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HERBERT GROSS: Hi.

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00:00:33,370 --> 00:00:35,230
Last time, you
may recall that we

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00:00:35,230 --> 00:00:37,810
were looking at
some of our examples

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00:00:37,810 --> 00:00:39,850
in three-dimensional
space and talking

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00:00:39,850 --> 00:00:41,770
about picking a
couple of vectors

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00:00:41,770 --> 00:00:45,600
in three-dimensional space, be
they i and j or alpha and beta.

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And we then talked
about the subspaces

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00:00:48,370 --> 00:00:53,590
spanned by alpha and beta or
the subspace span by i and j,

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00:00:53,590 --> 00:00:56,260
noticing that the
subspace could be smaller

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00:00:56,260 --> 00:00:57,520
than the entire space.

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00:00:57,520 --> 00:01:00,130
Or possibly, it could even
equal the entire space.

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00:01:00,130 --> 00:01:02,950
In no event could it be
greater than the entire space.

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00:01:02,950 --> 00:01:05,560
At any rate, what we
would like to do today

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00:01:05,560 --> 00:01:09,880
is to generalize this
idea into more than two

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00:01:09,880 --> 00:01:11,110
or three dimensions.

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00:01:11,110 --> 00:01:14,950
And also, to make
the generalization

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00:01:14,950 --> 00:01:18,052
go independently of
our n-tuple notation

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00:01:18,052 --> 00:01:19,510
that we used earlier
in the course.

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00:01:19,510 --> 00:01:22,570
At any rate, today's
lesson is entitled,

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00:01:22,570 --> 00:01:25,610
Spanning Vectors,
and the idea is this.

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00:01:25,610 --> 00:01:28,300
Let's suppose we have
a vector space, v,

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00:01:28,300 --> 00:01:33,440
and we now arbitrarily choose n
vectors, alpha 1 up to alpha n

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00:01:33,440 --> 00:01:35,650
and v. For example,
in the lecture

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00:01:35,650 --> 00:01:40,150
last time, some of our
examples involved v

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00:01:40,150 --> 00:01:43,590
being three-dimensional
space and n being two.

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00:01:43,590 --> 00:01:46,600
In other words, we picked
i and j, alpha and beta,

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00:01:46,600 --> 00:01:48,280
in three-dimensional space.

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00:01:48,280 --> 00:01:50,270
But at any rate,
what we do is this.

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00:01:50,270 --> 00:01:54,160
We take alpha 1 up to alpha
n, any n arbitrarily chosen,

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00:01:54,160 --> 00:01:58,210
vectors, n, v, and we look
at the set, which I write

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00:01:58,210 --> 00:02:01,090
s of alpha 1 up to alpha n.

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00:02:01,090 --> 00:02:04,510
We look at the set of all
linear combinations of alpha 1

40
00:02:04,510 --> 00:02:05,740
up to alpha n.

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00:02:05,740 --> 00:02:08,139
And written compactly,
that's written this way.

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00:02:08,139 --> 00:02:11,200
In other words, it's a c1 alpha
1 plus, et cetera, cn alpha n.

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00:02:11,200 --> 00:02:14,380
Where the cs are real
numbers, they belong to r.

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00:02:14,380 --> 00:02:18,610
Now, my claim is that this
set is more than a set.

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00:02:18,610 --> 00:02:20,530
It's a space.

46
00:02:20,530 --> 00:02:23,830
Namely, if you of two linear
combinations of alpha 1

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00:02:23,830 --> 00:02:27,130
up to alpha n, you get a
linear combination of alpha 1

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00:02:27,130 --> 00:02:28,780
up to alpha n.

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00:02:28,780 --> 00:02:32,260
If you take a scalar multiple of
a linear combination of alpha 1

50
00:02:32,260 --> 00:02:35,860
up to alpha n, you, again, get
a linear combination of alpha 1

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00:02:35,860 --> 00:02:36,940
up to alpha n.

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00:02:36,940 --> 00:02:38,920
Algebraically, this
is quite simple,

53
00:02:38,920 --> 00:02:41,020
because structurally
we mimic what's

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00:02:41,020 --> 00:02:42,975
happening in ordinary algebra.

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00:02:42,975 --> 00:02:44,350
And what we're
saying is, you can

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00:02:44,350 --> 00:02:46,780
add the two alpha 1
components together,

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00:02:46,780 --> 00:02:49,210
the alpha 2 components
together, et cetera.

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00:02:49,210 --> 00:02:52,940
I'll illustrate that in a
simple example in a moment,

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00:02:52,940 --> 00:02:54,790
but the idea is this.

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00:02:54,790 --> 00:02:57,592
What we say is, by definition--

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00:02:57,592 --> 00:02:59,800
and we've done this in some
of the exercises already,

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00:02:59,800 --> 00:03:01,090
you'll recall--

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00:03:01,090 --> 00:03:04,720
that the subspace consisting
of all linear combinations

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00:03:04,720 --> 00:03:07,990
of alpha 1 up to alpha n,
which is a subspace of v,

65
00:03:07,990 --> 00:03:11,830
it's called the space spanned
by alpha 1 up to alpha n.

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00:03:11,830 --> 00:03:13,240
That's all this thing means.

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00:03:13,240 --> 00:03:15,850
The space spanned by
alpha 1 up to alpha n

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00:03:15,850 --> 00:03:18,550
means the set of all linear
combinations of alpha 1

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00:03:18,550 --> 00:03:21,670
up to alpha n, and that
is a subspace of v.

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00:03:21,670 --> 00:03:23,950
For example, the easiest
case I can think of

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00:03:23,950 --> 00:03:28,450
is to pick n equals 1, given
a vector space, v, simply pick

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00:03:28,450 --> 00:03:32,080
any elements, say alpha
1, that belongs to v.

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00:03:32,080 --> 00:03:35,590
And all linear combinations
of alpha 1 are simply what?

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00:03:35,590 --> 00:03:37,480
All scalar multiples of alpha 1.

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00:03:37,480 --> 00:03:40,440
In other words, the
space spanned by alpha 1

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00:03:40,440 --> 00:03:43,330
is just a set of all
constant multiples of alpha.

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00:03:43,330 --> 00:03:46,970
So c alpha 1, where
c is a real number.

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00:03:46,970 --> 00:03:51,340
Now, my claim is that, clearly,
the sum of any two scalar

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00:03:51,340 --> 00:03:54,520
multiples of alpha 1 is a
scalar multiple of alpha 1.

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00:03:54,520 --> 00:03:57,100
And a scalar multiple of a
scalar multiple of alpha 1

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00:03:57,100 --> 00:03:59,320
is, again, a scalar
multiple of alpha 1.

