1
00:00:16,570 --> 00:00:24,099
in a combination set of vectors and so on
so i started by defining the set of vectors
2
00:00:24,099 --> 00:00:27,170
and i had a special notation for it
3
00:00:27,170 --> 00:00:39,989
so today we are going to look at specific
things like so basis and dimension of a vector
4
00:00:39,989 --> 00:00:49,370
space or a subspace and other than that just
giving definitions i want to give some examples
5
00:00:49,370 --> 00:00:56,719
for it so that why these ideas are being pursued
should become a little bit clear? why there
6
00:00:56,719 --> 00:01:03,190
we get these funny spaces look at set of all
continuous functions and then call it a space
7
00:01:03,190 --> 00:01:04,629
and why we look at set of polynomials and
call it a subspace
8
00:01:04,629 --> 00:01:17,190
so yesterday we had developed this notation
this notation was a set s let us say was defined
9
00:01:17,190 --> 00:01:25,430
as xi where i=1 2 this would be a finite set
in some situations this could be an infinite
10
00:01:25,430 --> 00:01:35,780
set i am currently defining a finite set of
n vectors which means there are n vectors
11
00:01:35,780 --> 00:02:01,719
x belongs to some vectors so x is my vector
space and s is the subset of x
12
00:02:01,719 --> 00:02:14,800
so one example that i gave was from r5 so
i said that you know 2 vectors say x1 this
13
00:02:14,800 --> 00:02:31,170
was 1 2 3 4 5 and x2 which is 5 4 3 2 1 and
putting some vectors from r5 so i can define
14
00:02:31,170 --> 00:02:52,850
the set s which is x1 and x2 these are 2 elements
in the set where s is subset of r5 okay
15
00:02:52,850 --> 00:03:07,490
in some cases i will define a set s to be
set of polynomials i define a set s which
16
00:03:07,490 --> 00:03:27,180
is set of polynomials so my first vector here
my first vector x1z is 1 x2z so in this particular
17
00:03:27,180 --> 00:03:36,579
set i have n+1 vectors which are denominated
like this this is the first function this
18
00:03:36,579 --> 00:03:43,090
is the second function x2z second vector okay
this is the third vector this is the n+1th
19
00:03:43,090 --> 00:03:52,910
vector and so on so the notation here to generate
could be used in a context of n-dimensional
20
00:03:52,910 --> 00:03:53,910
spaces
21
00:03:53,910 --> 00:03:58,210
it will be used in the context of function
space and it will be used in a context of
22
00:03:58,210 --> 00:04:04,300
polynomials depends upon what context they
are okay so that is first thing then yesterday
23
00:04:04,300 --> 00:04:13,500
the segment that we talked about linear combinations
so f is my finite and if i take some scalars
24
00:04:13,500 --> 00:04:22,930
say alpha 1 alpha 2 alpha 3 which are the
scalar set this belongs to f
25
00:04:22,930 --> 00:04:35,120
if this scalar set belongs to f then we could
find a set of vector which is linear combination
26
00:04:35,120 --> 00:04:39,180
which is linear combination of the original
vectors
27
00:04:39,180 --> 00:04:50,450
so i can construct the new vector x which
is alpha 1 x1+alpha 2 x2 and so on + alpha
28
00:04:50,450 --> 00:04:58,420
n xn so this linear combination in the later
example where we took vectors as polynomials
29
00:04:58,420 --> 00:05:08,870
would be a nth order polynomial okay then
for n-1 polynomial the vector to the n in
30
00:05:08,870 --> 00:05:15,230
the case of n dimension it will be a vector
in alpha n okay
31
00:05:15,230 --> 00:05:30,590
i can talk about a new vector x which is alpha
1 1 2 3 4 5 and alpha 2 5 4 3 2 1 when alpha
32
00:05:30,590 --> 00:05:41,690
1 and alpha 2 are any 2 arbitrary scalars
drawn from r okay this is linear combination
33
00:05:41,690 --> 00:05:45,940
and then we said that we have this notion
of span of a set
34
00:05:45,940 --> 00:05:55,990
the span of a set is set of all possible linear
combination span of s which is many times
35
00:05:55,990 --> 00:06:05,900
denoted as square bracket s so we take all
possible combinations of elements of s and
36
00:06:05,900 --> 00:06:13,770
then what i get is span of if i take all possible
linear combinations in this set then what
37
00:06:13,770 --> 00:06:18,110
i get is span of this particular set it is
also example in 3 dimensions i took 2 vectors
38
00:06:18,110 --> 00:06:20,710
and said span of these 2 vectors will be a
plane cutting through the origin
39
00:06:20,710 --> 00:06:28,320
if i take 3 vectors in the same plane span
will be again plane cutting through the origin
40
00:06:28,320 --> 00:06:36,090
that is not going to be different well this
sounds very abstract let us get on to some
41
00:06:36,090 --> 00:06:39,820
concrete examples which probably you have
seen in your undergraduate and then let us
42
00:06:39,820 --> 00:06:41,720
connect why do i need span? why do i need
to have of spaces?
