CHAPTER 02.21: VECTORS: In this segment, we’ll talk about if a set of
vectors is linearly independent, then if we take a subset of those linearly
independent, that subset is also linearly independent. So let’s suppose – let
us be given “m” vectors. SO if we are given “m” vectors and it is given that
this set is linearly independent, so you are given a set of “m” vectors and
people are saying that “Hey, these “m” vectors which are given to you are linearly
independent.” And then let’s suppose we take a subset. Let us take a subset
of “p” vectors. So out of these “m” vectors we are going to take a subset of “p”
vectors. So of course we’ll assume that “p” is less than “m” and we can also
assume that “p” is less than or equal to “m”, but we already know for “m”
vectors they’re linearly independent so we don’t need to prove it for exactly
“m” vectors because its already given to us. So I’m going to take a subset of “p” vectors, “p”
number of vectors, where “p” is less than “m”. And let this subset be as
follows: let it be Aa1, Aa2, all the way up to Aap.
I’m not calling then A1, A2, Ap because those are
not the vectors I’m necessarily taking. I’m taking some “p” vectors out of
all these “m” vectors which have been given to me. So that’s why I’m just
giving them different names than let’s suppose A1, A2, A3 and so forth. So
these are the vectors- let’s suppose – are a subset of these “m” vectors and I’ve
chosen “p” of them. Then a linear combination of these vectors will be what?
I’ll have Aa1 times k1, let’s suppose. Then I’ll have another linear
combination Aa2 all the way up to kp Aap. This will be a linear combination of these “p”
vectors. And let’s suppose if I put this equal to zero vector, what I need to
show is that k1, k2, up to kp equal to 0 is the only
solution possible for this linear combination we put equal to zero vector. Now how do I do that? Is by knowing that – hey –
if I have these “p” vectors which I have chosen, there are “m” minus “p”
vectors left over. So if I have chosen “p” vectors out of the “m” vectors, I
got “m” minus “p” vectors left over. So if I have chosen “p” vectors left
over which I am not including in this subset, and let’s suppose if I call
those vectors to be Aa(p+1) all the way up to Aam.
I have those – these are my “m” minus “p” vectors which I have leftover. Then
what I can do is I can write this: I can say –okay, hey – I’ve got k1, Aa1
vector plus k2, Aa2 vector plus kp Aap vector. Then what I’ll do is I’ll put zero plus this
thing plus zero times Aap plus 2 vector plus all
the way up to zero Aam vector and put that equal to
zero. So I’m basically doing is to this linear combination put equal to zero
vector, I’m adding a zero because all these extra vectors which I have these “m”
minus “p” which were leftover, not extra “m” minus “p” vectors which I had
leftover. They are multiplied by zero, so this is same as this. But if you look at this, this is a linear
combination of all the vectors which have all the “m” vectors which I have.
And I know that if I take the linear combination of all the “m” vectors and I
put that equal to a zero vector, the only possible solution is k1=0, k2=0,
all the way up to km=0. But I already assumed that the other ones are 0; I
already put them as 0. So that means that that’s the only solution possible.
SO if somebody told me “Hey, what is the solution corresponding to this
linear combination for a zero vector?” I’ll have to say that –hey, the only
combination possible is that it’s a trivial solution. That means that this
has to be 0, this has to be 0, this has to be 0, this would have been zero
whether I would have assumed it or not. All these are put 0 anyways, so these
ones also have to be 0. SO maybe I should put kp=0
right here. So that means that if that is the case, if this
is the solution to this as well – because this one is the same as this – so if
this is the solution to this as well that means that hey, the only possible
solution for this one is given through kp being
equal to 0. So if that’s the case that means that hey, this subset which you
are finding out here is linearly independent as well. And that’s the proof of
this theorem. And that’s the end of this segment. |