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00:03:59,320 --> 00:04:01,360
I'd like to show you that,
though, independently

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00:04:01,360 --> 00:04:03,190
of a geometric interpretation.

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00:04:03,190 --> 00:04:06,230
This follows clearly from
our axiomatic approach,

85
00:04:06,230 --> 00:04:08,770
namely, suppose
beta and gamma are

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00:04:08,770 --> 00:04:12,490
any two elements in the
space spanned by alpha 1.

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00:04:12,490 --> 00:04:15,160
That means beta is a
scalar multiple as alpha 1,

88
00:04:15,160 --> 00:04:16,720
say c1 alpha 1.

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00:04:16,720 --> 00:04:21,040
Gamma is a scalar multiple
of alpha 1, say c2 alpha 1.

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00:04:21,040 --> 00:04:26,260
Therefore, beta plus gamma is
c1 alpha 1 plus c2 alpha 1.

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00:04:26,260 --> 00:04:28,210
By our distributive
property, that's

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00:04:28,210 --> 00:04:30,910
c1 plus c2 times alpha 1.

93
00:04:30,910 --> 00:04:33,490
But since c1 and
c2 a real numbers,

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00:04:33,490 --> 00:04:36,190
and the sum of two real numbers
is, again, a real number,

95
00:04:36,190 --> 00:04:39,070
we see that beta plus
gamma is a real number

96
00:04:39,070 --> 00:04:42,160
times alpha 1, in other words,
a scalar multiple of alpha 1.

97
00:04:42,160 --> 00:04:44,170
And that means that
by definition it

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00:04:44,170 --> 00:04:46,840
belongs to the space
spanned by alpha 1.

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00:04:46,840 --> 00:04:50,350
Similarly, if we take beta and
multiply it by the real number

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00:04:50,350 --> 00:04:55,930
r, r times c1 alpha 1, by
the associated property

101
00:04:55,930 --> 00:04:59,230
for scalar multiplication,
is the scale of multiple r

102
00:04:59,230 --> 00:05:01,880
times c1 times alpha 1.

103
00:05:01,880 --> 00:05:04,880
And if r and c1 are real
number, so is their product.

104
00:05:04,880 --> 00:05:09,550
So consequently, r times beta
is a scalar times alpha 1,

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00:05:09,550 --> 00:05:11,800
which by definition means
it's in the space spanned

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00:05:11,800 --> 00:05:13,090
by alpha 1.

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00:05:13,090 --> 00:05:16,210
And geometrically, if you want
to see what this thing means,

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00:05:16,210 --> 00:05:18,520
notice that the space
spanned by alpha 1--

109
00:05:18,520 --> 00:05:21,190
if you want think of
alpha 1 as an arrow--

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00:05:21,190 --> 00:05:24,780
the space spanned by
alpha 1 is simply what?

111
00:05:24,780 --> 00:05:27,100
The set of all linear
combinations of alpha--

112
00:05:27,100 --> 00:05:31,120
all scalar multiples of alpha
1, and that is precisely

113
00:05:31,120 --> 00:05:33,863
the straight line
determined by alpha 1.

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00:05:33,863 --> 00:05:35,530
In other words, any
linear combination--

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00:05:35,530 --> 00:05:38,260
any scalar multiple
of alpha 1 has what?

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00:05:38,260 --> 00:05:40,938
The same direction as alpha 1.

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00:05:40,938 --> 00:05:42,730
In other words, lies
in the same direction,

118
00:05:42,730 --> 00:05:44,950
but may have a
different magnitude

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00:05:44,950 --> 00:05:46,450
and/or a different sense.

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00:05:46,450 --> 00:05:50,330
But certainly, the line
part doesn't change.

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00:05:50,330 --> 00:05:52,240
Now, the key point
is this, and this

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00:05:52,240 --> 00:05:55,150
is going to play a very key
role throughout the remainder

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00:05:55,150 --> 00:05:56,890
of our discussion.

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00:05:56,890 --> 00:06:00,310
One would believe
that the more vectors

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00:06:00,310 --> 00:06:03,940
you take linear combinations of,
the bigger your space becomes.

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00:06:03,940 --> 00:06:05,980
In other words, if you
take the space spanned

127
00:06:05,980 --> 00:06:09,130
by two vectors, then tack
onto those two vectors

128
00:06:09,130 --> 00:06:11,080
a third vector,
you might believe

129
00:06:11,080 --> 00:06:16,030
that the resulting space must be
larger than the original space.

130
00:06:16,030 --> 00:06:18,183
Well, obviously, it
can't be any smaller,

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00:06:18,183 --> 00:06:20,350
and I'll explain what
obviously means in the moment,

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00:06:20,350 --> 00:06:21,890
if it's not that obvious.

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00:06:21,890 --> 00:06:25,780
But the key point is, that
if I take not just alpha

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00:06:25,780 --> 00:06:27,820
1 up to alpha n, but
an additional vector,

135
00:06:27,820 --> 00:06:31,510
say, alpha 1 up to alpha n plus
the additional vector alpha

136
00:06:31,510 --> 00:06:34,960
n plus 1, and look at the
space spanned by alpha 1

137
00:06:34,960 --> 00:06:38,260
through alpha n plus
1, that space need not

138
00:06:38,260 --> 00:06:42,070
be larger than the space spanned
by alpha 1 up to alpha n.

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00:06:42,070 --> 00:06:45,250
In other words, merely by
tacking on a new vector,

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00:06:45,250 --> 00:06:47,740
you do not necessarily
tack on to the space

141
00:06:47,740 --> 00:06:49,180
spanned by the vectors.

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00:06:49,180 --> 00:06:53,470
By means of a trivial example,
let alpha 1 be i, let alpha 2

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00:06:53,470 --> 00:06:56,620
be j, an alpha 3 be i plus j.

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00:06:56,620 --> 00:06:59,770
Clearly, the space spanned
by alpha 1 and alpha 2

145
00:06:59,770 --> 00:07:02,230
is by definition this sort
of all linear combinations

146
00:07:02,230 --> 00:07:03,760
of alpha 1 alpha 2.

147
00:07:03,760 --> 00:07:05,650
And in terms of this
simple example, that

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00:07:05,650 --> 00:07:09,340
obviously is the x, y plane,
because alpha 1 is i, alpha 2

149
00:07:09,340 --> 00:07:10,420
as j.

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00:07:10,420 --> 00:07:13,390
Now, let's look at the space
spanned by alpha 1, alpha 2,

151
00:07:13,390 --> 00:07:14,590
and alpha 3.

152
00:07:14,590 --> 00:07:19,090
Again, geometrically, notice
that that space is still

153
00:07:19,090 --> 00:07:20,110
the x, y plane.