43
00:06:41,720 --> 00:06:50,080
why do i need them? and then we will move
on to this concept of linear independence
44
00:06:50,080 --> 00:06:53,610
and linear dependence and then talk about
basis before i do that let us get some concepts
45
00:06:53,610 --> 00:07:05,430
clear about this span so why do we need span?
why it is used? why it is important? so something
46
00:07:05,430 --> 00:07:13,180
that we actually keep using all the time these
concepts are used when you solve linear algebraic
47
00:07:13,180 --> 00:07:15,050
equations differential equations
48
00:07:15,050 --> 00:07:19,210
you will keep doing it unknowingly without
understanding what eventually happened because
49
00:07:19,210 --> 00:07:26,030
some of these concepts are not introduced
in undergraduate let me start with some example
50
00:07:26,030 --> 00:07:28,490
which is simple solving linear algebraic equations
51
00:07:28,490 --> 00:07:39,680
so my example 1 is i want to solve ax=b where
a is an invertible matrix so i am taking one
52
00:07:39,680 --> 00:08:12,699
single matrix here 1 1 0 0 1 1 1 0 1*x1 x2
x3 gives me b1 b2 b3 okay actually this equation
53
00:08:12,699 --> 00:08:25,699
i can rewrite slightly differently you will
start getting insights to why i am talking
54
00:08:25,699 --> 00:08:42,909
about this span i can rewrite this as 1 1
0*x1+0 1 1*x2+1 0 1*x3=b1 b2 b3 am i correct?
55
00:08:42,909 --> 00:08:50,580
i am just rewriting this equation
56
00:08:50,580 --> 00:09:03,990
here b is
a vector which is linear combination of first
57
00:09:03,990 --> 00:09:15,710
vector second column vector and third column
vector okay what is span of column vector
58
00:09:15,710 --> 00:09:25,770
1 column vector 2 column vector 3? what is
the span? the span of these 3 vectors i want
59
00:09:25,770 --> 00:09:35,230
to show is same as 3-dimensional vector space
what is span? all possible linear approximations
60
00:09:35,230 --> 00:09:37,760
are these vectors linearly independent?
61
00:09:37,760 --> 00:09:45,339
by inspection for this particular case you
know about linear algebra you can see that
62
00:09:45,339 --> 00:09:51,610
these 3 vectors are linearly independent what
will be span? what will be all possible linear
63
00:09:51,610 --> 00:09:59,580
combinations of 3 linearly independent vectors
in 3 dimensions okay
64
00:09:59,580 --> 00:10:10,930
so i can write the solution which is x this
is a inverse b well i am not right now worried
65
00:10:10,930 --> 00:10:24,140
about how inverse i suppose you know this
and this a inverse b comes out to be b1 times
66
00:10:24,140 --> 00:10:39,740
1/2 -1/2 1/2+b2 times 1/2 1/2 -1/2+b3 times
-1/2 1/2 1/2 if i actually compute a solution
67
00:10:39,740 --> 00:10:50,390
of this system of equations okay we can do
it very easily by gauss elimination gauss-legendre
68
00:10:50,390 --> 00:10:52,690
method and compute inverse okay
69
00:10:52,690 --> 00:11:00,530
i am just rewriting the solution in terms
of a inverse b except that i have chosen to
70
00:11:00,530 --> 00:11:10,630
write like this i have chosen to write like
this okay give me any b vector i can write
71
00:11:10,630 --> 00:11:22,839
the solution like this okay so where does
the solution belong to? the solution x belongs
72
00:11:22,839 --> 00:11:29,130
to span of this vector this vector and this
vector again from undergraduate experience
73
00:11:29,130 --> 00:11:34,540
will know that these 3 vectors are linearly
independent okay
74
00:11:34,540 --> 00:11:52,370
so the solution actually belongs to the span
of these vectors okay now this is a well behaved
75
00:11:52,370 --> 00:12:00,960
example of matrix a is invertible i am going
to do another example if matrix a is non-invertible
76
00:12:00,960 --> 00:12:09,339
i still want to find a solution and then we
will get this concept of the solution belonging
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00:12:09,339 --> 00:12:16,280
to a space which is not in this particular
case what is the solution space?
