:39:05.780 --> 00:39:09.790 Until next time, goodbye. 00:39:09.790 --> 00:39:12.190 Funding for the publication of this video 00:39:12.190 --> 00:39:17.050 was provided by the Gabriella and Paul Rosenbaum Foundation. 00:39:17.050 --> 00:39:21.220 Help OCW continue to provide free and open access to MIT 00:39:21.220 --> 00:39:26.655 courses by making a donation at ocw.mit.edu/donate.