154
00:07:20,110 --> 00:07:23,650
Because after all, alpha
1, alpha 2, and alpha 3,

155
00:07:23,650 --> 00:07:27,190
being i, j, and i plus
j, respectively, all

156
00:07:27,190 --> 00:07:28,840
lie in the x, y plane.

157
00:07:28,840 --> 00:07:30,940
Consequently, any linear
combination of them

158
00:07:30,940 --> 00:07:32,560
will lie in the x, y plane.

159
00:07:32,560 --> 00:07:34,660
But let's pretend
we didn't know that.

160
00:07:34,660 --> 00:07:37,840
Let's, first of all, observe
that from the pure mathematics

161
00:07:37,840 --> 00:07:41,830
of this alone, that the space
spanned by alpha 1 alpha 2

162
00:07:41,830 --> 00:07:45,280
must be a subspace of the space
spanned by alpha 1, alpha 2,

163
00:07:45,280 --> 00:07:46,360
an alpha 3.

164
00:07:46,360 --> 00:07:48,580
And the proof, by the
way, is quite trivial.

165
00:07:48,580 --> 00:07:50,980
Namely, by definition,
the space spanned

166
00:07:50,980 --> 00:07:53,620
by alpha 1, alpha
2, alpha 3 is a set

167
00:07:53,620 --> 00:07:57,130
of all linear combinations
of alpha 1, alpha 2, alpha 3.

168
00:07:57,130 --> 00:08:00,730
In particular, if I
choose c3 to equal 0,

169
00:08:00,730 --> 00:08:03,040
notice that this term drops out.

170
00:08:03,040 --> 00:08:04,900
And in particular,
what I'm left with

171
00:08:04,900 --> 00:08:07,760
is c1 alpha 1 plus c2 alpha 2.

172
00:08:07,760 --> 00:08:11,380
In other words, every linear
combination of alpha 1

173
00:08:11,380 --> 00:08:15,940
and alpha 2 is, in particular,
a linear combination of alpha 1,

174
00:08:15,940 --> 00:08:19,090
alpha 2, and alpha 3.

175
00:08:19,090 --> 00:08:21,470
But my claim is, just as
we saw geometrically--

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00:08:21,470 --> 00:08:23,540
let me now prove
this algebraically--

177
00:08:23,540 --> 00:08:28,210
that in particular, if beta
is any vector belonging

178
00:08:28,210 --> 00:08:31,510
to the space spanned by alpha
1, alpha 2, and alpha 3,

179
00:08:31,510 --> 00:08:35,139
axiomatically, it must belong
to the space spanned by alpha 1

180
00:08:35,139 --> 00:08:36,220
and alpha 2.

181
00:08:36,220 --> 00:08:37,330
We see that geometrically.

182
00:08:37,330 --> 00:08:40,390
We already know that the
space is still the x, y plane.

183
00:08:40,390 --> 00:08:42,340
How would we prove
this axiomatically?

184
00:08:42,340 --> 00:08:45,690
Well, if beta belongs to
this space, by definition

185
00:08:45,690 --> 00:08:49,420
it's a linear combination as
alpha 1, alpha 2, alpha 3.

186
00:08:49,420 --> 00:08:52,810
Noticing, that alpha 3
is by definition alpha 1

187
00:08:52,810 --> 00:08:54,130
plus alpha 2.

188
00:08:54,130 --> 00:08:57,520
I can replace alpha 3
by alpha 1 plus alpha 2.

189
00:08:57,520 --> 00:09:00,880
I can now distribute c3
with alpha 1 and alpha 2.

190
00:09:00,880 --> 00:09:04,120
I can now combine the alpha
1 terms and the alpha 2 terms

191
00:09:04,120 --> 00:09:04,650
together.

192
00:09:04,650 --> 00:09:12,430
In other words, c1 plus c3
alpha 1 plus c2 plus c3 alpha 2.

193
00:09:12,430 --> 00:09:14,890
Notice, that since
c1, c2, and c3

194
00:09:14,890 --> 00:09:19,450
are real numbers, c1 plus c3
is the real number. c2 plus c3

195
00:09:19,450 --> 00:09:20,500
is a real number.

196
00:09:20,500 --> 00:09:23,785
Consequently, beta is a
linear combination of alpha 1

197
00:09:23,785 --> 00:09:24,910
and alpha 2.

198
00:09:24,910 --> 00:09:28,700
Therefore, beta is in the space
spanned by alpha 1 and alpha 2.

199
00:09:28,700 --> 00:09:32,740
Therefore, the space spanned by
alpha 1, alpha 2, and alpha 3

200
00:09:32,740 --> 00:09:36,580
is a subspace of the space
spanned by alpha 1 and alpha 2.

201
00:09:36,580 --> 00:09:39,580
And since each of the subspaces
is a subspace of the other,

202
00:09:39,580 --> 00:09:41,770
we see that these
two spaces happen

203
00:09:41,770 --> 00:09:43,545
to be equal in this sense.

204
00:09:43,545 --> 00:09:44,920
And by the way,
if you're looking

205
00:09:44,920 --> 00:09:48,340
for a very quick way of seeing
how this transcends such

206
00:09:48,340 --> 00:09:50,740
a simple application,
notice that when

207
00:09:50,740 --> 00:09:53,830
we were dealing with
differential equations,

208
00:09:53,830 --> 00:09:57,280
we looked for solutions that
were linearly independent.

209
00:09:57,280 --> 00:09:59,440
In other words, we
wanted solutions

210
00:09:59,440 --> 00:10:02,980
that could not be expressed
as linear combinations

211
00:10:02,980 --> 00:10:04,660
of the previous solutions.

212
00:10:04,660 --> 00:10:06,160
And the reason for
that was, if they

213
00:10:06,160 --> 00:10:09,370
could be expressed this
way, the arbitrary constants

214
00:10:09,370 --> 00:10:13,060
weren't independent in the sense
that they could be amalgamated.

215
00:10:13,060 --> 00:10:17,110
At any rate, whichever
interpretation you want here,

216
00:10:17,110 --> 00:10:20,650
notice that this leads to
a generalized definition.

217
00:10:20,650 --> 00:10:25,180
Namely, given the n vectors,
alpha 1 up to alpha n and v,

218
00:10:25,180 --> 00:10:28,210
they are called
linearly dependent.

219
00:10:28,210 --> 00:10:29,980
Now, this may look
like a messy expression

220
00:10:29,980 --> 00:10:31,180
that might frighten you.

221
00:10:31,180 --> 00:10:35,700
All this says is they're called
linearly dependent if one

222
00:10:35,700 --> 00:10:39,660
of the vectors can be written
as a linear combination

223
00:10:39,660 --> 00:10:40,740
of the others.