78
00:12:16,280 --> 00:12:33,089
so x belongs to span of 1/2 -1/2 1/2*1*1/2
1/2 -1/2*1*-1/2 1/2 1/2 the solution actually
79
00:12:33,089 --> 00:12:46,550
belongs to span of this all possible linear
combinations right if you specify b i will
80
00:12:46,550 --> 00:12:58,560
get one particular x how many ways i can specify
b? infinite possible ways b1 b2 b3 can be
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00:12:58,560 --> 00:13:06,870
chosen all vectors are put here i will get
one particular solution so this is an example
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00:13:06,870 --> 00:13:12,490
where you have to look at the span
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00:13:12,490 --> 00:13:22,370
now let us look another example then the things
will become clearer okay i am just going to
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00:13:22,370 --> 00:13:38,720
change this matrix to slightly different matrix
1 2 -4 -1 -2 4 and 2 4 -8 and then i am going
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00:13:38,720 --> 00:13:51,730
to look for a solution which is for 0 0 0
i am going to look for a solution which is
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00:13:51,730 --> 00:14:10,620
0 0 0 well so basically i want to solve for
1 -1 2 3 -2 4 and -4 4 -8 this becomes 0 0
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00:14:10,620 --> 00:14:20,800
0 is this matrix invertible first of all?
“professor - student conversation starts”
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00:14:20,800 --> 00:14:23,350
no sir why?
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00:14:23,350 --> 00:14:31,470
the columns are linearly dependent “professor
- student conversation ends” i just multiplied
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00:14:31,470 --> 00:14:33,570
this column by a vector and just multiplied
this column by another vector i will get this
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00:14:33,570 --> 00:14:42,070
column okay is this problem solving? “professor
- student conversation starts” this is going
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00:14:42,070 --> 00:14:57,850
to 0? yes why? i told why not what 3 unknowns
3 equations yeah linearly dependence
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00:14:57,850 --> 00:15:10,960
so they are linearly dependent and i should
be able to get infinite solutions “professor
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00:15:10,960 --> 00:15:22,840
- student conversation ends” so this matrix
is non-invertible so as the columns are linearly
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00:15:22,840 --> 00:15:30,310
dependent this matrix is non-invertible can
i find the solution? what are the possible
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00:15:30,310 --> 00:15:47,050
solutions to this problem? so what about 2
i am not interested in 0 0 0 i am not interested
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00:15:47,050 --> 00:15:54,560
in trivial solution what i want to know is
are there nonlinear solutions
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00:15:54,560 --> 00:16:01,679
“professor - student conversation starts”
yeah so tell me the solution “professor
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00:16:01,679 --> 00:16:12,570
- student conversation ends” so what about
x2=2 x1=-1 and x3=0 will this form a solution?
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00:16:12,570 --> 00:16:22,740
yeah right multiply this by 2 multiply this
by -1 okay i will get 0 the last column is
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00:16:22,740 --> 00:16:31,890
multiplied by 0 other candidate solution of
course is multiplied by 4 multiply this by
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00:16:31,890 --> 00:16:39,220
0 and multiply this by 1 right i get 2 possible
candidate solutions
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00:16:39,220 --> 00:16:46,399
here my first solution well remember these
are linearly dependent columns which means
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00:16:46,399 --> 00:16:51,730
there are same equations repeated again multiplied
by scalars okay i am not taking the equation
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00:16:51,730 --> 00:16:55,660
view point i am taking the column view point
i am not taking the row view point what is
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00:16:55,660 --> 00:17:00,790
different between row viewpoint and column
viewpoint will come to that little later okay
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00:17:00,790 --> 00:17:09,390
so now i have 2 possible solutions i have
1 solution which is x1 so i would write this
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00:17:09,390 --> 00:17:26,579
as 2 -1 0 and have x2 which is 4 0 1 okay
i found 2 possible equations that are infinite
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00:17:26,579 --> 00:17:32,879
solutions how do i get 2 different solutions?
“professor - student conversation starts”
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00:17:32,879 --> 00:17:38,509
any linear combination “professor - student
conversation ends” so if i construct the
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00:17:38,509 --> 00:17:58,649
vector which is alpha times 2 -1 0 and beta
times 4 0 1 just check this also will be a
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00:17:58,649 --> 00:17:59,649
solution
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00:17:59,649 --> 00:18:06,630
because a times this vector is 0 +beta times
this vector 0 so you know the solutions at
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00:18:06,630 --> 00:18:17,429
0 so a vector which is what do i mean by any
vector which is span of these so what i want
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00:18:17,429 --> 00:18:29,379
to say here is that a solution lies in the
span of these true vectors are these linearly
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00:18:29,379 --> 00:18:38,739
independent? yes are linearly independent
and the solution lies in the span on these
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00:18:38,739 --> 00:18:44,700
2 vectors what is the dimension of space created
by the span of these?
