1
00:00:07,065 --> 00:00:07,920
PROFESSOR: Hi there.
2
00:00:07,920 --> 00:00:09,540
Welcome back to recitation.
3
00:00:09,540 --> 00:00:12,870
In lecture, you've been learning
about when vectors are linearly
4
00:00:12,870 --> 00:00:17,550
independent, when they span
the space, what a basis is,
5
00:00:17,550 --> 00:00:19,790
what a dimension of
a vector space is.
6
00:00:19,790 --> 00:00:23,630
And the problem for today
is exactly about that.
7
00:00:23,630 --> 00:00:26,500
We have a vector
space that is given.
8
00:00:26,500 --> 00:00:28,780
It's spanned by
these four vectors.
9
00:00:28,780 --> 00:00:32,009
And we're asked to find the
dimension of that vector space
10
00:00:32,009 --> 00:00:33,190
and the basis for it.
11
00:00:35,700 --> 00:00:39,270
Well, why don't you hit pause
on the video, and work on it
12
00:00:39,270 --> 00:00:39,770
for a while.
13
00:00:39,770 --> 00:00:42,510
And I'll come back in a little
bit to help you out with it.
14
00:00:51,120 --> 00:00:51,900
We're back.
15
00:00:51,900 --> 00:00:54,900
Let's work on it.
16
00:00:54,900 --> 00:01:00,310
So we need to find the
dimension and the basis.
17
00:01:00,310 --> 00:01:01,970
Remember what the dimension is?
18
00:01:01,970 --> 00:01:03,570
It's simply the
number of vectors
19
00:01:03,570 --> 00:01:05,590
in a basis for the vector space.
20
00:01:05,590 --> 00:01:07,680
So actually, the
problem is backwards.
21
00:01:07,680 --> 00:01:09,460
We want to find the
basis for the space
22
00:01:09,460 --> 00:01:11,840
first, and then
find the dimension.
23
00:01:11,840 --> 00:01:19,700
I'll write "first"
here and "second" here.
24
00:01:19,700 --> 00:01:22,270
So we want to find a basis
for the vector space spanned
25
00:01:22,270 --> 00:01:24,040
by these four vectors.
26
00:01:24,040 --> 00:01:26,040
So you might be
tempted to just say
27
00:01:26,040 --> 00:01:27,680
that a basis for
this vector space
28
00:01:27,680 --> 00:01:31,410
is those four vectors, because
they span the vector space.
29
00:01:31,410 --> 00:01:34,350
But there's another thing
that a basis has to satisfy.
30
00:01:34,350 --> 00:01:36,920
And it is the elements
of the basis have
31
00:01:36,920 --> 00:01:38,760
to be linearly independent.
32
00:01:38,760 --> 00:01:40,340
We don't know that these are.
33
00:01:40,340 --> 00:01:42,636
So we have to check.
34
00:01:42,636 --> 00:01:44,010
How do we check
that four vectors
35
00:01:44,010 --> 00:01:45,670
are linearly independent?
36
00:01:45,670 --> 00:01:47,800
Well, there's a couple
of different ways.
37
00:01:47,800 --> 00:01:50,610
But here's what
we're going to do.
38
00:01:50,610 --> 00:01:54,340
We're going to put these
vectors as rows of a matrix.
39
00:01:54,340 --> 00:01:57,070
And then we'll do elimination.
40
00:01:57,070 --> 00:02:00,710
And when we get to the end,
the rows that have pivots
41
00:02:00,710 --> 00:02:02,570
are the independent ones.
42
00:02:02,570 --> 00:02:04,220
So let's do that.
43
00:02:06,760 --> 00:02:25,740
1, 1, -2, 0, -1; 1, 2, 0, -4, 1;
0, 1, 3, -3, 2; 2, 3, 0, -2, 0.
44
00:02:30,480 --> 00:02:32,730
By now you must have done
elimination a million times,
45
00:02:32,730 --> 00:02:36,310
so I'll go a little bit faster.