224
00:10:40,740 --> 00:10:43,170
And by the way, without
proving it here, but leaving

225
00:10:43,170 --> 00:10:48,060
that for the exercises, it
can be shown that not only

226
00:10:48,060 --> 00:10:50,730
when this can be done,
but when it can be done,

227
00:10:50,730 --> 00:10:53,790
the vector can be written
as a linear combination

228
00:10:53,790 --> 00:10:55,870
of the vectors that came before.

229
00:10:55,870 --> 00:10:57,540
In other words, if
any vector, when you

230
00:10:57,540 --> 00:11:00,930
list these in any order, if any
one of the vectors in the given

231
00:11:00,930 --> 00:11:06,150
order can be expressed in
terms of a linear combination

232
00:11:06,150 --> 00:11:08,630
of the preceding
ones, then the set

233
00:11:08,630 --> 00:11:11,340
is called linearly dependent.

234
00:11:11,340 --> 00:11:13,200
If this can't
happen, then they are

235
00:11:13,200 --> 00:11:15,240
called linearly independent.

236
00:11:15,240 --> 00:11:17,400
Now, the reason I give this
definition first-- it's

237
00:11:17,400 --> 00:11:19,290
a bad definition,
because it seems

238
00:11:19,290 --> 00:11:21,780
to depend on the order in
which I list the vectors.

239
00:11:21,780 --> 00:11:23,640
The reason I give
this definition first

240
00:11:23,640 --> 00:11:25,530
is because, I think,
intuitively, it's

241
00:11:25,530 --> 00:11:26,640
easier to see.

242
00:11:26,640 --> 00:11:28,950
In a moment, I will give
you a different definition

243
00:11:28,950 --> 00:11:30,270
that's equivalent to this one.

244
00:11:30,270 --> 00:11:33,540
But first, let me exploit this
fairly intuitive definition.

245
00:11:33,540 --> 00:11:36,630
By means of an
example, suppose I'm

246
00:11:36,630 --> 00:11:42,990
given the four vectors, i plus
j, i minus j, 5i plus j, and i

247
00:11:42,990 --> 00:11:45,420
plus j plus k.

248
00:11:45,420 --> 00:11:47,400
My claim is that
these four vectors

249
00:11:47,400 --> 00:11:50,670
form a linearly-dependent
set, because, in particular,

250
00:11:50,670 --> 00:11:54,150
in the given order,
5i plus j can

251
00:11:54,150 --> 00:11:57,600
be expressed as a linear
combination of i plus j and i

252
00:11:57,600 --> 00:11:59,190
minus j.

253
00:11:59,190 --> 00:12:03,870
Namely, 5i plus j is three
times the first vector here, i

254
00:12:03,870 --> 00:12:07,560
plus j, plus 2 times the
second vector, i minus j.

255
00:12:07,560 --> 00:12:09,660
By the way, in the
next lecture, we'll

256
00:12:09,660 --> 00:12:12,700
show how you pick off these
coefficients very quickly,

257
00:12:12,700 --> 00:12:14,430
but we'll let that
go until next time.

258
00:12:14,430 --> 00:12:16,290
For now, all that's
important is notice

259
00:12:16,290 --> 00:12:19,213
that this vector is a linear
combination of these two.

260
00:12:19,213 --> 00:12:20,880
Notice, by the way,
you could have said,

261
00:12:20,880 --> 00:12:24,890
but couldn't you express i minus
j as a linear combination of i

262
00:12:24,890 --> 00:12:27,420
plus j and 5i plus j?

263
00:12:27,420 --> 00:12:28,410
The answer is, yes.

264
00:12:28,410 --> 00:12:33,510
In other words, if I had written
the 5i plus j and the i minus j

265
00:12:33,510 --> 00:12:35,790
in an interchanged
order, notice that it

266
00:12:35,790 --> 00:12:37,830
would be i minus
j that would have

267
00:12:37,830 --> 00:12:41,710
been a linear combination
of i plus j and 5i plus j.

268
00:12:44,400 --> 00:12:46,650
That's why this is kind
of a hazy definition.

269
00:12:46,650 --> 00:12:50,100
It seems to depend on the
order in which the vectors are

270
00:12:50,100 --> 00:12:50,910
written.

271
00:12:50,910 --> 00:12:54,210
A more conventional
definition is the following.

272
00:12:54,210 --> 00:12:57,210
And just to show you that
dependence and independence are

273
00:12:57,210 --> 00:13:00,600
themselves independent, meaning
you can start with either one,

274
00:13:00,600 --> 00:13:02,640
let me give you an
equivalent definition,

275
00:13:02,640 --> 00:13:06,110
but now stressing linear
independence instead.

276
00:13:06,110 --> 00:13:10,170
Namely, notice that, if I
pick all 0 coefficients,

277
00:13:10,170 --> 00:13:14,040
in other words, 0 alpha 1 plus
a set with 0 alpha n, obviously,

278
00:13:14,040 --> 00:13:18,130
that linear combination
will add up to 0.

279
00:13:18,130 --> 00:13:20,890
If that's the only way you can
make a linear combination add

280
00:13:20,890 --> 00:13:22,990
up to 0, then the
vectors are said

281
00:13:22,990 --> 00:13:24,820
to be linearly independent.

282
00:13:24,820 --> 00:13:27,310
In other words,
alpha 1 up to alpha n

283
00:13:27,310 --> 00:13:31,030
are called linearly
independent if the only way

284
00:13:31,030 --> 00:13:33,880
that a linear combination
of them can equal 0

285
00:13:33,880 --> 00:13:36,950
is if every coefficient is 0.

286
00:13:36,950 --> 00:13:39,970
Again, keep in mind that
the converse is trivial.

287
00:13:39,970 --> 00:13:43,930
If every coefficient is 0,
then, obviously, this will be 0.

288
00:13:43,930 --> 00:13:46,870
For linearly-independent
vectors what we're saying is,

289
00:13:46,870 --> 00:13:51,400
the only way this can be 0 is
if all the coefficients are 0.

290
00:13:51,400 --> 00:13:54,200
And again, to emphasize
linear dependence,

291
00:13:54,200 --> 00:13:55,740
what that would mean is what?

292
00:13:55,740 --> 00:13:58,270
That the vectors would
be linearly dependent

293
00:13:58,270 --> 00:14:00,400
if you could find a
linear combination,

294
00:14:00,400 --> 00:14:04,680
c1 alpha 1 plus, et cetera,
cn alpha n equal to 0,

295
00:14:04,680 --> 00:14:08,230
where at least one of the
cs was not equal to 0.

296
00:14:08,230 --> 00:14:10,780
Now, I'm going to stress
that in the exercises.

