1
00:00:07,442 --> 00:00:08,650
DAVID SHIROKOFF: Hi everyone.
2
00:00:08,650 --> 00:00:10,060
I'm Dave.
3
00:00:10,060 --> 00:00:11,840
Now today, I'd like
to tackle a problem
4
00:00:11,840 --> 00:00:13,990
in orthogonal subspaces.
5
00:00:13,990 --> 00:00:18,260
So the problem we'd like to
tackle: given a subspace S,
6
00:00:18,260 --> 00:00:22,220
and suppose S is spanned by
two vectors, [1, 2, 2, 3]
7
00:00:22,220 --> 00:00:24,530
and [1, 3, 3, 2].
8
00:00:24,530 --> 00:00:28,630
We have a question here which
is to find a basis for S perp--
9
00:00:28,630 --> 00:00:32,409
S perp is another subspace
which is orthogonal to S.
10
00:00:32,409 --> 00:00:35,870
And then secondly, can every
vector in R^4 be uniquely
11
00:00:35,870 --> 00:00:39,260
written in terms
of S and S perp.
12
00:00:39,260 --> 00:00:41,282
So I'll let you think
about this for now,
13
00:00:41,282 --> 00:00:42,573
and I'll come back in a minute.
14
00:00:53,419 --> 00:00:54,620
Hi everyone.
15
00:00:54,620 --> 00:00:56,060
Welcome back.
16
00:00:56,060 --> 00:00:58,180
OK, so why don't we
tackle this problem?
17
00:01:02,880 --> 00:01:04,440
OK, so first off,
what does it mean
18
00:01:04,440 --> 00:01:06,760
for a vector to be in S perp?
19
00:01:06,760 --> 00:01:16,530
Well, if I have a
vector x, and S perp,
20
00:01:16,530 --> 00:01:20,540
and x is in S perp,
what this means is
21
00:01:20,540 --> 00:01:24,450
x is going to be orthogonal
to every vector in S. Now
22
00:01:24,450 --> 00:01:27,640
specifically, S is spanned
by these two vectors.
23
00:01:27,640 --> 00:01:31,660
So it's sufficient that x be
perpendicular to the two basis
24
00:01:31,660 --> 00:01:33,250
vectors in S.
25
00:01:33,250 --> 00:01:45,720
So specifically, I can take
[1, 2, 2, 3] and dot it with x,
26
00:01:45,720 --> 00:01:47,230
and it's going to be 0.
27
00:01:47,230 --> 00:01:52,210
So I'm treating x as
a column vector here.
28
00:01:52,210 --> 00:01:59,340
In addition, x must also be
orthogonal to [1, 3, 2, 2].
29
00:02:03,600 --> 00:02:05,880
So any vector x
that's an S perp must
30
00:02:05,880 --> 00:02:08,750
be orthogonal to both
of these vectors.
31
00:02:08,750 --> 00:02:10,280
So what we can do
is we can write
32
00:02:10,280 --> 00:02:11,680
this as a matrix equation.
33
00:02:16,140 --> 00:02:18,790
And we do this by
combining these two vectors
34
00:02:18,790 --> 00:02:19,720
as rows of the matrix.
35
00:02:31,470 --> 00:02:33,700
So if we step back and take
a look at this equation,
36
00:02:33,700 --> 00:02:35,200
we see that what
we're really asking
37
00:02:35,200 --> 00:02:40,410
is to find all x that are in
the null space of this matrix.
38
00:02:40,410 --> 00:02:42,990
So how do we find x in the
null space of a matrix?
39
00:02:42,990 --> 00:02:46,330
Well what we can do is we
can row reduce this matrix
40
00:02:46,330 --> 00:02:49,270
and try and find a basis
for the null space.
41
00:02:49,270 --> 00:02:51,660
So I'm going to just
row reduce this matrix.
42
00:02:51,660 --> 00:02:53,160
And notice that
by row reduction,
43
00:02:53,160 --> 00:02:56,850
we don't actually change
the null space of a matrix.
44
00:02:56,850 --> 00:03:00,619
So if I'm only interested
in the null space,
45
00:03:00,619 --> 00:03:02,160
this system is going
to be equivalent
46
00:03:02,160 --> 00:03:04,530
to-- I can keep the
top row the same.