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00:18:44,700 --> 00:18:49,380
“professor - student conversation starts”
two because there are 2 linearly independent
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00:18:49,380 --> 00:18:53,590
vectors all possible linear combinations of
these 2 vectors will give rise to a 2-dimensional
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00:18:53,590 --> 00:19:02,160
subspace of 3 dimensions okay “professor
- student conversation ends” will give rise
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00:19:02,160 --> 00:19:10,529
to a 2-dimensional subspace of all 3 okay
so span is something which you have been using
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00:19:10,529 --> 00:19:12,080
without knowing what is happening now
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00:19:12,080 --> 00:19:22,520
let us graduate from these examples to differential
equations okay so that we get a feel of how
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00:19:22,520 --> 00:19:31,710
we can generalize these concepts to some other
spaces that what we have been looking for
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00:19:31,710 --> 00:19:32,710
okay
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00:19:32,710 --> 00:19:41,120
so my third example i have already covered
2 examples so third example is not ax=b well
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00:19:41,120 --> 00:19:44,410
conceptually it is not different to ax=b but
you know take some time to realize that what
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00:19:44,410 --> 00:20:06,019
i am talking about is 1 and the same thing
okay let us look at this example so now i
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00:20:06,019 --> 00:20:12,809
want to look at i want a transition to example
which is different from linear algebraic equations
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00:20:12,809 --> 00:20:18,280
i am looking at differential equations while
this equation is not too different from ax=b
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00:20:18,280 --> 00:20:28,210
a here would be this differential operator
operating on vector u and giving me 0 vector
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00:20:28,210 --> 00:20:55,979
okay so this is exactly my ax=0 except i have
to write as d3/dz cube+6 d2/dz square+11 d/dz+6
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00:20:55,979 --> 00:21:07,000
this operator operating on uz is=0 actually
0 is the 0 vector in this space what is this
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00:21:07,000 --> 00:21:12,700
funny space? i have not talked about it earlier
this is set of twice differentiable continuous
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00:21:12,700 --> 00:21:16,269
functions
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00:21:16,269 --> 00:21:30,269
all intervals 0 to 1 so z belongs to this
is the set of 0 to 1 and this particular set
137
00:21:30,269 --> 00:21:33,850
is thrice differential continuous function
c3 0 to 1 is thrice differentiable continuous
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00:21:33,850 --> 00:21:39,460
function while you know how to solve this
problem i have given the initial conditions
139
00:21:39,460 --> 00:21:46,500
i want to get a general solution i want to
get a general solution for this problem how
140
00:21:46,500 --> 00:22:01,509
do you do that? you write a calculus in equation
mode and then you write the solution
141
00:22:01,509 --> 00:22:09,870
so let us write all what you know the characteristic
equation is will be p3+6p square+11 p+6=0
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00:22:09,870 --> 00:22:20,450
and then i can write p+1 p+2 p+3=0 so i have
3 routes -1 -2 -3 and all of you know how
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00:22:20,450 --> 00:22:25,320
to write a solution but what i want to point
out here is something different i am going
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00:22:25,320 --> 00:22:32,259
to talk about this span okay so the general
solution to this particular problem is given
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00:22:32,259 --> 00:22:48,210
by uz=alpha e to the power –z+beta e to
the power -2z+gamma e to the power -3z okay
146
00:22:48,210 --> 00:23:06,990
look at these 3 vectors where are the vectors
coming here? the vectors are 1 2 3 so this
147
00:23:06,990 --> 00:23:24,309
is my x1z this is my x2z and this is my x3z
okay so a general solution to this problem
148
00:23:24,309 --> 00:23:39,210
it belongs to span of all possible linear
combinations what is this alpha beta and gamma?
149
00:23:39,210 --> 00:23:41,129
the moment we specify initial conditions alpha
beta gamma will get this
150
00:23:41,129 --> 00:23:50,030
well if you have not done this we will be
doing these solution little later this is
151
00:23:50,030 --> 00:23:52,799
the linear combination of these 3 vectors
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00:23:52,799 --> 00:24:05,409
so i can say that uz actually belongs to span
of e to the power –z e to the power -2z
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00:24:05,409 --> 00:24:18,830
and e to the power -3z okay so span is something
which i need to know; something that i need
154
00:24:18,830 --> 00:24:28,799
to know; if you remember there are 2 solutions
to our problem of this side if there is something
155
00:24:28,799 --> 00:24:36,309
on the right hand side; right if there is
something on the right hand side it is a forcing
156
00:24:36,309 --> 00:24:37,309
function
157
00:24:37,309 --> 00:24:42,499
then you have a particular solution and you
have general solution which is; the general
158
00:24:42,499 --> 00:24:48,739
solution will be this then we have a particular
solution depending upon the forcing function
159
00:24:48,739 --> 00:24:54,720
the general solution is actually governed
by the initial condition and span of these
160
00:24:54,720 --> 00:24:55,720