297
00:14:10,780 --> 00:14:14,500
This stuff is so new to
many of you, I believe,

298
00:14:14,500 --> 00:14:17,680
that it's very, very difficult
to teach it in great depth

299
00:14:17,680 --> 00:14:18,730
during a lecture.

300
00:14:18,730 --> 00:14:22,630
I think it has to be kept on
a fairly superficial level,

301
00:14:22,630 --> 00:14:25,360
leaving additional
details to the exercises.

302
00:14:25,360 --> 00:14:28,060
Rather than hammer home
these exercises now,

303
00:14:28,060 --> 00:14:31,330
let's simply point out
that linear dependence

304
00:14:31,330 --> 00:14:33,070
implies redundancy.

305
00:14:33,070 --> 00:14:35,080
In other words, if
one of the vectors

306
00:14:35,080 --> 00:14:38,470
can be expressed as linear
combination of the others,

307
00:14:38,470 --> 00:14:42,310
it can be deleted when you're
talking about a set of spanning

308
00:14:42,310 --> 00:14:43,180
vectors.

309
00:14:43,180 --> 00:14:45,190
In other words, going
back to our example

310
00:14:45,190 --> 00:14:48,250
of the space spanned by
alpha 1 up to alpha n,

311
00:14:48,250 --> 00:14:51,010
there are no redundancies
if the alphas

312
00:14:51,010 --> 00:14:53,350
happen to be independent.

313
00:14:53,350 --> 00:14:55,330
But if they're
dependent, it means

314
00:14:55,330 --> 00:14:57,640
that any vector which
can be expressed

315
00:14:57,640 --> 00:14:59,950
as a linear combination
of the others

316
00:14:59,950 --> 00:15:03,580
can be deleted without
losing anything in the space.

317
00:15:03,580 --> 00:15:06,353
And I'll show you that in
terms of other examples later.

318
00:15:06,353 --> 00:15:08,770
But let me show what that means
from another point of view

319
00:15:08,770 --> 00:15:11,750
also that I want to bring
out later in the lecture,

320
00:15:11,750 --> 00:15:13,480
and that is another key point.

321
00:15:13,480 --> 00:15:17,140
That if the alphas happen to
be linearly independent, then

322
00:15:17,140 --> 00:15:20,560
if beta belongs to the
space spanned by the alphas,

323
00:15:20,560 --> 00:15:24,100
not only can beta be written
as a linear combination

324
00:15:24,100 --> 00:15:26,680
of the alphas, but
it can be done so

325
00:15:26,680 --> 00:15:29,620
in one and only one way.

326
00:15:29,620 --> 00:15:32,980
In other words, to say
this in a different form,

327
00:15:32,980 --> 00:15:36,910
if the alphas are linearly
independent, if beta belongs

328
00:15:36,910 --> 00:15:39,550
to the space spanned
by the alphas,

329
00:15:39,550 --> 00:15:43,990
it has a unique representation
as a linear combination

330
00:15:43,990 --> 00:15:45,220
of the alphas.

331
00:15:45,220 --> 00:15:47,380
And what that means from
another point of view

332
00:15:47,380 --> 00:15:50,140
is, that if the alphas
are linearly independent,

333
00:15:50,140 --> 00:15:53,590
the only way that two linear
combinations can be equal

334
00:15:53,590 --> 00:15:56,770
is if they are equal
coefficient by coefficient.

335
00:15:56,770 --> 00:15:59,560
By the way, that was a
technique that we used

336
00:15:59,560 --> 00:16:01,220
in undetermined coefficients.

337
00:16:01,220 --> 00:16:03,340
And again, that shows
you why functions,

338
00:16:03,340 --> 00:16:06,220
like the exponential, sine,
and cosine, et cetera,

339
00:16:06,220 --> 00:16:09,340
tie-in with this generalized
definition of a vector space.

340
00:16:09,340 --> 00:16:11,440
That all of these
things have applications

341
00:16:11,440 --> 00:16:15,550
far beyond any interpretation
due to arrows alone.

342
00:16:15,550 --> 00:16:17,170
But the proof of
something like this,

343
00:16:17,170 --> 00:16:19,810
given that these two linear
combinations are equal,

344
00:16:19,810 --> 00:16:22,940
simply transpose and
collect like terms,

345
00:16:22,940 --> 00:16:25,480
which we can do
algebraically by the axioms

346
00:16:25,480 --> 00:16:27,220
that we're permitted
to work with here.

347
00:16:27,220 --> 00:16:30,520
If we group these
terms, we get this.

348
00:16:30,520 --> 00:16:32,680
Since the alphas are
linearly independent,

349
00:16:32,680 --> 00:16:35,410
the only way a linear
combination can be 0

350
00:16:35,410 --> 00:16:37,570
is if each of the
coefficients is 0.

351
00:16:37,570 --> 00:16:40,090
That means, in particular,
that a1 minus b1

352
00:16:40,090 --> 00:16:44,140
must be 0, et cetera, up
to an minus bn being 0.

353
00:16:44,140 --> 00:16:45,730
And that says, in
addition, what?

354
00:16:45,730 --> 00:16:49,630
That a1 must equal b1,
et cetera, an equals bn,

355
00:16:49,630 --> 00:16:52,330
and that verifies the claim,
that if the alphas are

356
00:16:52,330 --> 00:16:54,640
independent,
linearly independent,

357
00:16:54,640 --> 00:16:57,880
then two different
sets of coefficients

358
00:16:57,880 --> 00:17:00,430
must lead to two
different vectors.

359
00:17:00,430 --> 00:17:03,070
In terms of example
two, you see, notice

360
00:17:03,070 --> 00:17:07,839
that alpha 3 was a linear
combination of alpha 1

361
00:17:07,839 --> 00:17:08,410
and alpha 2.

362
00:17:08,410 --> 00:17:11,490
Remember, alpha 3 was
alpha 1 plus alpha 2.

363
00:17:11,490 --> 00:17:15,099
Notice that, therefore,
I can express alpha 3

364
00:17:15,099 --> 00:17:18,400
in more than one way as a
linear combination of alpha 1,

365
00:17:18,400 --> 00:17:19,750
alpha 2, and alpha 3.

366
00:17:19,750 --> 00:17:24,010
For example, certainly alpha
3 is 0 alpha 1 plus 0 alpha 2

367
00:17:24,010 --> 00:17:25,569
plus 1 alpha 3.

368
00:17:25,569 --> 00:17:26,890
That's just alpha 3.

369
00:17:26,890 --> 00:17:29,170
On the other hand, since
alpha 1 plus alpha 2

370
00:17:29,170 --> 00:17:32,800
is alpha 3, and 1 alpha 1
is alpha 1 and 1 alpha 2

371
00:17:32,800 --> 00:17:35,650
is alpha 2, notice
that alpha 3 can also

372
00:17:35,650 --> 00:17:40,120
be written as 1 alpha 1 plus
1 alpha 2 plus 0 alpha 3.

