1 00:00:05,760 --> 00:00:06,820 ANA RITA PIRES: Hi there. 2 00:00:06,820 --> 00:00:08,510 Welcome to recitation. 3 00:00:08,510 --> 00:00:10,990 In lecture, you've been learning about vector 4 00:00:10,990 --> 00:00:15,040 spaces whose vectors are actually matrices or functions, 5 00:00:15,040 --> 00:00:17,640 and this is what our problem today is about. 6 00:00:17,640 --> 00:00:22,020 We have a set of 2 by 3 matrices whose null space 7 00:00:22,020 --> 00:00:24,250 contains the vector [2, 1, 1]. 8 00:00:24,250 --> 00:00:27,610 And I want you to show that this set is actually 9 00:00:27,610 --> 00:00:32,970 a vector subspace of the space of all 2 by 3 matrices. 10 00:00:32,970 --> 00:00:36,590 And then, I want you to find a basis for it. 11 00:00:36,590 --> 00:00:39,150 When you're done, here is an additional question. 12 00:00:39,150 --> 00:00:44,500 What about the set of those 2 by 3 matrices whose column space 13 00:00:44,500 --> 00:00:47,760 contains the vector [2, 1]? 14 00:00:47,760 --> 00:00:48,430 All right. 15 00:00:48,430 --> 00:00:50,670 Hit pause and work on it yourself, 16 00:00:50,670 --> 00:00:53,824 and when you're ready, I'll come back and show you how I did it. 17 00:01:01,390 --> 00:01:01,890 Hi. 18 00:01:01,890 --> 00:01:05,090 I hope you managed to solve it. 19 00:01:05,090 --> 00:01:07,070 Let's do it. 20 00:01:07,070 --> 00:01:11,360 So, how do we show that something is a vector subspace? 21 00:01:11,360 --> 00:01:13,920 Well, there are only two things that we need to check. 22 00:01:13,920 --> 00:01:18,420 One is that if two vectors, in this case two matrices, 23 00:01:18,420 --> 00:01:21,670 are in that space, then their sum is in that space. 24 00:01:21,670 --> 00:01:24,640 And if you take a vector, in this case a matrix, 25 00:01:24,640 --> 00:01:28,440 and you multiply by a scalar you'll still be in the space. 26 00:01:28,440 --> 00:01:37,230 So, suppose that the matrices A and B are in this set 27 00:01:37,230 --> 00:01:39,590 that we want to prove is a subspace. 28 00:01:39,590 --> 00:01:46,570 So that means that A times the vector [2, 1, 1] 29 00:01:46,570 --> 00:01:48,710 is equal to the vector [0, 0]. 30 00:01:48,710 --> 00:01:52,050 Notice that the dimensions are right: A is 2 by 3, 31 00:01:52,050 --> 00:01:53,260 so this 3 by 1. 32 00:01:53,260 --> 00:01:55,470 I should get a 2 by 1. 33 00:01:55,470 --> 00:01:59,170 Suppose that [2, 1, 1] is in the null space of A, 34 00:01:59,170 --> 00:02:07,310 and that [2, 1, 1] is also in the null space of B. Then what 35 00:02:07,310 --> 00:02:15,190 is A plus B times [2, 1, 1]? 36 00:02:15,190 --> 00:02:18,480 Is [2, 1, 1] in the null space of A plus B? 37 00:02:18,480 --> 00:02:20,680 Well, if you think about what this means, 38 00:02:20,680 --> 00:02:23,000 you're just adding entry by entry. 39 00:02:23,000 --> 00:02:27,710 And you can do it slowly on your own 40 00:02:27,710 --> 00:02:35,900 and just check that this is what happens. 41 00:02:35,900 --> 00:02:38,530 But by now you are familiar enough with matrices 42 00:02:38,530 --> 00:02:40,810 that this should not be a surprise. 43 00:02:40,810 --> 00:02:43,110 Well, this is [0, 0], and this is [0, 0], 44 00:02:43,110 --> 00:02:46,220 so their sum is [0, 0]. 45 00:02:46,220 --> 00:02:50,230 So, indeed, [2, 1, 1] is in the null space of A plus B. 46 00:02:50,230 --> 00:02:56,260 So if A and B are in the set, A plus B is in the set, as well. 47 00:02:56,260 --> 00:02:58,040 Let's check the other thing. 