characteristic vectors
161
00:24:55,720 --> 00:25:04,450
any solution; any solution which is where
forcing function in 0 will always lie in the
162
00:25:04,450 --> 00:25:09,999
span of these 3 vectors; will always lie in
the span of these 3 vectors you cannot leave
163
00:25:09,999 --> 00:25:15,869
this; solution of this cannot leave this sub
space what is this? this is a; what is this
164
00:25:15,869 --> 00:25:21,059
is the subspace right is it a subspace? it
is a sub space
165
00:25:21,059 --> 00:25:34,330
all possible linear combinations of these
3 is a 3 dimensional subspace
166
00:25:34,330 --> 00:25:40,279
what is the underlying space which we are
talking about? the underlying space which
167
00:25:40,279 --> 00:25:45,110
we are talking about is thrice differentiable
continuous functions each one of them is thrice
168
00:25:45,110 --> 00:25:55,190
differentiable; do you agree? all of them
are thrice differentiable okay first of all
169
00:25:55,190 --> 00:26:01,549
i wanted to understand that a operating on
a finite dimensional vector x will be 0 zero
170
00:26:01,549 --> 00:26:08,970
and this operator operating on this vector
belonging to this space or not wonder will
171
00:26:08,970 --> 00:26:12,950
be different ideas they are same things; geometrically
same things nothing different if you start
172
00:26:12,950 --> 00:26:15,849
doing; “professor-student conversation starts”
sir how can we visualize these are vectors;
173
00:26:15,849 --> 00:26:23,720
i told you you can visualize these spaces
you can visualise only in 3 dimensions but
174
00:26:23,720 --> 00:26:31,470
believe me that if you understand in 3 dimension
what is happening very well okay
175
00:26:31,470 --> 00:26:38,729
we talked about a 2 dimensional subspace of
r3 just now right okay in the same sense see
176
00:26:38,729 --> 00:26:46,809
what is the 2 dimensional subspaces r3 is
it a very thin set? there are too many vectors
177
00:26:46,809 --> 00:26:50,609
in r3 right and then r2 is the plane passing
through the origin so very thin set pure vectors
178
00:26:50,609 --> 00:26:54,029
relatively smaller infinity than this bigger
infinity right now in the same sense well
179
00:26:54,029 --> 00:27:02,080
i can construct a vectors in this space which
are you know -2 -3 -4 -5 -7 all of them belong
180
00:27:02,080 --> 00:27:03,080
to continuous functions right
181
00:27:03,080 --> 00:27:04,700
actually we will come to this idea of basis
and what will show is that this particular
182
00:27:04,700 --> 00:27:12,500
space is an infinite dimensional space there
are infinite vectors in the spaces; linearly
183
00:27:12,500 --> 00:27:19,399
dependent vectors that we can find in this
particular space are infinite let the right
184
00:27:19,399 --> 00:27:34,359
hand side not be 0 this will not be a sub
space? which one? the solution; no no no you
185
00:27:34,359 --> 00:27:38,659
will get a specific solution who might i specify
an initial condition
186
00:27:38,659 --> 00:27:50,549
if i specify initial condition u0 is = something
say alpha no no take a and then u dash 0 = b
187
00:27:50,549 --> 00:27:57,259
and u double prime 0 is = some c moment i
give you these 3 initial conditions these
188
00:27:57,259 --> 00:28:02,659
are some specific numbers know you can take
-1 2 -3 i have taken some arbitrary numbers
189
00:28:02,659 --> 00:28:12,049
okay moment i fix this okay i will get one
specific value of alpha beta gamma; i will
190
00:28:12,049 --> 00:28:19,879
get 1 specific value alpha beta gamma which
is one element okay now if i change the initial
191
00:28:19,879 --> 00:28:20,879
condition what will happen?
192
00:28:20,879 --> 00:28:30,080
i will get another value of alpha beta gamma
okay but all of them have to belong to; sir
193
00:28:30,080 --> 00:28:37,820
but if the right hand side of the main ode
had not been 0 then zero vector would not
194
00:28:37,820 --> 00:28:45,399
have been in the span then it would not have
been a subspace no no when you have something
195
00:28:45,399 --> 00:28:50,979
here f of z how do you solve? there are 2
solutions; one is the characteristics solution
196
00:28:50,979 --> 00:28:58,210
this part will anywhere appear plus something
which is appear here
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00:28:58,210 --> 00:29:06,269
so this part is invariant with there is something
on the right hand side or nothing on the right
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hand side i can say it can be zero vector
can be non-zero vector this will appear all
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the time so some part of the solution will
always belong to this sub space you cannot
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escape it this is characteristic of this particular
equation okay so these 3 vectors are kind
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of to this equation okay this can change;
right hand side can change
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this can be sin cos or whatever forcing function
can be done okay but this is something like
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inherent nature of these differential equation
you cannot change this let me change if you
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change these coefficients okay you change
if you change the coefficients okay ”professor-student
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00:29:54,399 --> 00:30:01,040
conversation ends” so this is fine let us
move on to another example which is again
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differential equation and it will sort of
what i am talking about as dimension
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now i am going to introduce what is the dimension
of the vector; vector space but this particular
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00:30:13,059 --> 00:30:18,070
example we knew that idea of dimension well
there is one question; one question; these
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three vectors are they linearly independent?
