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