46
00:02:36,310 --> 00:02:48,864
1, 1, -2, 0, -1; 0,
1, 0, +2, -4, 2--
47
00:02:48,864 --> 00:02:56,670
this one's already done-- 1,
3, -3, 2; 2 minus 2, 3 minus 2,
48
00:02:56,670 --> 00:03:02,170
0 plus 4, -2, and 2.
49
00:03:05,457 --> 00:03:06,040
One more step.
50
00:03:08,740 --> 00:03:21,146
1, 1, -2, 0, -1-- all these
are done-- 0, 1, 2, -4, 2; 0,
51
00:03:21,146 --> 00:03:30,491
1 minus 1 is 0, 3 minus 2 is 1,
-3 plus 4 is 1, 2 minus 2 is 0,
52
00:03:30,491 --> 00:03:37,402
1 minus 1 is 0,
2, 2 again, and 0.
53
00:03:37,402 --> 00:03:39,280
Well, you can see
where this is going.
54
00:03:39,280 --> 00:03:41,760
In the next step, this
row is going to disappear.
55
00:03:44,748 --> 00:03:59,120
1, 1, -2, 0 -1; 0, 1, 2, -4, 2;
0, 0, 1, 1, 0; 0, 0, 0, 0, 0.
56
00:03:59,120 --> 00:03:59,954
All right.
57
00:03:59,954 --> 00:04:01,120
We're done with elimination.
58
00:04:03,890 --> 00:04:05,300
So let's circle our pivots.
59
00:04:08,934 --> 00:04:10,470
All right, here are our pivots.
60
00:04:10,470 --> 00:04:14,040
We have three pivots.
61
00:04:14,040 --> 00:04:17,340
And so these three rows
are linearly independent.
62
00:04:19,950 --> 00:04:24,140
And in fact, these rows
still span the same space
63
00:04:24,140 --> 00:04:25,801
that the initial four rows did.
64
00:04:25,801 --> 00:04:28,050
Because when you do elimination,
all that you're doing
65
00:04:28,050 --> 00:04:31,380
is recombining
your rows by doing
66
00:04:31,380 --> 00:04:32,560
linear combinations of them.
67
00:04:32,560 --> 00:04:36,310
So, for example, your first
row stayed the same throughout.
68
00:04:36,310 --> 00:04:41,850
Your second row was replaced
by row 2 minus row 1.
69
00:04:41,850 --> 00:04:45,310
But it's really still
spanning the same space.
70
00:04:45,310 --> 00:04:46,330
And it goes on.
71
00:04:46,330 --> 00:04:48,640
And then the fourth row,
it turns out, was useless.
72
00:04:48,640 --> 00:04:50,590
You only needed the first three.
73
00:04:50,590 --> 00:04:57,295
So the elements for a basis--
well, it will be these three.
74
00:04:57,295 --> 00:04:58,170
So let me write that.
75
00:05:01,170 --> 00:05:18,640
Basis [1, 1, -2, 0, -1],
2, 2, and [0, 0, 1, 1, 0].
76
00:05:18,640 --> 00:05:22,440
Could you have used
the first three rows?
77
00:05:22,440 --> 00:05:24,230
Yes, you could.
78
00:05:24,230 --> 00:05:25,640
You can't always do that.
79
00:05:25,640 --> 00:05:28,390
Sometimes in elimination,
you have to switch rows,
80
00:05:28,390 --> 00:05:30,750
because there's a 0
where a pivot should be.
81
00:05:30,750 --> 00:05:33,380
When that happens, you
have to use these three,
82
00:05:33,380 --> 00:05:35,310
or you have to keep
track of which row
83
00:05:35,310 --> 00:05:39,200
you switched to go back
and use the initial ones.
84
00:05:39,200 --> 00:05:41,110
But it's really safe
to use these ones.
85
00:05:41,110 --> 00:05:43,880
And also they're simpler than
the ones that you started with,
86
00:05:43,880 --> 00:05:45,820
so why not?
87
00:05:45,820 --> 00:05:47,840
The other question
that we had was:
88
00:05:47,840 --> 00:05:49,740
what is the dimension
of the vector space?
89
00:05:49,740 --> 00:05:52,780
Well, this is the easy part.