373
00:17:40,120 --> 00:17:43,160
See, the alphas were linearly
dependent in this example.

374
00:17:43,160 --> 00:17:45,880
Notice that this allows
me to express alpha 3

375
00:17:45,880 --> 00:17:49,150
as two different linear
combinations of alpha 1,

376
00:17:49,150 --> 00:17:51,550
alpha 2, and alpha 3.

377
00:17:51,550 --> 00:17:53,670
By the way, it's
not just these two.

378
00:17:53,670 --> 00:17:56,470
There are instantly many
different ways I can do this.

379
00:17:56,470 --> 00:17:59,950
Namely, since alpha 3
is alpha 1 plus alpha 2,

380
00:17:59,950 --> 00:18:02,260
pick c to be an
arbitrary constant,

381
00:18:02,260 --> 00:18:06,580
and look at the expression c
alpha 1 plus c alpha 2 times

382
00:18:06,580 --> 00:18:08,770
1 minus c alpha 3.

383
00:18:08,770 --> 00:18:13,210
If I expand this, this becomes
c alpha 1 plus c alpha 2

384
00:18:13,210 --> 00:18:14,560
minus c alpha 3.

385
00:18:14,560 --> 00:18:17,100
That's minus c alpha
1 plus alpha 2.

386
00:18:17,100 --> 00:18:20,280
That cancels, and all
I have left is alpha 3.

387
00:18:20,280 --> 00:18:23,250
In other words, alpha 3 can be
written in infinitely many ways

388
00:18:23,250 --> 00:18:26,610
as a linear combination of
alpha 1, alpha 2, and alpha 3,

389
00:18:26,610 --> 00:18:29,070
because the alphas were
linearly dependent.

390
00:18:29,070 --> 00:18:31,080
At any rate, again,
let me emphasize,

391
00:18:31,080 --> 00:18:33,150
I'll drill on this
with the exercises.

392
00:18:33,150 --> 00:18:36,360
What I'd like to do now
is revisit the concept

393
00:18:36,360 --> 00:18:38,880
of dimension, which
we treated in terms

394
00:18:38,880 --> 00:18:42,030
of n-tuples in terms
of spanning vectors

395
00:18:42,030 --> 00:18:44,220
and linear independence.

396
00:18:44,220 --> 00:18:47,250
Namely, let's suppose I
have any vector space now,

397
00:18:47,250 --> 00:18:50,850
not necessarily one written
in terms of n-tuples.

398
00:18:50,850 --> 00:18:53,473
Let's suppose that this other
vectors and the 0 vector

399
00:18:53,473 --> 00:18:55,140
in the vector space,
because, after all,

400
00:18:55,140 --> 00:18:58,500
if the only vector in the
vector space is the 0 vector,

401
00:18:58,500 --> 00:19:00,687
it's not a very
exciting vector space.

402
00:19:00,687 --> 00:19:02,520
So let me assume that
I have something other

403
00:19:02,520 --> 00:19:04,020
than a 0 vector in here.

404
00:19:04,020 --> 00:19:08,790
Let me pick any non-zero vector
in v. Let me call it alpha 1.

405
00:19:08,790 --> 00:19:11,940
And I now-- look at the
space-- span by alpha 1.

406
00:19:11,940 --> 00:19:15,300
Clearly, that's a subspace of v.

407
00:19:15,300 --> 00:19:16,860
On the other hand--

408
00:19:16,860 --> 00:19:18,160
or not on the other hand.

409
00:19:18,160 --> 00:19:20,280
Well, on the other
hand, this might

410
00:19:20,280 --> 00:19:25,590
be all of v. If the space
spanned by alpha 1 is all of v,

411
00:19:25,590 --> 00:19:29,280
we stop at that point and
say, v is one-dimensional,

412
00:19:29,280 --> 00:19:33,390
and it's spanned by the
single vector, alpha 1.

413
00:19:33,390 --> 00:19:36,870
On the other hand, suppose s,
the space spanned by alpha 1,

414
00:19:36,870 --> 00:19:40,210
it's not all of v. Well,
what does that mean?

415
00:19:40,210 --> 00:19:42,160
It means there must
be a vector, say alpha

416
00:19:42,160 --> 00:19:45,540
2, which belongs to v, but
not to the space spanned

417
00:19:45,540 --> 00:19:47,100
by alpha 1.

418
00:19:47,100 --> 00:19:52,830
Well, what we do now is
augment alpha 1 by alpha 2.

419
00:19:52,830 --> 00:19:56,850
And we now look at the space
spanned by alpha 1 and alpha 2.

420
00:19:56,850 --> 00:20:00,750
The first thing I claim is that
the space spanned by alpha 1

421
00:20:00,750 --> 00:20:06,680
and alpha 2 is a proper
superspace of the space spanned

422
00:20:06,680 --> 00:20:07,490
by alpha 1.

423
00:20:07,490 --> 00:20:09,950
In other words, it's a
space spanned by alpha 1.

424
00:20:09,950 --> 00:20:12,920
It's contained in but not equal
to the space spanned by alpha 1

425
00:20:12,920 --> 00:20:14,030
and alpha 2.

426
00:20:14,030 --> 00:20:15,230
Why is this?

427
00:20:15,230 --> 00:20:18,260
Well, for one thing,
alpha 2 by definition

428
00:20:18,260 --> 00:20:21,530
didn't belong to the
space spanned by alpha 1.

429
00:20:21,530 --> 00:20:23,870
See, we picked alpha 2 not
to be in the space spanned

430
00:20:23,870 --> 00:20:25,010
by alpha 1.

431
00:20:25,010 --> 00:20:27,920
On the other hand, alpha 2 does
belong to the space spanned

432
00:20:27,920 --> 00:20:30,490
by alpha 1 and alpha 2.

433
00:20:30,490 --> 00:20:34,010
You see, notice that the space
spanned by alpha 1 and alpha 2

434
00:20:34,010 --> 00:20:37,060
consists of all linear
combinations of alpha 1

435
00:20:37,060 --> 00:20:39,770
and alpha 2, and in
particular, alpha 2

436
00:20:39,770 --> 00:20:42,430
can be written as 0
alpha 1 plus 1 alpha

437
00:20:42,430 --> 00:20:45,035
2, which makes it a linear
combination of alpha 1

438
00:20:45,035 --> 00:20:46,010
and alpha 2.

439
00:20:46,010 --> 00:20:49,340
So alpha 2 does not belong to
the space spanned by alpha 1,

440
00:20:49,340 --> 00:20:54,470
but it does belong to the space
spanned by alpha 1 and alpha 2.

