 So, in the previous lecture we have introduced this notion of what is the dimension of a vector space and I hope that even after all of that if you encounter some n-tuple of numbers you would not just naively assume that n would be the dimension of the vector space right that is at least the bare minimum level of sophistication with which we will expect you to deal with this idea of dimension not just counting the tuples right, but it is also counting the rather instead counting the number of elements in the basis set for the given vector space. We have also shown you that if multiple bases do exist as the most often would in that case the number of elements in the basis is still nonetheless going to be unique right. So, we will just take a quick example I would not really call it a new proposition it is some example or an illustration of the ideas we have seen the other day most particularly this idea this amazing idea that we saw the other day right at the end which is that if you start with any linearly independent set inside a vector space or vector space which admits a finite basis then you can always extend this linearly independent set of vectors to form a basis eventually right and we showed you constructive way of doing that. So, today we shall see an illustration of that to demonstrate another result concerning subspaces and operations thereof which you are already familiar with. So, let us take w1 and w2 to be subspaces of v right and what we shall claim to show today is. So, this v is a finite dimensional vector space. So, by the way if v is a finite dimensional vector space and w1 and w2 are subspaces thereof can we agree that w1 and w2 must also be finite dimensional yeah it is easy right if they were not then neither could this fellow be finite dimensional it is obvious because this is sitting inside this. So, that is taken care of. So, now once we have admitted that v is finite dimensional these are obviously finite dimensional and therefore it makes sense to talk about equalities of finite numbers we do not want to equate infinities that is tricky business. So, we have already defined the sum of subspaces couple of lectures back probably. Now, notice very carefully everything that we are talking about in terms of dimension must be a subspace. So, w1 is a subspace as is w2 as is w1 intersection w2 right. So, all of them make sense at least that is the first thing we should check that whatever we are claiming should make sense this is a subspace. So, its dimension make sense these two are by definition subspaces and this also we have seen earlier the intersections are indeed subspaces. So, we shall now go about showing this just to refresh your memory I will draw that lattice which had drawn earlier right the containment lattice. So, what do you what did we start with we had w1 intersection w2 is contained inside both w1 and w2 right and finally these were both in turn contained inside w1 plus w2. So, you start out with something some claims such as this it is very important to recall this lattice because we are now going to use this lattice the picture of this lattice in our minds and the idea that we can extend any linearly independent set to form a basis for the overall vector space. So, how do we go about this obviously this is a subspace of v. So, this also must be finite dimensional. So, suppose b1 is a basis for w1 intersection w2 given by let us say I take the dimension to be k which is. So, let us say this dimension v is equal to n okay given by this one let me call it v1 v2 till vk which is just another way of saying that the dimension of w1 intersection w2 is k yeah that is straight forward is in turn because it is the cardinality of the basis of course, with k less than or equal to n. Now, if I look at this as a bunch of vectors contained inside w1 intersection w2 it is also a bunch of vectors contained inside both w2 and w1 yeah. So, therefore, this can further be extended because by being a linearly independent basis it is obviously a linearly independent set sitting inside either of those two subspaces. So, I can extend this linearly independent set to form a basis for w1 I can also extend the same set to form a basis for w2 yeah. So, let us say extend b1 to form a basis for w1 given by bw1 is equal to of course it must contain these first k fellows from here and then I add some arbitrary number let us say l1 all right. So, let us say w1 1 w1 1 w1 2 until w1 l1 which basically means that the subspace w1 has a dimension which is given by k plus l1 right just keep count of these just a bit of bookkeeping is necessary and also to form a basis for w2 given by w2 bw2 given by again the first k fellows must be inherited from this because this is the linearly independent set I started with. So, v1 v2 till vk and then w21 w22 until w2 l2 which essentially means the w2 has a dimension k plus l2 right. So, this is the construction that I am doing here ok. So, now if I have to prove this proposition how do I go about this how do I pose this claim based on these constructions that I have made can you guess what I am going to do now I am trying to prove that the dimension of this is equal to the dimension of this plus this minus that of this and I have constructed basis in fact for this I have also had for this and for this sorry union of these 2 sets and then what should I claim right but how do you subtract I mean we are trying to prove something about dimensions right. So, you are on the right track but we have to do something so that they add up in the sense of dimensions sorry yes yeah. So, what I have to show is that if I take the union of bw1 and bw2 and if I manage to prove that this is indeed a basis for w1 plus w2 then look at the number if you take the union obviously you are not