90
00:05:52,780 --> 00:05:58,100
The dimension of the
vector space is 1, 2, 3.
91
00:05:58,100 --> 00:05:59,620
And that solves the problem.
92
00:05:59,620 --> 00:06:03,530
But there's one more thing
that I want to tell you.
93
00:06:03,530 --> 00:06:05,210
I told you that there's
a couple of ways
94
00:06:05,210 --> 00:06:08,950
to find out which of
those four vectors
95
00:06:08,950 --> 00:06:10,480
are linearly independent.
96
00:06:10,480 --> 00:06:12,990
And the one that I used
was I put them into rows
97
00:06:12,990 --> 00:06:16,920
and performed elimination
and picked out the rows
98
00:06:16,920 --> 00:06:19,050
that have pivots on them.
99
00:06:19,050 --> 00:06:23,380
Another way to do it is to
write the initial vectors
100
00:06:23,380 --> 00:06:26,840
as columns of the matrix and
then perform elimination.
101
00:06:26,840 --> 00:06:28,880
That also works
and, as you know,
102
00:06:28,880 --> 00:06:31,870
because you're only working with
the transpose of this matrix,
103
00:06:31,870 --> 00:06:34,990
you get the same
number of pivots.
104
00:06:34,990 --> 00:06:40,250
Let's go over here where I have
the same-- well, the transpose
105
00:06:40,250 --> 00:06:42,260
of that matrix-- I
have the same vectors,
106
00:06:42,260 --> 00:06:44,100
but now written as columns.
107
00:06:44,100 --> 00:06:46,740
My four initial vectors
written as columns.
108
00:06:46,740 --> 00:06:49,440
And here I have
performed elimination.
109
00:06:49,440 --> 00:06:55,466
And this is the final result.
Let me circle the pivots again.
110
00:06:55,466 --> 00:06:57,870
Here they are,
three, which is going
111
00:06:57,870 --> 00:07:01,210
to give me three linearly
independent vectors
112
00:07:01,210 --> 00:07:04,060
and hence, three elements
of the basis, and hence,
113
00:07:04,060 --> 00:07:06,350
dimension 3 for
the vector space.
114
00:07:06,350 --> 00:07:11,040
But I could no longer
use these three columns
115
00:07:11,040 --> 00:07:15,040
as elements of the basis,
because when I did elimination,
116
00:07:15,040 --> 00:07:17,500
I changed the column space.
117
00:07:17,500 --> 00:07:20,410
So the column space
of this matrix
118
00:07:20,410 --> 00:07:22,650
is not the same as
the column space
119
00:07:22,650 --> 00:07:24,290
of the eliminated version.
120
00:07:24,290 --> 00:07:27,410
So I cannot use these.
121
00:07:27,410 --> 00:07:29,550
In fact, if you
look at them, you
122
00:07:29,550 --> 00:07:31,550
can probably understand
that they're not
123
00:07:31,550 --> 00:07:34,790
going to span the
same space as these.
124
00:07:34,790 --> 00:07:37,040
Because all that I
have down here are 0's,
125
00:07:37,040 --> 00:07:39,960
and I get a lot more than
just 0's in the last two
126
00:07:39,960 --> 00:07:41,840
entries of the vectors.
127
00:07:41,840 --> 00:07:45,260
So what I need to
do is-- the pivots
128
00:07:45,260 --> 00:07:48,310
are in the first, second,
and third columns.
129
00:07:48,310 --> 00:07:53,550
I need to use these three
columns as my basis.
130
00:07:53,550 --> 00:07:57,420
I'll just write basis down here.
131
00:07:57,420 --> 00:07:58,790
And that will work too.
132
00:07:58,790 --> 00:08:01,200
So see, I have produced
two different bases
133
00:08:01,200 --> 00:08:04,190
for the same vector space,
which is totally fine.
134
00:08:04,190 --> 00:08:06,420
You can pick the
basis that you prefer.
135
00:08:06,420 --> 00:08:07,100
All right.
136
00:08:07,100 --> 00:08:08,040
We're done.
137
00:08:08,040 --> 00:08:09,398
Thank you.