47
00:03:09,030 --> 00:03:11,120
And then just to
simplify our lives,
48
00:03:11,120 --> 00:03:13,597
we can take the second
row and subtract
49
00:03:13,597 --> 00:03:14,680
one copy of the first row.
50
00:03:18,590 --> 00:03:23,080
Now, if I do that, I
obtain 0, 1, 1, -1.
51
00:03:29,450 --> 00:03:33,210
Now, to parameterize the null
space, what I'm going to do
52
00:03:33,210 --> 00:03:36,640
is I'm going to write
x out as components.
53
00:03:36,640 --> 00:03:48,550
So if I write x with components
x_1, x_2, x_3 and x_4,
54
00:03:48,550 --> 00:03:52,790
we see here that this
matrix has a rank of 2.
55
00:03:52,790 --> 00:03:56,970
Now, we're looking at
vectors which live in R^4,
56
00:03:56,970 --> 00:04:00,610
so we know that the null space
is going to have a dimension
57
00:04:00,610 --> 00:04:02,362
which is 4 minus 2.
58
00:04:02,362 --> 00:04:04,820
So that means there should be
two vectors in the null space
59
00:04:04,820 --> 00:04:07,610
of this matrix.
60
00:04:07,610 --> 00:04:10,350
To parameterize these
two-dimensional vectors, what
61
00:04:10,350 --> 00:04:14,040
I'm going to do is I'm going
to let x_4 equal some constant,
62
00:04:14,040 --> 00:04:17,250
and x_3 equal another constant.
63
00:04:17,250 --> 00:04:25,020
So specifically, I'm going
to let x_4 equal b, and x_3
64
00:04:25,020 --> 00:04:25,838
equal a.
65
00:04:29,190 --> 00:04:33,160
Now what we do is we take a
look at these two equations,
66
00:04:33,160 --> 00:04:35,990
and this bottom
equation will say
67
00:04:35,990 --> 00:04:42,030
that x_2 is equal
to negative x_3 plus
68
00:04:42,030 --> 00:04:51,230
x_4, which is going to equal
negative a-- x_4-- plus b.
69
00:04:55,090 --> 00:04:59,210
And then the top equation says
that x_1 is equal to negative
70
00:04:59,210 --> 00:05:07,700
2*x_2 minus 2*x_3 minus 3*x_4.
71
00:05:12,200 --> 00:05:18,252
And if I substitute
in, x_2 is -a plus b.
72
00:05:22,500 --> 00:05:25,608
x_3 is a.
73
00:05:25,608 --> 00:05:26,780
And x_4 is b.
74
00:05:30,020 --> 00:05:32,220
So when the dust
settles, the a's cancel
75
00:05:32,220 --> 00:05:33,930
and I'm left with minus 5b.
76
00:05:38,030 --> 00:05:40,560
So we can combine
everything together
77
00:05:40,560 --> 00:05:56,290
and we end up obtaining [x_1,
x_2, x_3, x_4] equals -5b,
78
00:05:56,290 --> 00:06:09,759
x_2 is minus a plus b,
x_3 is a, and x_4 is b.
79
00:06:09,759 --> 00:06:11,800
And now what we can do is
we can take this vector
80
00:06:11,800 --> 00:06:13,780
and we can decompose
it into pieces
81
00:06:13,780 --> 00:06:16,190
which are a multiplied
by a vector,
82
00:06:16,190 --> 00:06:19,630
and b multiplied by a vector.
83
00:06:19,630 --> 00:06:29,610
So you'll note that this is
actually a times [0, -1, 1, 0]
84
00:06:29,610 --> 00:06:40,110
plus b times [-5, 1, 0, 1].
85
00:06:40,110 --> 00:06:42,100
OK?
86
00:06:42,100 --> 00:06:45,260
So we have successfully
achieved a parameterization
87
00:06:45,260 --> 00:06:49,000
of the null space of this
matrix as some constant a
88
00:06:49,000 --> 00:06:52,980
times a vector
[0, -1, 1, 0] plus b
89
00:06:52,980 --> 00:06:56,700
times a vector [-5, 1, 0, 1].
