1
00:00:05,176 --> 00:00:06,300
MARTINA BALAGOVIC: Welcome.

2
00:00:06,300 --> 00:00:10,840
Today's problem actually
appeared in a quiz.

3
00:00:10,840 --> 00:00:16,440
It appeared in quiz one in
fall of 1999 as question four.

4
00:00:16,440 --> 00:00:20,420
The problem puts the usual solve
the following system upside

5
00:00:20,420 --> 00:00:24,130
down by saying we have some
matrix and we know that all

6
00:00:24,130 --> 00:00:29,500
the solutions to A*x equals
this vector here, [1, 4, 1, 1],

7
00:00:29,500 --> 00:00:33,750
all the solutions to this
problem are given by x equals

8
00:00:33,750 --> 00:00:38,820
[0, 1, 1] plus any
number c times [1, 2, 1].

9
00:00:38,820 --> 00:00:40,400
And we're asked
to say everything

10
00:00:40,400 --> 00:00:43,180
that we can about the
columns of the matrix A.

11
00:00:43,180 --> 00:00:46,067
So I'm going to let you pretend
that you are on an exam,

12
00:00:46,067 --> 00:00:47,900
try to solve it yourself,
and then come back

13
00:00:47,900 --> 00:00:49,400
and compare your
solution with mine.

14
00:00:57,210 --> 00:00:58,610
OK, welcome back.

15
00:00:58,610 --> 00:01:00,350
So the first thing
that you should

16
00:01:00,350 --> 00:01:02,350
think about in this
sort of situation

17
00:01:02,350 --> 00:01:04,140
is what is the size of A?

18
00:01:04,140 --> 00:01:10,710
Well, we want to multiply A with
an x that has three entries,

19
00:01:10,710 --> 00:01:14,734
so A should have three columns.

20
00:01:14,734 --> 00:01:25,300
Let me call those columns
c_1, c_2, and c_3.

21
00:01:25,300 --> 00:01:27,700
And when I take some linear
combinations of c_1, c_2

22
00:01:27,700 --> 00:01:32,420
and c_3, I'm going to get this
vector here, [1, 4, 1, 1].

23
00:01:32,420 --> 00:01:43,380
So all the c_i's, c_1, c_2,
and c_3 are vectors in R_4.

24
00:01:46,150 --> 00:01:49,130
Now, if you know
about, if you learned

25
00:01:49,130 --> 00:01:55,064
about particular solutions
and special solutions, then

26
00:01:55,064 --> 00:01:56,730
my notation here
shouldn't surprise you.

27
00:01:56,730 --> 00:01:59,410
I'm going to call
this vector here x_p,

28
00:01:59,410 --> 00:02:01,990
and this vector here x_s.

29
00:02:01,990 --> 00:02:05,210
And I'm going to
use the fact that

30
00:02:05,210 --> 00:02:15,990
x_p plus c times x_s
satisfies A times this

31
00:02:15,990 --> 00:02:22,020
equals b-- I will call this
vector b-- for any number c.

32
00:02:26,580 --> 00:02:28,850
In particular, what
I'm going to conclude

33
00:02:28,850 --> 00:02:40,030
is that when c equals 0 we
get A times x_p equals b.

34
00:02:40,030 --> 00:02:42,680
But also that when
c equals 1, we

35
00:02:42,680 --> 00:02:53,750
get A times x_p plus
A times x_s equals b.

36
00:02:53,750 --> 00:02:57,830
Replacing this by b, we
get that this implies

37
00:02:57,830 --> 00:03:04,820
that A times x_s equals 0.

38
00:03:04,820 --> 00:03:07,230
So in trying to find
what are the columns

39
00:03:07,230 --> 00:03:10,840
c_1, c_2, and c_3 of the matrix
A, let's look at these two

40
00:03:10,840 --> 00:03:11,770
equations.

41
00:03:11,770 --> 00:03:15,370
x_p satisfies A
times x_p equals b,

42
00:03:15,370 --> 00:03:19,330
and x_s satisfies A
times x_s equals 0.

43
00:03:19,330 --> 00:03:21,440
Again, if you know what
particular and special

44
00:03:21,440 --> 00:03:23,610
solutions are this
shouldn't surprise you.

45
00:03:23,610 --> 00:03:26,660
But we also know
what x_p and x_s are,

46
00:03:26,660 --> 00:03:32,340
so let's use them to try to
calculate c_1, c_2, and c_3.

47
00:03:32,340 --> 00:03:38,630
A times x_p equals b means that
the linear combination of c_1,

48
00:03:38,630 --> 00:03:44,990
c_2, and c_3 encoded in the
vector x_p, which is [0, 1, 1],

49
00:03:44,990 --> 00:03:46,730
gives the vector b.