“professor-student conversation starts”
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yes why? you have to prove it so i should
prove that alpha e to the power –z + beta
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00:30:36,619 --> 00:30:44,639
e to the power 2z + gamma e to the power -3z
is = 0
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00:30:44,639 --> 00:30:53,169
actually it means zero function; zero does
not mean 0; 0 to 0 function can be found only
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00:30:53,169 --> 00:31:04,190
when alpha beta gamma are 0 so these are linearly
independent okay can you prove it well i leave
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00:31:04,190 --> 00:31:12,879
you this is one example which is here i do
not know whether i get time to do it but we
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00:31:12,879 --> 00:31:25,659
should look at this example the example in
my notes how we prove that three vectors are
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linearly independent?
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00:31:29,710 --> 00:31:36,179
it is not that state of it; you have to do
some; some amount of thinking okay “professor-student
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00:31:36,179 --> 00:31:48,940
conversation ends” let us move on this fourth
example so my example 4 is; so bounded value
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00:31:48,940 --> 00:31:56,899
problem now; for i have 2 boundary conditions
this is the differential equation which is
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satisfied in the domain say going from 0 to
1 okay then i have this boundary condition
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at z is = 0
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00:32:07,419 --> 00:32:21,899
so this u0 = 0 and there are i have second
boundary condition z = 1 so this is u1 is
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00:32:21,899 --> 00:32:28,870
= 0 and this kind of problems will appear
when you start solving partial differential
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00:32:28,870 --> 00:32:36,879
equations and so called which will be covered
in the other course so or; we might have seen
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00:32:36,879 --> 00:32:50,700
this in your undergraduate these kind of problems
we may see it in once in a while
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00:32:50,700 --> 00:33:07,929
uz = alpha 1 sin pi z + alpha 2 sin 2pi z
well well well the general solution here cannot
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be expressed in terms of finite number vectors
for this particular problem a general solution
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00:33:17,909 --> 00:33:31,279
cannot be arrived at by finite number of vectors
what are vectors here? so this is my x1z this
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00:33:31,279 --> 00:33:46,979
is my x2z this is my x3z and so on how many
such vectors i have? infinite i have infinite
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vectors okay
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00:33:48,029 --> 00:34:01,360
where does this solution belong to? span off
what is this; what is this? this is linear
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combination of vectors; linear combination
of vectors right same idea which was we talked
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about earlier like i said as okay what is
the underlying space here? somebody said it
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c2 0 1 okay now underlying space here is c2
well you have to be careful while writing
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this notation it has to be square values
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00:34:33,650 --> 00:34:38,919
because two ends are included there are 2
boundary conditions not just; so twice differentiable
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00:34:38,919 --> 00:34:50,510
continuous functions all interval 0 to 1 that
is underlying space okay so where does this
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00:34:50,510 --> 00:34:59,350
solution belong to? solutions belongs to the
span of; these are vectors; these are vectors;
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00:34:59,350 --> 00:35:05,050
these vectors belong to twice differentiable
continuous functions okay a linear combination
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00:35:05,050 --> 00:35:13,500
of that gives me a solution to this problem
okay infinite linear combination not just
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one value
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00:35:14,500 --> 00:35:21,750
these; well in a partial differential equation
these coefficients; alpha 1 alpha 2 alpha
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00:35:21,750 --> 00:35:29,510
3 to be fixed by some other particular solution
which this comes as a couple problem in the
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00:35:29,510 --> 00:35:32,940
partial differential equation so there will
be one part of the partial differential equation
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00:35:32,940 --> 00:35:37,830
is solution to this problem and the other
part will be particular solution so that is;
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00:35:37,830 --> 00:35:49,040
you can look at book by professor pushapavanam
mathematical methods in chemical engineering
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00:35:49,040 --> 00:35:57,230
but the point that i want to make here is
that uz actually belongs to the span of; it
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belongs to span of this set of vectors; belongs
to span of this set of vectors so span is
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00:36:10,470 --> 00:36:15,630
a very very important concept and we keep
require everywhere not been introduce just
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00:36:15,630 --> 00:36:26,571
for the sake of it okay in this particular
case the space or the subspace of; subspace
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00:36:26,571 --> 00:36:33,410
of twice differentiable continuous function
which is fine by these vectors okay is equally
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dimension is infinite dimension okay
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00:36:37,650 --> 00:36:44,110
because these are linearly independent vectors
