CHAPTER 02.21: VECTORS: Prove that if a set of vectors is linearly independent, then a subset of it is also linearly independent

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.