50
00:03:46,730 --> 00:04:05,850
So c_1, c_2, c_3 times
[0, 1, 1] gives us [1, 4, 1, 1].

51
00:04:08,890 --> 00:04:15,110
In other words, c_2
plus c_3 equal b.

52
00:04:17,649 --> 00:04:20,959
Let's turn our attention
to A times x_s equals 0.

53
00:04:20,959 --> 00:04:30,040
This says that c_1,
c_2, c_3 times--

54
00:04:30,040 --> 00:04:37,960
x_s was defined to be
[0, 2,  1]-- equals 0.

55
00:04:37,960 --> 00:04:47,550
In other words, 2 times
c_2 plus c_3 equals 0.

56
00:04:47,550 --> 00:04:51,170
Now solving this system where
the unknowns are vectors

57
00:04:51,170 --> 00:04:54,100
but it's still just a
linear system, we can see,

58
00:04:54,100 --> 00:04:58,490
for example, from the second
equation that c_3 equals minus

59
00:04:58,490 --> 00:05:00,350
2*c_2.

60
00:05:00,350 --> 00:05:02,870
And plugging it back into
the original equation,

61
00:05:02,870 --> 00:05:11,310
getting c_2 minus
2*c_2 equals b,

62
00:05:11,310 --> 00:05:16,120
from which it follows that
c_2 is equal to minus b,

63
00:05:16,120 --> 00:05:20,280
and that c3 is
equal to 2 times b.

64
00:05:23,290 --> 00:05:25,620
So from this tiny
amount of information--

65
00:05:25,620 --> 00:05:29,870
we just knew the solutions to
this one particular equation

66
00:05:29,870 --> 00:05:32,980
involving A-- we got
the second column of A

67
00:05:32,980 --> 00:05:35,310
and the third column of
A completely explicitly

68
00:05:35,310 --> 00:05:36,540
calculated.

69
00:05:36,540 --> 00:05:39,980
Now, what can we say
about the first column?

70
00:05:39,980 --> 00:05:46,760
I said before that all the
solutions of A*x equals b are

71
00:05:46,760 --> 00:05:51,880
of the form a particular
solution plus some number times

72
00:05:51,880 --> 00:05:54,660
a special solution.

73
00:05:54,660 --> 00:05:58,340
And the information that we have
is that there's just one number

74
00:05:58,340 --> 00:05:58,840
here.

75
00:05:58,840 --> 00:06:03,640
So we said everything, once
we remove this vector here,

76
00:06:03,640 --> 00:06:09,380
everything that we get here
will satisfy A times x equals 0.

77
00:06:09,380 --> 00:06:12,010
And the fact that
everything that

78
00:06:12,010 --> 00:06:16,050
satisfies A times x equals 0 is
a multiple of this one vector

79
00:06:16,050 --> 00:06:21,360
that was given to us means
that the null space of A

80
00:06:21,360 --> 00:06:23,390
has dimension one.

81
00:06:23,390 --> 00:06:26,000
There is just one
special solution.

82
00:06:26,000 --> 00:06:34,080
So dimension of the
null space of A is 1.

83
00:06:37,700 --> 00:06:43,880
So rank of A is the
number of columns

84
00:06:43,880 --> 00:06:49,850
minus this dimension of null
space, and it's equal to 2.

85
00:06:49,850 --> 00:06:52,560
As rank of A is equal
to 2, the number

86
00:06:52,560 --> 00:06:57,350
of linearly independent
columns needs to be 2 as well.

87
00:06:57,350 --> 00:06:59,740
So the only thing that
we can say at this point

88
00:06:59,740 --> 00:07:03,090
is if the first column
was also a multiple of b,

89
00:07:03,090 --> 00:07:07,860
as the second and the third are,
then the rank would be smaller

90
00:07:07,860 --> 00:07:08,990
than 2.

91
00:07:08,990 --> 00:07:11,240
So that is the only
thing that cannot happen.

92
00:07:11,240 --> 00:07:22,400
So c_1 is not a multiple of b.

93
00:07:22,400 --> 00:07:26,000
Not any multiple, including
not a zero multiple.

94
00:07:26,000 --> 00:07:29,160
And that's pretty much
everything we can say.

95
00:07:29,160 --> 00:07:30,999
Yes, if it was some
other multiple of it,

96
00:07:30,999 --> 00:07:33,165
then we would be able to
find some other vector here

97
00:07:33,165 --> 00:07:34,760
and we would have
two parameters.

98
00:07:34,760 --> 00:07:38,040
But it's not, and
this is everything

99
00:07:38,040 --> 00:07:40,190
that we can say about it.