well coming to the point where we have define
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00:36:44,110 --> 00:36:51,250
what are the linear dependence and then we
have also worry about the basis business is
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00:36:51,250 --> 00:36:57,520
this clear; clear? somebody has any doubt
please; please feel free to stop me at any
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00:36:57,520 --> 00:37:11,790
point “professor-student conversation starts”
is it possible? well if you take initial condition
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it depends upon the initial condition so if
you take initial condition to be 0 0 0 only
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possible solution that you get this
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00:37:25,120 --> 00:37:33,200
depends upon what is the initial condition
of u0 u prime 0 u double prime 0 all the 3
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00:37:33,200 --> 00:37:39,550
initial conditions are 0 0 0 yeah no they
are the functions; they are functions there
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00:37:39,550 --> 00:37:46,780
are not values of the functions 0 to 1 according
to the problem this particular problem would
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00:37:46,780 --> 00:37:54,490
be heat transfer problem; it will be something
by conduction in a rod 0 to 1 comes because
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00:37:54,490 --> 00:38:02,300
non dimensionalize the space variable that
one is; it could be 0 to n if you want does
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00:38:02,300 --> 00:38:03,300
not matter
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00:38:03,300 --> 00:38:10,570
well if you know; no no these are different
spaces if you have; you may have a situation
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see there are some problems in chemical engineering
where you have some infinite dimension of
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00:38:18,330 --> 00:38:29,240
a rod and all so you may have such scenario
where you know which is c2 - infinity to infinity
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00:38:29,240 --> 00:38:41,070
or c2 0 to infinity but these are different
spaces this space is different than; no no
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00:38:41,070 --> 00:38:47,940
a function defined over 0 to 1 is not same
as function defined over 0 to pi
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00:38:47,940 --> 00:38:58,900
or –pi to pi there are different functions
maybe some part which is not included here
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00:38:58,900 --> 00:39:03,350
okay they are not same; exactly that is another
problem and it cannot be continues it might
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00:39:03,350 --> 00:39:10,280
be continuous in some region but not continuous
in some other region so all kinds of okay
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00:39:10,280 --> 00:39:16,530
“professor-student conversation ends”
okay let us move on to this basis and dimension
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00:39:16,530 --> 00:39:30,990
business so now what is basis what is dimension
of the space or subspace okay
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00:39:30,990 --> 00:39:37,931
the formal definition of linear dependence
that will find in the notes so we again define
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00:39:37,931 --> 00:39:56,170
abstract terms we start with this set s which
is the; we start with set s okay zero vector
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00:39:56,170 --> 00:40:12,440
well we are writing 0 on the right hand side
it means zero vector in that particular space
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00:40:12,440 --> 00:40:17,980
it is reverse now if these vectors are n dimensional
what will be this 0? well let me qualify by
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00:40:17,980 --> 00:40:24,000
put a bar over it; 0 bars and 0 vectors 0
bar is 0 vectors
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00:40:24,000 --> 00:40:30,950
so if this is = 0 bar that is zero vector
okay if this is function space; set of continuous
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00:40:30,950 --> 00:40:36,310
function so 0 to 1 what is the 0 vector? zero
function everywhere okay fz = 0 on 0 to 1
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00:40:36,310 --> 00:40:44,790
that is the 0 function okay that is what it
means it does not mean liberally 0 at one
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00:40:44,790 --> 00:40:50,070
point it depends upon what space okay so if
this is a linear combination such that; okay
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00:40:50,070 --> 00:40:58,660
so if there exists a linear combination such
that okay if there exists a nonzero linear
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00:40:58,660 --> 00:41:06,680
combination there exists a nonzero linear
combination which gives 0 vector
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00:41:06,680 --> 00:41:19,370
which means; if i can find some nonzero elements;
alpha 1 alpha 2 which will give the zero vector
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00:41:19,370 --> 00:41:29,920
when these elements are linearly dependent
else there if you cannot find; if you cannot
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00:41:29,920 --> 00:41:33,460
find a nonzero combination which means only
when to get a zero vector is to put alpha
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00:41:33,460 --> 00:41:44,770
1 = 0 alpha 2 = 0 alpha 3 = 0 all of them
are 0 okay then they are linearly; then they
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00:41:44,770 --> 00:42:03,100
are linearly independent then the set of vectors
is linearly independent okay
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00:42:03,100 --> 00:42:24,280
and what is spaces; yeah basis of the vector
is nothing but the set of linearly independent
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00:42:24,280 --> 00:42:45,860
vectors which generate the entire space; minimal
set okay which generates the entire space
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00:42:45,860 --> 00:42:52,280
so the minimal linearly independent set okay
minimal linearly independent set which generates
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00:42:52,280 --> 00:42:56,410
the entire space yeah; “professor-student