48 00:02:58,040 --> 00:03:03,340 The other thing is A times [2, 1, 1] is [0, 0]. 49 00:03:03,340 --> 00:03:06,050 So A is in the set, because [2, 1, 1] 50 00:03:06,050 --> 00:03:10,675 is in the null space of A. And also, let's let c be a scalar. 51 00:03:13,550 --> 00:03:16,690 That just means that c is a number. 52 00:03:16,690 --> 00:03:21,670 Then we want to check that [2, 1, 53 00:03:21,670 --> 00:03:24,840 1] is in the null space of the matrix c*A. 54 00:03:24,840 --> 00:03:27,950 That matrix is just the matrix A except every entry is 55 00:03:27,950 --> 00:03:30,360 multiplied by the number c. 56 00:03:30,360 --> 00:03:35,790 Well, again, this is how matrices work. 57 00:03:35,790 --> 00:03:37,770 You can just pull out the constant 58 00:03:37,770 --> 00:03:40,400 and do A times [2, 1, 1] first. 59 00:03:40,400 --> 00:03:43,240 Now this is the vector [0, 0]. 60 00:03:43,240 --> 00:03:48,270 So, this will simply be c times [0, 0], which is [0, 0]. 61 00:03:48,270 --> 00:03:53,370 So the matrix c*A is also contained in this set. 62 00:03:53,370 --> 00:03:55,760 So the set is closed under addition 63 00:03:55,760 --> 00:03:58,040 and under multiplication by scalar, 64 00:03:58,040 --> 00:04:01,206 so the set is, indeed, a vector subspace. 65 00:04:01,206 --> 00:04:05,200 Well that takes care of the first part of the question. 66 00:04:05,200 --> 00:04:08,880 The second part was: find a basis for the subspace. 67 00:04:08,880 --> 00:04:12,710 So let's work on that now. 68 00:04:12,710 --> 00:04:16,110 So the condition for a matrix to be in this subspace 69 00:04:16,110 --> 00:04:20,760 is that the vector [2, 1, 1] is in the null space. 70 00:04:20,760 --> 00:04:27,130 So I must have A times [2, 1, 1] is equal to [0, 0]. 71 00:04:27,130 --> 00:04:28,210 So what is happening? 72 00:04:28,210 --> 00:04:31,650 Well A is a 2 by 3 matrix. 73 00:04:31,650 --> 00:04:34,410 So you can actually think about what happens in each row 74 00:04:34,410 --> 00:04:35,520 separately. 75 00:04:35,520 --> 00:04:40,450 You'll have the first row of A times [2, 1, 1] is equal to 0. 76 00:04:40,450 --> 00:04:44,397 And the second row of A times [2, 1, 1] is equal to 0. 77 00:04:44,397 --> 00:04:45,605 So let's see what that means. 78 00:04:49,230 --> 00:05:01,660 Each row of the matrix A in this vector subspace must be [a, b, 79 00:05:01,660 --> 00:05:06,410 c] [2; 1; 1] equal to 0. 80 00:05:06,410 --> 00:05:10,220 This is not really a good sentence, but you understand. 81 00:05:10,220 --> 00:05:15,030 Which means that-- well let's see-- 2a 82 00:05:15,030 --> 00:05:19,460 plus b plus c is equal to 0. 83 00:05:19,460 --> 00:05:21,565 So I can actually write it in this format. 84 00:05:25,320 --> 00:05:34,810 It must be of the form a, b, and then 85 00:05:34,810 --> 00:05:42,380 c must be equal to -2a minus b. 86 00:05:42,380 --> 00:05:43,810 Right? 87 00:05:43,810 --> 00:05:51,000 So, furthermore, we can say that it-- well, 88 00:05:51,000 --> 00:05:54,570 one thing that you can do-- let me write this here to help you. 89 00:05:54,570 --> 00:06:02,920 It will be [a, 0, -2a] plus [0, b, -bl. 90 00:06:02,920 --> 00:06:03,420 See? 91 00:06:03,420 --> 00:06:06,080 So what I'm doing is I'm splitting this 92 00:06:06,080 --> 00:06:09,220 into the linear combination of two vectors. 