441
00:20:54,470 --> 00:20:58,280
Notice, therefore, that we
have now enlarged our space.

442
00:20:58,280 --> 00:21:02,750
Notice also that alpha 1 alpha
2 must be linearly independent.

443
00:21:02,750 --> 00:21:05,540
In other words, we
chose any alpha 2,

444
00:21:05,540 --> 00:21:08,730
provided only that it belonged
to v but not to the space

445
00:21:08,730 --> 00:21:10,340
spanned by alpha 1.

446
00:21:10,340 --> 00:21:13,190
But that alpha 2, no
matter how we choose it,

447
00:21:13,190 --> 00:21:15,980
must be linearly
independent of alpha 1,

448
00:21:15,980 --> 00:21:18,650
because, after all, if they
were linearly dependent,

449
00:21:18,650 --> 00:21:22,070
that would mean that alpha 2 is
a scalar multiple as alpha 1.

450
00:21:22,070 --> 00:21:25,005
But that's a contradiction,
because if alpha 2 were

451
00:21:25,005 --> 00:21:27,680
a scalar multiple of
alpha 1, by definition

452
00:21:27,680 --> 00:21:30,470
it would belong to the
space spanned by alpha 1.

453
00:21:30,470 --> 00:21:34,100
But we specifically chose it
not to belong to that space.

454
00:21:34,100 --> 00:21:36,560
At any rate, we
continue on in this way.

455
00:21:36,560 --> 00:21:40,580
Namely, the space spanned
by alpha 1 and alpha 2

456
00:21:40,580 --> 00:21:43,130
might be all of v,
in which case we then

457
00:21:43,130 --> 00:21:47,000
say v is two-dimensional and
that it's spanned by alpha 1

458
00:21:47,000 --> 00:21:48,170
and alpha 2.

459
00:21:48,170 --> 00:21:52,100
Or it may be that the space
spanned by alpha 1 and alpha 2

460
00:21:52,100 --> 00:21:55,700
is a proper subset of v and
doesn't give me all of v.

461
00:21:55,700 --> 00:21:58,520
In this event, I
now pick an alpha 3,

462
00:21:58,520 --> 00:22:02,120
which belongs to v, but not to
the space spanned by alpha 1

463
00:22:02,120 --> 00:22:04,070
and alpha 2.

464
00:22:04,070 --> 00:22:07,370
Again, notice that alpha
1, alpha 2, and alpha 3

465
00:22:07,370 --> 00:22:10,970
must be linearly independent
since if alpha 3 could

466
00:22:10,970 --> 00:22:12,440
be expressed as a linear--

467
00:22:12,440 --> 00:22:14,420
see, we know alpha
2 can't be expressed

468
00:22:14,420 --> 00:22:16,580
as a linear
combination of alpha 1,

469
00:22:16,580 --> 00:22:20,030
because alpha 2 was linearly
independent of alpha 1.

470
00:22:20,030 --> 00:22:21,560
Now we tack on alpha 3.

471
00:22:21,560 --> 00:22:24,860
The only possibility is is
alpha 3 a linear combination

472
00:22:24,860 --> 00:22:26,570
of alpha 1 and alpha 2?

473
00:22:26,570 --> 00:22:28,400
And the answer is,
it couldn't be.

474
00:22:28,400 --> 00:22:31,370
Because if alpha 3 were a
linear combination of alpha 1

475
00:22:31,370 --> 00:22:34,040
and alpha 2, by
definition it would

476
00:22:34,040 --> 00:22:37,760
belong to the space spanned
by alpha 1 and alpha 2.

477
00:22:37,760 --> 00:22:42,050
But by construction, alpha 3 was
chosen not to be in that space.

478
00:22:42,050 --> 00:22:44,630
Consequently, if alpha
3 were in that space,

479
00:22:44,630 --> 00:22:46,760
that would obviously
be a contradiction.

480
00:22:46,760 --> 00:22:50,180
In other words, alpha 3
cannot be a linear combination

481
00:22:50,180 --> 00:22:51,650
of alpha 1 and alpha 2.

482
00:22:51,650 --> 00:22:54,710
Therefore, these three vectors
are linearly independent.

483
00:22:54,710 --> 00:22:57,550
And similarly, alpha 3,
not belonging to the space

484
00:22:57,550 --> 00:23:01,010
spanned by alpha 1 and
alpha 2, enlarges the space

485
00:23:01,010 --> 00:23:03,350
spanned by alpha 1 and alpha 2.

486
00:23:03,350 --> 00:23:05,750
In terms of a
rough picture here,

487
00:23:05,750 --> 00:23:08,930
what we're saying is, we start
with some vector, alpha 1.

488
00:23:08,930 --> 00:23:11,600
s of alpha 1 is some subspace.

489
00:23:11,600 --> 00:23:14,660
We put an alpha 2, which doesn't
belong to the space spanned

490
00:23:14,660 --> 00:23:16,250
by alpha 1.

491
00:23:16,250 --> 00:23:19,100
Then the space spanned
by alpha 1 and alpha 2

492
00:23:19,100 --> 00:23:22,280
is this larger
subspace, which contains

493
00:23:22,280 --> 00:23:24,590
the space spanned by alpha 1.

494
00:23:24,590 --> 00:23:28,850
If that isn't all of v, we
pick a third vector, alpha 3,

495
00:23:28,850 --> 00:23:32,150
which isn't in the space
spanned by alpha 1 and alpha 2.

496
00:23:32,150 --> 00:23:35,030
And we then form the space
spanned by alpha 1, alpha 2,

497
00:23:35,030 --> 00:23:38,885
and alpha 3, which will contain
the space spanned by alpha 1

498
00:23:38,885 --> 00:23:41,000
and alpha 2 as a subspace.

499
00:23:41,000 --> 00:23:45,890
And to make a long story short,
we simply continue in this way

500
00:23:45,890 --> 00:23:50,482
until we find alpha
1, say, up to alpha n.

501
00:23:50,482 --> 00:23:53,480
In other words, we continue
for n steps this way,

502
00:23:53,480 --> 00:23:57,470
such that v will eventually be
the space spanned by alpha 1

503
00:23:57,470 --> 00:23:58,680
up to alpha n.

504
00:23:58,680 --> 00:24:01,260
And let me emphasize that
this need not happen.

505
00:24:01,260 --> 00:24:05,210
In other words, v might be
such a huge vector space,

506
00:24:05,210 --> 00:24:08,840
that continuing in this way,
we could go on forever and not

507
00:24:08,840 --> 00:24:10,940
exhaust all of v this way.

