1 00:00:00,000 --> 00:00:07,810 2 00:00:07,810 --> 00:00:08,650 DAVID SHIROKOFF: Hi everyone. 3 00:00:08,650 --> 00:00:10,060 I'm Dave. 4 00:00:10,060 --> 00:00:11,840 Now today, I'd like to tackle a problem 5 00:00:11,840 --> 00:00:13,990 in orthogonal subspaces. 6 00:00:13,990 --> 00:00:18,260 So the problem we'd like to tackle, given a subspace S, 7 00:00:18,260 --> 00:00:22,400 and suppose S is spanned by two vectors, 1, 2, 2, 3, and 8 00:00:22,400 --> 00:00:24,530 1, 3, 3, 2. 9 00:00:24,530 --> 00:00:26,920 We have a question here which is to find 10 00:00:26,920 --> 00:00:28,630 a basis for S perp-- 11 00:00:28,630 --> 00:00:32,540 S perp is another subspace which is orthogonal to S. And 12 00:00:32,540 --> 00:00:35,870 then secondly, can every vector in R4 be uniquely 13 00:00:35,870 --> 00:00:39,260 written in terms of S and S perp. 14 00:00:39,260 --> 00:00:41,800 So I'll let you think about this for now, and I'll come 15 00:00:41,800 --> 00:00:43,050 back in a minute. 16 00:00:43,050 --> 00:00:53,419 17 00:00:53,419 --> 00:00:54,620 Hi everyone. 18 00:00:54,620 --> 00:00:56,060 Welcome back. 19 00:00:56,060 --> 00:00:58,180 OK, so why don't we tackle this problem? 20 00:00:58,180 --> 00:01:02,880 21 00:01:02,880 --> 00:01:04,650 OK, so first off, what does it mean for a 22 00:01:04,650 --> 00:01:06,760 vector to be an S perp? 23 00:01:06,760 --> 00:01:18,470 Well if I have a vector x, and S perp, and x is in S perp, 24 00:01:18,470 --> 00:01:22,530 what this means is x is going to be orthogonal to every 25 00:01:22,530 --> 00:01:25,870 vector in S. Now specifically, S is spanned 26 00:01:25,870 --> 00:01:27,640 by these two vectors. 27 00:01:27,640 --> 00:01:31,660 So it's sufficient that x be perpendicular to the two bases 28 00:01:31,660 --> 00:01:33,250 vectors in S. 29 00:01:33,250 --> 00:01:45,720 So specifically, I can take 1, 2, 2, 3, and dot it with x, 30 00:01:45,720 --> 00:01:47,230 and it's going to be 0. 31 00:01:47,230 --> 00:01:52,210 So I'm treating x as a column vector here. 32 00:01:52,210 --> 00:01:59,340 In addition, x must also be orthogonal to 1, 3, 2, 2. 33 00:01:59,340 --> 00:02:03,600 34 00:02:03,600 --> 00:02:06,570 So any vector x that's an S perp must be orthogonal to 35 00:02:06,570 --> 00:02:08,750 both of these vectors. 36 00:02:08,750 --> 00:02:10,280 So what we can do is we can write 37 00:02:10,280 --> 00:02:11,680 this as a matrix equation. 38 00:02:11,680 --> 00:02:16,140 39 00:02:16,140 --> 00:02:19,290 And we do this by combining these two vectors as rows of 40 00:02:19,290 --> 00:02:20,540 the matrix. 41 00:02:20,540 --> 00:02:31,470 42 00:02:31,470 --> 00:02:34,070 So if we step back and take a look at this equation, we see 43 00:02:34,070 --> 00:02:37,700 that what we're really asking is to find all x that are in 44 00:02:37,700 --> 00:02:40,410 the null space of this matrix. 45 00:02:40,410 --> 00:02:42,990 So how do we find x in the null space of a matrix? 46 00:02:42,990 --> 00:02:46,740 Well what we can do is we can row reduce this matrix and try 47 00:02:46,740 --> 00:02:49,270 and find a basis for the null space. 48 00:02:49,270 --> 00:02:52,140 So I'm going to just row reduce this matrix, and notice 49 00:02:52,140 --> 00:02:54,380 that by row reduction, we don't actually change the null 50 00:02:54,380 --> 00:02:56,850 space of a matrix. 