93 00:06:09,220 --> 00:06:11,750 I can pull out the a out of this one, 94 00:06:11,750 --> 00:06:16,340 and pull the b out of this one, and it 95 00:06:16,340 --> 00:06:20,660 must be a linear combination-- that's 96 00:06:20,660 --> 00:06:24,570 what this means, linear combination-- of the following: 97 00:06:24,570 --> 00:06:31,460 [1, 0, -2] and [0, 1, -1]. 98 00:06:31,460 --> 00:06:33,000 Does that make sense? 99 00:06:33,000 --> 00:06:36,820 So this is what each row of a must satisfy. 100 00:06:36,820 --> 00:06:39,710 So we can now put everything together 101 00:06:39,710 --> 00:06:43,720 into what a basis for the vector space has to be. 102 00:06:43,720 --> 00:06:50,650 The basis will be, well it's 2 by 3 matrix, 103 00:06:50,650 --> 00:06:53,650 and each row must be a linear combination of these two 104 00:06:53,650 --> 00:06:54,820 vectors. 105 00:06:54,820 --> 00:06:59,860 So let's write that-- 1, 0, -2. 106 00:06:59,860 --> 00:07:04,740 I'll keep the second row with zeros-- 0, 1, -1, 107 00:07:04,740 --> 00:07:06,865 and I'll keep the second row of zeros. 108 00:07:06,865 --> 00:07:12,560 And now the same, but keeping the first row with zeros, 109 00:07:12,560 --> 00:07:19,710 I'm taking these vectors on the second row. 110 00:07:19,710 --> 00:07:23,640 So this is a basis for my vector space. 111 00:07:23,640 --> 00:07:26,220 One, two, three, four; that also means 112 00:07:26,220 --> 00:07:30,560 that the dimension of the subspace is 4. 113 00:07:30,560 --> 00:07:32,760 There was one further question, which 114 00:07:32,760 --> 00:07:38,600 was what can you say about the set of those matrices that 115 00:07:38,600 --> 00:07:41,430 contain the vector [2, 1] in their column space? 116 00:07:44,610 --> 00:07:48,180 What about the set of those 2 by 3 matrices whose column space 117 00:07:48,180 --> 00:07:50,147 contains the vector [2, 1]. 118 00:07:50,147 --> 00:07:51,230 Is that a vector subspace? 119 00:07:54,600 --> 00:07:58,020 Well one quick check that you can always do 120 00:07:58,020 --> 00:08:03,070 is check that the zero vector, in this case the zero matrix, 121 00:08:03,070 --> 00:08:05,390 belongs to the set. 122 00:08:05,390 --> 00:08:13,180 Does the zero 2 by 3 matrix-- [0, 0, 0; 0, 0, 0]. 123 00:08:13,180 --> 00:08:15,800 Does this matrix belong to this set? 124 00:08:15,800 --> 00:08:18,840 Does this matrix contain the vector [2, 1] 125 00:08:18,840 --> 00:08:20,540 in its column space? 126 00:08:20,540 --> 00:08:25,490 It does not, so this set cannot be a vector subspace. 127 00:08:25,490 --> 00:08:29,210 If you want to think about how this 0 belonging in the set 128 00:08:29,210 --> 00:08:32,570 has anything to do with the two conditions being closed 129 00:08:32,570 --> 00:08:35,770 under sum and closed under multiplication by scalar, 130 00:08:35,770 --> 00:08:38,179 simply think that you should always 131 00:08:38,179 --> 00:08:41,590 be able to multiply a matrix by the scalar 0 132 00:08:41,590 --> 00:08:44,210 and have it still be in the set. 133 00:08:44,210 --> 00:08:46,040 That will be your zero matrix. 134 00:08:46,040 --> 00:08:49,250 Well the zero matrix is not in the set, 135 00:08:49,250 --> 00:08:51,722 so the set is not a vector subspace. 136 00:08:51,722 --> 00:08:52,430 All right? 137 00:08:52,430 --> 00:08:53,310 We're done. 138 00:08:53,310 --> 00:08:54,881 Thank you.