counting this twice what is the total number of fellows then you end up with k plus l1 plus l2. So, if I show that bw1 union bw2 is a basis for w1 plus w2 then I would have shown that the dimension of w1 plus w2 is k plus l1 plus l2 and what is this k plus l1 k plus l2 minus k right. So, that is k plus l1 plus l2 I would have shown the equality. So, now based on this construction I already formulate the problem to showing that S given by v1 v2 vk w11 w12 w1 l1 w21 w22 w2 l2 is so this is what I am required to show now is a basis for w1 plus w2 right. If I can show this then I will be done what is it that defines a basis the defining properties of a basis it must be a generating set it must be a linearly independent set. So, it must be a generating set for w1 plus w2 and it must be linearly independent these are the two things that I need to show. So, first let us try and show that this set is indeed going to be a generating set for which I pick any arbitrary element from this subspace and if I show that it can be written as a linear combination of the fellows sitting here in I am done but that is pretty straightforward. So, consider any w belonging to w1 plus w2 implies w can be written as small w1 plus small w2 with w1 coming from the subspace w1 and w2 coming from the subspace w2 yeah that is a defining property of the sum of subspaces. So, this means this w1 if it belongs to the subspace it can be written as a linear combination of which fellows of all the fellows that constitute the basis for w1 right. So, this can be written as summation alpha i v i i going from 1 through k plus summation beta i w1 i i going from 1 through l1 that is my little w1 plus little w2 is going to be say summation gamma i i going from 1 through k gamma i v i plus summation i going from 1 through l2 delta w2 i delta i w2 i yeah that is right what does that mean in order to show express this w as a linear combination of a certain bunch of vectors the only vectors I needed are k of these. So, I can just combine alpha i and gamma i into one term and w1 i is and w2 i is right. So, therefore it clearly means that this means that w is equal to summation alpha i plus gamma i times v i I am not even writing the indices now you can just fill that up for yourself plus beta i w1 i plus summation delta i w2 i which means it belongs to the span of. So, therefore indeed this set S actually I can write probably S whichever way you like it is just S it does not matter it is a matter of taste you could have chosen either. So, anyway it belongs to the span of S which means that S is indeed a generating set for w1 plus w2. So, now what I need to show is that S is also linearly independent all right. So, consider summation alpha ok let us choose some value a i a i v i plus summation b i w1 i plus summation c i w2 i is equal to 0. In order to prove linear independence I must show that each of the k a i's must be 0 each of the l1 b i's must be 0 and each of the l2 c i's must also be 0 yeah and there is no other possibility that can take it to 0 that is linear independence. So, now let us do something let us write this down as summation a i v i plus summation b i w1 i is equal to minus summation c i w2 i is equal to say some vector p nothing non-trivial non-trivial being said here right it is straight forward, but based on this what can I say look at look at this what can you say about this object this one I marked in purple this only uses vectors in w1. So, it because of the closure and closure under scalar multiplication and vector addition this part must belong to w1 on the other hand what I have here must belong to what w2 this uses fellows that solely belong to w2, but both of these are equal to this vector p. So, then p must belong to w1 because it is equal to this and p must belong to w2 because it is equal to this which means that this p. So, please write this out in your own words it must belong to w1 intersection w2 is that clear please ask if it is not clear all that I have done is I have just written the same expression push this one of these terms to the other side and then call that newly constructed vector as p. And now by equating the p with the first term I come to the conclusion that p must belong to w1 by equating with the second term I come to the conclusion that it must belong to w2 if something belongs to both at the same time then it must belong to the intersection of those two subspaces right. If this belongs to the intersection of these two subspaces can I not write it down in turn as a linear combination of the basis of w1 intersection w2 right. So, now this implies p is equal to summation di vi what does that tell me so p is equal to summation di vi p is equal to summation minus ci w2i right which means that minus summation ci w2i is equal to summation di vi let us push this to one side shall we and we get 0, but what about this w2i is along with vi is they form a basis for the subspace w2 and then they must be therefore linearly independent. So, therefore each of the ci's and di's must be 0, but di's we really do not care because we were initially inclined to show that ai vi and ci are 0 and we already seen from this that ci is equal to 0 for all i and di is equal to 0 for all i, but this is not of much consequence to us. So, we can just ignore this, but at least we have shown ci is equal to 0 for all i. Now go back to this result here this original let us mark that equation with a blue star yeah. So, from this blue star equation what can we now write we know that ci is equal to 0 for all i. So, we therefore have summation ai vi plus summation vi w1i is equal to 0 again the vi is along with the w1i's form a basis for the subspace w1 and therefore they must be linearly independent. So, any linear combination thereof being equal to 