51 00:02:56,850 --> 00:03:01,390 So if I'm only interested in the null space, this system is 52 00:03:01,390 --> 00:03:04,530 going to be equivalent to, I can keep the top row the same. 53 00:03:04,530 --> 00:03:09,030 54 00:03:09,030 --> 00:03:12,560 And then just to simplify our lives, we can take the second 55 00:03:12,560 --> 00:03:14,680 row and subtract one copy of the first row. 56 00:03:14,680 --> 00:03:18,590 57 00:03:18,590 --> 00:03:23,080 Now, if I do that, I obtain 0, 1, 1, -1. 58 00:03:23,080 --> 00:03:29,450 59 00:03:29,450 --> 00:03:33,310 Now, to parameterize the null space, what I'm going to do is 60 00:03:33,310 --> 00:03:36,640 I'm going to write x out as components. 61 00:03:36,640 --> 00:03:48,550 So if I write x with components x1, x2, x3 and x4, 62 00:03:48,550 --> 00:03:52,790 we see here that this matrix has a rank of 2. 63 00:03:52,790 --> 00:03:57,570 Now, we're looking at vectors which live in R4, so we know 64 00:03:57,570 --> 00:04:00,610 that the null space is going to have a dimension 65 00:04:00,610 --> 00:04:02,620 which is 4 minus 2. 66 00:04:02,620 --> 00:04:04,820 So that means there should be two vectors in the null space 67 00:04:04,820 --> 00:04:07,610 of this matrix. 68 00:04:07,610 --> 00:04:10,440 To parameterize these two dimensional vectors, what I'm 69 00:04:10,440 --> 00:04:14,040 going to do is I'm going to let x4 equals some constant, 70 00:04:14,040 --> 00:04:17,250 and x3 equal another constant. 71 00:04:17,250 --> 00:04:23,530 So specifically, I'm going to let x4 equal b, 72 00:04:23,530 --> 00:04:25,838 and x3 equal a. 73 00:04:25,838 --> 00:04:29,190 74 00:04:29,190 --> 00:04:33,160 Now what we do is we take a look at these two equations, 75 00:04:33,160 --> 00:04:39,770 and this bottom equation will say that x2 is equal to 76 00:04:39,770 --> 00:04:51,230 negative x3 plus x4, which is going to equal -a, x4 plus b. 77 00:04:51,230 --> 00:04:55,090 78 00:04:55,090 --> 00:05:01,500 And then the top equation says that x1 is equal to -2, x2 79 00:05:01,500 --> 00:05:16,680 minus 2x3 minus 3x4, And if I substitute in, 80 00:05:16,680 --> 00:05:18,252 x2 is -a plus b. 81 00:05:18,252 --> 00:05:22,500 82 00:05:22,500 --> 00:05:25,608 x3 is a. 83 00:05:25,608 --> 00:05:26,858 And x4 is b. 84 00:05:26,858 --> 00:05:30,020 85 00:05:30,020 --> 00:05:32,380 So when the dust settles, the a's cancel and 86 00:05:32,380 --> 00:05:33,930 I'm left with -5b. 87 00:05:33,930 --> 00:05:38,030 88 00:05:38,030 --> 00:05:42,430 So we can combine everything together and we end up 89 00:05:42,430 --> 00:05:56,290 obtaining x1, x2, x3, x4 equals -5b. 90 00:05:56,290 --> 00:05:58,566 x2 is -a plus b. 91 00:05:58,566 --> 00:06:03,692 92 00:06:03,692 --> 00:06:10,130 x3 is a, and x4 is b. 93 00:06:10,130 --> 00:06:11,980 And now what we can do is we can take this vector and we 94 00:06:11,980 --> 00:06:15,840 can decompose it into pieces which are a multiplied by a 95 00:06:15,840 --> 00:06:19,630 vector, and b multiplied by a vector. 96 00:06:19,630 --> 00:06:29,610 So you'll note that this is actually a times 0, -1, 1, 0 97 00:06:29,610 --> 00:06:40,110 plus b times -5, 1, 0, 1. 98 00:06:40,110 --> 00:06:42,100 OK? 99 00:06:42,100 --> 00:06:45,390 So we have successfully achieved a parameterization of 100 00:06:45,390 --> 00:06:49,600 the null space of this matrix as some constant a times a 101 00:06:49,600 --> 00:06:56,700 vector 0, -1, 1, 0 plus b times a vector -5, 1, 0, 1. 