conversation starts” in 3 dimensions i can
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00:42:56,410 --> 00:43:03,330
take 4 vectors sir but they would not be linearly
independent one will be dependent
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00:43:03,330 --> 00:43:10,100
yeah minimal set okay the minimal is not really
required the linearly independent set which
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00:43:10,100 --> 00:43:15,760
is; which generates the entire space; when
you entire space actually the minimal set
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00:43:15,760 --> 00:43:30,290
“professor-student conversation ends”
so when do you say that a space as finite
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00:43:30,290 --> 00:43:40,030
dimension the number of elements in the set
is finite okay; number of elements in the
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00:43:40,030 --> 00:43:42,660
set is finite it is called a finite dimensional
space
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00:43:42,660 --> 00:43:48,870
if the number of elements in the basis side
with number of linearly independent vectors
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00:43:48,870 --> 00:43:54,820
is the basis set is infinite that is called
infinite dimension space; we saw one example
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00:43:54,820 --> 00:44:02,200
of infinite dimension space just now linear
combination of sin sin sin pi x sin 2 pi x
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00:44:02,200 --> 00:44:13,280
sin 3 pi x sin 4 pi x infinity okay all these
vectors actually form a basis; all these vectors
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00:44:13,280 --> 00:44:20,980
actually form a basis and we can show that
it generate the entire subspace okay
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00:44:20,980 --> 00:44:27,000
the dimension of that particular subspace
is why? because number of linearly independent
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00:44:27,000 --> 00:44:33,920
vector; you cannot find the combination which
you cannot find the nonzero combination which
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00:44:33,920 --> 00:44:40,850
will give the sum of all those vectors equal
to 0 linear combination of 0 so that means
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00:44:40,850 --> 00:44:54,470
the dimension of the space generated by sin
pi x sin 2pi x sin 3pi x sin 4pi x is infinity
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00:44:54,470 --> 00:44:55,470
okay
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00:44:55,470 --> 00:45:01,180
so we started with; till i mention this space
all that we are doing is just trying to know
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00:45:01,180 --> 00:45:09,550
generalize ideas from 3 dimensions just remember
that i am just trying to generalize the ideas
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00:45:09,550 --> 00:45:17,680
in 3 dimensions so visualization geometric
ideas you should be clear in 3 dimensions
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00:45:17,680 --> 00:45:25,540
you understand 3 dimension very well no difficulty
understanding all these infinite dimension
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00:45:25,540 --> 00:45:26,540
spaces okay
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00:45:26,540 --> 00:45:36,040
so just be clear about this 3 dimensional
geometry and then that is enough all visualization
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00:45:36,040 --> 00:45:44,500
can be done in 3 dimensions the concepts are
just be generalised for this grand generalizations
318
00:45:44,500 --> 00:45:51,830
we will apply mathematics in a period you
know 1850 to 1950 these 100 year period and
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00:45:51,830 --> 00:45:57,820
with these grand generalizations you can actually
view you can have a very different view of
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00:45:57,820 --> 00:45:58,820
engineering
321
00:45:58,820 --> 00:46:07,930
so you will see that you are just working
with the; if you start having this viewpoint
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00:46:07,930 --> 00:46:13,710
then working with ax = b finite dimensions
or working with some boundary value problem
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00:46:13,710 --> 00:46:18,360
some differential equation is not different
operators change what we call is a vector
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00:46:18,360 --> 00:46:24,680
change what we call this b on the right hand
side change actually the problem that we talked
325
00:46:24,680 --> 00:46:31,100
about this problem of; d3/d z cube + something
something 6 * uz
326
00:46:31,100 --> 00:46:51,340
i put this = some fz okay this is fundamentally
not different from ax = b solving for ax = b
327
00:46:51,340 --> 00:46:59,460
in 3 dimensional spaces is same as solving
for this equation in thrice differentiable
328
00:46:59,460 --> 00:47:07,200
continuous functions fundamentally not different
this same equation different space spaces
329
00:47:07,200 --> 00:47:13,100
is different we are not doing something fundamentally
different when you solving this and solving
330
00:47:13,100 --> 00:47:20,840
okay the specific solution which i wrote was
f of z = 0 that is a specific solution
331
00:47:20,840 --> 00:47:31,330
and then to write; if you take some fz whether
you can solve this problem or not that they
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00:47:31,330 --> 00:47:39,320
depends upon the span of; you know the characteristic
equation and so on the same idea which will
333
00:47:39,320 --> 00:47:44,180
apply here when you saw the x = b the same
idea will apply so we will continue in the
334
00:47:44,180 --> 00:47:48,530
next lecture; this is not enough; this is
not enough we are just defining the space;
335
00:47:48,530 --> 00:47:50,020
we define the span linear dependence and basis
336
00:47:50,020 --> 00:47:57,290
we need to go further let us say now when
in 3 dimensions you know i have something
337
00:47:57,290 --> 00:48:02,490
more important i had length of vector now
how do i can generate the length of vector
338
00:48:02,490 --> 00:48:08,090
and why generalise the length of vector that
this how it because we will need to define
339
00:48:08,090 --> 00:48:11,930
something called conversions limit and so
on that is why we have to define structures
340
00:48:11,930 --> 00:48:13,519
like norm we will talk about it in the next
lecture