0 means each of those coefficients must be equal to 0 which means ai is equal to 0 for all i and vi is equal to 0 for all i. In other words I have shown that all the ci's all the bi's all the ai's must be 0 and therefore in this linear combination the only way you can get a 0 here is to have all of those coefficients equal to 0 which means that the set S indeed is linearly independent. So, we have a set S which is a generating set for w1 plus w2 which is also linearly independent and therefore is a legitimate candidate for a basis and because the number of elements in any basis is the same for finite dimensional vector spaces. So, the dimension of w1 plus w2 must be equal to the cardinality of S and the cardinality of S is k plus l1 plus l2 which obviously equals the dimension of w1 k plus l1 plus the dimension of w2 k plus l2 minus the dimension of w1 intersection w2 which is minus k right. Yeah, by extending v1 through vk to be a basis for w2 I added those w2 i's does not matter it is still w2 deletion w1 is still a part of w2 it is exclusively part of w2 you might say, but it is still part of w2 in fact that deletion may not be a subspace, but w2 is a subspace that came as a consequence of the fact I am not imposing anything. The fact that I am able to write it only in terms of v i's is a consequence of the fact that this P by dint of this very expression suggests to me that P must belong to w1 and w2 that is my argument I am not just saying that P must be written in this fashion P has to be written in this fashion because of this conclusion I am not imposing it that let us write P like this I can write it I have already proved that this can be written like this I am not just imposing it as an assumption that let us write P like this I am saying that P always can be written like this because P belongs to the intersection of those two see whatever what is P equal to P is equal to this entire thing is 0 now I have split it into so you have 5 plus 3 minus 8 3 numbers so I take 5 minus 8 on one side and minus 3 on the other side and let us say P is equal to minus 3 that is all that I am doing here the equivalence of that and then that minus 3 becomes like P. So I am splitting up a 0 quantity into 2 non-zero quantities presumably 2 non-zero vectors and then calling that non-zero vector P now that because that can be written as a linear combination of fellows only in w1 not only in w1 as in fellows in w1 which form a basis for w1 and also fellows that only belong to w2. So therefore, this must belong to both w1 and w2 because w1 and w2 by because their subspaces they must be closed under addition and scalar multiplication. So if I have taken up only fellows from w2 and cooked up a linear combination it must belong to w2 if I have taken up fellows only from w1 and cooked up a linear combination it must also belong to only w1 I mean must belong to w1. So if it belongs to both w1 and w2 it must belong to the intersection yeah that is that is all that I am saying here any other questions. So ci is equal to 0 from this this you are clear with okay. So w2i is a what look at this set w2i I mean this set bw2 this is what this is the basis for the subspace w2. So if it is a basis it is a linearly independent set if it is a linearly independent set then any linear combination of these fellows can be 0 if and only if all the coefficients are 0. But that is precisely what this is it is a linear combination of the fellows that form the basis for w2 if that vanishes then every coefficient must be 0 and because ci's are only some of those coefficients they must also be 0. So that is why the ci is at 0 once the ci is at 0 you go back to the original equation we started from plug it back in here. So this part is already 0. So then you are left with only these first two terms here. So these two terms there some must be 0 then this also forms a basis for w1 and repeat the same argument right yeah. I do not know I do not know if there is any non-trivial that is what I have to check when I am investigating whether a particular set of vectors is linearly independent I have to try out and check that how can I make this 0 is there any way that some of those coefficients are not equal to 0 and yet I end up with a 0. If it is possible then it is definitely not linearly independent if I make it 0 obviously it is 0. So there is certainly one sure shot way of getting a 0 here which is to choose all of these to be 0. But apart from that is there any other way to get there if I manage to find any other way then definitely that set is not linearly independent. The question here is can I show that this is linearly independent in other words can I discount the possibility that I might get a 0 apart from the trivial choice of making all the ai's bi's equal to 0 and now I show that there is no other way that is the only way I can get a 0 by choosing each of those coefficients to be 0 yeah because w2 exclusion w1 is not a subspace yeah so that really doesn't have yeah it is sure it is part of w but you have to have you have to talk about subspaces right we want to talk about subspaces so this yeah if you want to be very defined it is definitely exclusively like coming from each of these terms is exclusively coming from w2 deletion w1 but the sum may not come from w2 deletion w1 because w2 deletion w1 is not a subspace for one if you are taking away everything that is there in w1 from w2 you are taking away the 0 and any subspace must contain the 0. So w2 deletion w1 cannot be a subspace so therefore it doesn't make any I mean make our job any easier if you were to just say this is part of w2 deletion w1 instead of just w2 which is why I just took the subspace that I can surely claim this to be a part of instead of the set that this is a part of okay