90
00:06:56,700 --> 00:07:00,170
And now we claim that this
is the entire space, S perp.
91
00:07:04,190 --> 00:07:10,060
So S perp is going to be spanned
by this vector and this vector.
92
00:07:10,060 --> 00:07:14,870
Now notice how, if I were
to take either of these two
93
00:07:14,870 --> 00:07:20,230
vectors in S and dot it with
any vector in the null space,
94
00:07:20,230 --> 00:07:22,425
by construction, it
automatically vanishes.
95
00:07:25,190 --> 00:07:28,190
So this concludes part one.
96
00:07:28,190 --> 00:07:30,330
Now for part two.
97
00:07:30,330 --> 00:07:33,510
Can every vector v in R^4 be
written uniquely in terms of S
98
00:07:33,510 --> 00:07:35,020
and S perp?
99
00:07:35,020 --> 00:07:35,800
The answer is yes.
100
00:07:42,440 --> 00:07:44,460
So how do we see this?
101
00:07:44,460 --> 00:07:48,670
Well, if I have a
vector v, what I can do
102
00:07:48,670 --> 00:07:53,500
is I can try and write
it as some constant c_1
103
00:07:53,500 --> 00:08:02,880
times the vector [1, 2, 2, 3]
plus c_2 times the vector [1,
104
00:08:02,880 --> 00:08:21,880
3, 3, 2] plus the vector
c_3 [0, -1, 1, 0] plus c4
105
00:08:21,880 --> 00:08:25,810
[-5, 1, 0, 1].
106
00:08:25,810 --> 00:08:26,870
OK?
107
00:08:26,870 --> 00:08:31,360
So c_1 and c_2 are
multiplying the vectors in S,
108
00:08:31,360 --> 00:08:35,450
and c_3 and c_4 are multiplying
the vectors in S perp.
109
00:08:35,450 --> 00:08:38,090
So the question is,
given any v, can I
110
00:08:38,090 --> 00:08:41,600
find constants c_1,
c_2, c_3, c_4, such
111
00:08:41,600 --> 00:08:44,370
that this equation holds?
112
00:08:44,370 --> 00:08:45,540
And the answer is yes.
113
00:08:48,680 --> 00:08:51,050
Just to see why it's
yes, what we can do
114
00:08:51,050 --> 00:08:54,660
is we can rewrite this
in matrix notation,
115
00:08:54,660 --> 00:08:56,175
and there's kind
of a handy trick.
116
00:09:13,860 --> 00:09:16,684
What I can do is I
can take these columns
117
00:09:16,684 --> 00:09:18,350
and write them as
columns of the matrix.
118
00:09:22,640 --> 00:09:25,020
And this whole
expression is actually
119
00:09:25,020 --> 00:09:28,510
equivalent to this
matrix multiplied
120
00:09:28,510 --> 00:09:32,100
by the constant,
c_1, c_2, c_3, c_4.
121
00:09:32,100 --> 00:09:36,760
And on the right-hand
side, we have the vector v.
122
00:09:36,760 --> 00:09:39,930
Now, by construction,
these vectors
123
00:09:39,930 --> 00:09:41,540
are linearly independent.
124
00:09:41,540 --> 00:09:43,370
And we know from linear
algebra that if we
125
00:09:43,370 --> 00:09:45,880
have a matrix with linearly
independent columns,
126
00:09:45,880 --> 00:09:48,280
the matrix is invertible.
127
00:09:48,280 --> 00:09:51,330
What this means is, for any
v on the right-hand side,
128
00:09:51,330 --> 00:09:54,730
we can invert this matrix and
obtain unique coefficients,
129
00:09:54,730 --> 00:09:57,590
c_1, c_2, c_3, c_4.
130
00:09:57,590 --> 00:10:00,640
This then gives us a
unique decomposition for v
131
00:10:00,640 --> 00:10:05,180
in terms of a
piece which is in S
132
00:10:05,180 --> 00:10:06,690
and a piece which is in S perp.
133
00:10:10,880 --> 00:10:14,630
And in general this can be
done for any vector space.
134
00:10:14,630 --> 00:10:17,250
Well I'd like to
conclude this problem now
135
00:10:17,250 --> 00:10:19,360
and I hope you had a good time.