508
00:24:10,940 --> 00:24:12,890
But suppose, for the
sake of argument,

509
00:24:12,890 --> 00:24:16,250
that this does happen,
that we find n alphas such

510
00:24:16,250 --> 00:24:21,410
that v is spanned by
alpha 1 up to alpha n,

511
00:24:21,410 --> 00:24:23,330
where the alphas
are constructed,

512
00:24:23,330 --> 00:24:25,650
as I've indicated previously.

513
00:24:25,650 --> 00:24:31,040
In this case, we say that v has
dimension n written this way,

514
00:24:31,040 --> 00:24:33,860
and the set of vectors,
alpha 1 up to alpha n,

515
00:24:33,860 --> 00:24:36,970
is called a basis
for v. And the reason

516
00:24:36,970 --> 00:24:39,860
that it's called the basis
for v is that not only can

517
00:24:39,860 --> 00:24:42,800
every vector be written
as a linear combination

518
00:24:42,800 --> 00:24:45,080
of the alphas, every
vector in v written

519
00:24:45,080 --> 00:24:47,270
as a linear combination
of the alphas,

520
00:24:47,270 --> 00:24:49,850
but it has a unique
representation

521
00:24:49,850 --> 00:24:52,460
in terms of the alphas,
because the alphas were

522
00:24:52,460 --> 00:24:53,630
linearly independent.

523
00:24:53,630 --> 00:24:58,460
In other words, once you have a
basis, you see, any vector in v

524
00:24:58,460 --> 00:25:02,000
can be written uniquely
as an n-tuple where

525
00:25:02,000 --> 00:25:06,240
the coefficients stand for
the coefficients of alpha 1

526
00:25:06,240 --> 00:25:07,950
up to alpha n.

527
00:25:07,950 --> 00:25:09,720
You see, in other
words, each v and v

528
00:25:09,720 --> 00:25:14,250
has a unique representation,
unique representation,

529
00:25:14,250 --> 00:25:17,370
as a linear combination
of alpha 1 up to alpha n.

530
00:25:17,370 --> 00:25:20,400
By the way, if this process
doesn't come to an end,

531
00:25:20,400 --> 00:25:24,330
then v is set to be an
infinite-dimensional vector

532
00:25:24,330 --> 00:25:25,830
space.

533
00:25:25,830 --> 00:25:29,310
Now, what I'd like to do is
take a few minutes to summarize

534
00:25:29,310 --> 00:25:30,750
this particular lecture.

535
00:25:30,750 --> 00:25:32,790
Because what we're going
to be doing between now

536
00:25:32,790 --> 00:25:36,090
and the next lecture is going
to involve a lot of work

537
00:25:36,090 --> 00:25:37,230
on your part.

538
00:25:37,230 --> 00:25:39,990
And that is, that in going
through this overview of what

539
00:25:39,990 --> 00:25:42,480
you mean by dimension, and
spanning sets, and what

540
00:25:42,480 --> 00:25:45,210
have you, and linear
dependence and independence,

541
00:25:45,210 --> 00:25:47,620
many subtleties have occurred.

542
00:25:47,620 --> 00:25:51,060
For example, if I have an
n-dimensional vector space

543
00:25:51,060 --> 00:25:53,820
going through this bit of
constructing the space spanned

544
00:25:53,820 --> 00:25:54,750
by alpha--

545
00:25:54,750 --> 00:25:57,660
alpha 1 and alpha 2, et
cetera-- questions come up.

546
00:25:57,660 --> 00:25:59,850
What if I had started
with different vectors?

547
00:25:59,850 --> 00:26:02,940
What if I had started with
beta 1 instead of alpha 1,

548
00:26:02,940 --> 00:26:06,540
and then picked a new vector,
beta 2 instead of alpha 2?

549
00:26:06,540 --> 00:26:11,520
Could I have spanned v
in fewer than n vectors?

550
00:26:11,520 --> 00:26:14,190
In other words,
does the dimension

551
00:26:14,190 --> 00:26:17,610
depend on the process by
which I span the space by how

552
00:26:17,610 --> 00:26:18,600
I choose the vectors?

553
00:26:18,600 --> 00:26:23,040
Obviously, if this happens to
be true, we're in dire trouble.

554
00:26:23,040 --> 00:26:25,350
Because we would like
to believe intuitively,

555
00:26:25,350 --> 00:26:27,870
based on our old
definition of dimension,

556
00:26:27,870 --> 00:26:29,370
that the dimension
of a vector space

557
00:26:29,370 --> 00:26:31,470
should not depend on
the representative

558
00:26:31,470 --> 00:26:32,430
vectors that you pick.

559
00:26:32,430 --> 00:26:35,580
This is what's bad about
the n-tuple representation.

560
00:26:35,580 --> 00:26:39,480
So you see, what we're going
to do in the exercises is show

561
00:26:39,480 --> 00:26:42,660
what other ingredients
come up in dimension

562
00:26:42,660 --> 00:26:44,280
and what we have to worry about.

563
00:26:44,280 --> 00:26:47,340
In our next lecture, we
will summarize the results

564
00:26:47,340 --> 00:26:49,350
that we will prove
in the exercises.

565
00:26:49,350 --> 00:26:50,880
So you can begin
the next lecture

566
00:26:50,880 --> 00:26:53,850
even if you're a little bit
hazy on some of the exercises.

567
00:26:53,850 --> 00:26:56,650
And then we will show, from
a practical point of view,

568
00:26:56,650 --> 00:26:59,040
once the theory is
developed, how do we actually

569
00:26:59,040 --> 00:27:01,980
go about constructing a basis.

570
00:27:01,980 --> 00:27:04,750
How do we find what a
dimension is in real life?

571
00:27:04,750 --> 00:27:08,670
How do we figure out how to
construct the alphas if we're

572
00:27:08,670 --> 00:27:11,370
not given the abstract
situation mentioned

573
00:27:11,370 --> 00:27:12,630
in this particular example?

574
00:27:12,630 --> 00:27:16,020
At any rate, all I want to
do now is review the lecture,

575
00:27:16,020 --> 00:27:19,260
do the exercises, start to
feel at home with the concepts,

576
00:27:19,260 --> 00:27:22,890
and we will drill more on this
and review in our next lecture.

577
00:27:22,890 --> 00:27:24,760
At any rate, then,
until next time.

578
00:27:24,760 --> 00:27:25,260
Goodbye.

579
00:27:28,350 --> 00:27:30,750
Funding for the
publication of this video

580
00:27:30,750 --> 00:27:35,640
was provided by the Gabriella
and Paul Rosenbaum Foundation.

581
00:27:35,640 --> 00:27:39,780
Help OCW continue to provide
free and open access to MIT

582
00:27:39,780 --> 00:27:45,215
courses by making a donation
at ocw.mit.edu/donate.