102 00:06:56,700 --> 00:07:00,170 And now we claim that this is the entire space, S perp. 103 00:07:00,170 --> 00:07:04,190 104 00:07:04,190 --> 00:07:07,740 So S perp is going to be spanned by this vector and 105 00:07:07,740 --> 00:07:10,060 this vector. 106 00:07:10,060 --> 00:07:14,870 Now notice how if I were to take either of these two 107 00:07:14,870 --> 00:07:20,230 vectors in S and dot it with any vector in the null space, 108 00:07:20,230 --> 00:07:22,425 by construction it automatically vanishes. 109 00:07:22,425 --> 00:07:25,190 110 00:07:25,190 --> 00:07:28,190 So this concludes part one. 111 00:07:28,190 --> 00:07:30,330 Now for part two. 112 00:07:30,330 --> 00:07:33,510 Can every vector v in R4 be written uniquely in terms of S 113 00:07:33,510 --> 00:07:35,020 and S perp? 114 00:07:35,020 --> 00:07:36,270 The answer is yes. 115 00:07:36,270 --> 00:07:42,440 116 00:07:42,440 --> 00:07:44,460 So how do we see this? 117 00:07:44,460 --> 00:07:49,300 Well if I have a vector v, what I can do is I can try and 118 00:07:49,300 --> 00:07:58,080 write it as some constant c1 times the vector 1, 2, 2, 3 119 00:07:58,080 --> 00:08:15,700 plus c2 times the vector 1, 3, 3, 2 plus the vector c3, 0, 120 00:08:15,700 --> 00:08:25,810 -1, 1, 0 plus c4, -5, 1, 0, 1. 121 00:08:25,810 --> 00:08:26,870 OK? 122 00:08:26,870 --> 00:08:32,190 So c1 and c2 are multiplying the vectors in S, and c3 and 123 00:08:32,190 --> 00:08:35,450 c4 are multiplying the vectors in S perp. 124 00:08:35,450 --> 00:08:39,760 So the question is, given any v, can I find constants c1, 125 00:08:39,760 --> 00:08:44,370 c2, c3, c4, such that this equation holds? 126 00:08:44,370 --> 00:08:45,620 And the answer is yes. 127 00:08:45,620 --> 00:08:48,680 128 00:08:48,680 --> 00:08:51,950 Just to see why it's yes, what we can do is we can rewrite 129 00:08:51,950 --> 00:08:56,175 this in matrix notation, and there's kind of a handy trick. 130 00:08:56,175 --> 00:09:13,860 131 00:09:13,860 --> 00:09:17,410 What I can do is I can take these columns and write them 132 00:09:17,410 --> 00:09:18,660 as columns of the matrix. 133 00:09:18,660 --> 00:09:22,640 134 00:09:22,640 --> 00:09:27,560 And this whole expression is actually equivalent to this 135 00:09:27,560 --> 00:09:32,100 matrix multiplied by the constant, c1, c2, c3, c4. 136 00:09:32,100 --> 00:09:36,760 And on the right-hand side, we have the vector v. 137 00:09:36,760 --> 00:09:40,510 Now, by construction, these vectors are linearly 138 00:09:40,510 --> 00:09:41,540 independent. 139 00:09:41,540 --> 00:09:43,520 And we know from linear algebra that if we have a 140 00:09:43,520 --> 00:09:45,990 matrix with linearly independent columns, the 141 00:09:45,990 --> 00:09:48,280 matrix is invertible. 142 00:09:48,280 --> 00:09:51,650 What this means is for any v on the right-hand side, we can 143 00:09:51,650 --> 00:09:54,080 invert this matrix and obtain unique 144 00:09:54,080 --> 00:09:57,590 coefficients, c1, c2, c3, c4. 145 00:09:57,590 --> 00:10:01,220 This then gives us a unique decomposition for v in terms 146 00:10:01,220 --> 00:10:06,690 of a piece which is in S, and the piece which is in S perp. 147 00:10:06,690 --> 00:10:10,880 148 00:10:10,880 --> 00:10:14,630 And in general this can be done for any vector space. 149 00:10:14,630 --> 00:10:18,060 Well I'd like to conclude this problem now and I hope you had 150 00:10:18,060 --> 00:10:19,310 a good time. 151 00:10:19,310 --> 00:10:19,860