 Because I end up writing the statement in such a manner that I leave very little room for the proof, okay. Again it's the same age old trick I know I mean it's only two days but it's been ages it seems to me. Same trick pick out two objects and test and show that for any arbitrary object that you pick from the field say alpha, alpha v1 plus v2 also belongs to this and you are done, right. So suppose v1 and v2 belong to this set as of now w1 plus w2. It means that there exist w1, w2 such that v1 is equal to w1 plus w2 with w1 belonging to big w1 and w2 belonging to big w2 and w1 hat, w2 hat such that v2 is equal to w1 hat plus w2 hat with w1 hat coming from the subspace w1 and w2 hat coming from the subspace w2. Again look at v which is defined by alpha v1 plus v2, yeah. So this is given by alpha w1 plus w2 plus w1 hat plus w2 hat. Of course the associativity of the addition and all and distributivity over to the scalar multiplication tells you that this in turn leads to alpha w1 plus w1 hat just a rearrangement of those terms plus alpha w2 plus w2 hat, yeah which is nothing but w1 till day plus w2 till day. What is so special about w1 till day and w2 till day? Look at w1 till day, w1 till day is a sum of alpha w1 plus w1 hat, both of those elements because of the closure under scalar multiplication alpha w1 belongs to big w1, w1 hat of course belongs to w1, their sum therefore because of closure under vector addition must also belong to w1 therefore w1 till day which I am defining like so must also belong to the subspace w1 and similarly w2 till day which is this object here must also belong to the subspace w2 therefore this object v can be written as a sum of two vectors one of which comes from w1 the other comes from w2 implies v belongs to w1 plus w2 which was to be proved right. So therefore this is a subspace right the way I have defined the sum of two subspaces turns out it has to be nothing but a subspace right. So these are all new kinds of definitions and objects that you might be encountering now but this tells you that there are so many different ways of cooking up subspaces given that you start with some subspace right what their utilities will be will come across gradually in this course. Now since we are already talking about some subspaces let us put forth another claim okay I might just as we will leave the first line probably. So this is w1, w2 subspaces of subspaces of v right suppose w1 intersection w2 what can it be what is the smallest possible intersection that you can have can it be an empty set look I have already shown you that this object has to be a vector space. So if this object is a vector space at least it should contain 0. So the smallest is probably the 0 of the original vector space yeah that is a minimal intersection that they can have because both of them must at least contain the 0 in common otherwise they cease to be subspaces themselves right. So suppose this is true then for every w belonging to w1 plus w2 there exist unique w1 in big w1 and w2 in big w2 such that small w is w1 plus w2 which means that earlier we had just shown that this is a subspace but it might so happen that you pick out two objects from one from w1 and one from w2 to lead to a resultant vector which is w. Somebody else might pick out another two vectors or different vector from w1 and a different vector from w2 to yield that same vector w. But now I am saying if those two subspaces are such that they have nothing but the additive identity in common then it does not matter who picks what they will invariably end up picking the same choice. So if you and you do not communicate with each other I tell you look this is my w1 and this is my w2 and this is the w that I have in mind okay and I am telling you it belongs to w1 plus w2 can you tell me what are the w1 and w2 those little w1 and w2 vectors that you picked up you do not need to cheat or communicate with each other if you have done your job correctly in a mathematically correct manner you will end up with the same w1 and the same w2 provided this intersection is just this okay. So standard way of proving this is to assume the contrary but suppose despite this condition there are indeed two different ways of representing a vector w in the sum of two subspaces such that you manage to end up with a different w1 and w1 hat and a different w2 and w2 hat which add up to give you the same w right. So what we will do is we will assume the contrary assume contrary that w belonging to w1 plus w2 can be written as w is equal to w1 plus w2 and w is equal to w1 hat plus w2 hat with w1 w1 hat plus w1 hat plus w1 hat plus w1 hat plus w1 hat coming from w1 and w2 hat coming from w2 where w i is not equal to w i hat for sum i is equal to 1, 2. What does that mean we have from these two this must equal this right and we are saying that at least one of these do not agree because if both of them agree then it is a unique representation. So at least one of them must disagree with the other yeah. So what we will do is we will do this we will do this we will do this we will do this we will do this we will so what happens this means that we have w1 plus w2 equal to w1 hat plus w2 hat which with a bit of jugglery and adding the additive inverses on both sides of suitable terms we have w1 minus w1 hat is equal to w2 minus w2 hat is equal to p some vector to w1 hat is equal to p. So look at p closely what can we say about this vector p from the expression here w1 and w1 hat must both come from the subspace w1 sorry oh it is the opposite is it yeah w1 minus w1 yeah that is right. Anyway the point still remains that this object here belongs to what w1 and this object here belongs to what w2 right. So therefore this object p which is equal to this and therefore it belongs to w1 and it belongs to and it belongs to w2 because it is also equal to this. So therefore p belongs to w1 intersection w1 intersection w2 is it not right what is that immediately when you tell us because what is the only object living inside w1 intersection w2 by our proposition it is 0. So this implies that p is equal to 0 but if p is 0 then each of these objects are in turn equal to 0 this means w1 is equal to w1 hat and w2 is equal to w2 hat which is a contradiction because we assumed that these are indeed different at least one of them has to be different in order for our claim to hold but contradicts assumption right. So therefore in general if you are cooking up a sum of subspaces every vector inside the sum of subspaces may be represented in multiple different ways right but the moment I tell you that those two subspaces are chosen in such a manner that they are as close as possible to being disjoint. So they are connected only at the origin in some sense right that is the only thing that they share they only share the additive identity in the vector space nothing else is common. In that case this takes up a beautiful form where the sum any object in the sum is uniquely representable as a sum of two objects one coming from each of those subspaces right. So that is an important assertion and you try to notice the pattern we are using to prove these things right in each case that very important result we convinced you about the other day about how to establish whether something is a subspace and then you know things like assuming the contrary and proving these are handy proof techniques that you will encounter every now and then. Before we proceed on to another important result I am just going to make an observation not really going to prove anything but we will try to convince you through some reasoning that it is true. I am going to write down a few subspaces which I have shown are definitely known to be subspaces. Suppose you have W1 and W2 which are subspaces of vector space V so this is W1 intersection W2 definitely a subspace right observe that this object is contained inside W1 as well as inside W2. So if I am drawing up a diagram for the containment relationship I can safely say that this object at the bottom is contained in this bigger subspace presumably bigger subspace as well as this and finally all of them end up being contained inside this. Why? I mean look every subspace must contain the 0 vector at least. So every time you pick out an object in W1 alone just write that object as W1 plus 0. Now that 0 vector is contained in every subspace so it is also contained in W2 yeah. So definitely any object that is here is also contained here isn't it by the same token this. So this is a containment sort of a relation diagram that you can look at this as. So this sort of an idea has profound impacts for example I will not solve this entirely but just as an example of showing you how this sort of things are useful in inserting certain claims. Suppose you have again W1, W2 so this is like an exercise I will not solve it in its entirety but give you hints about how to do this but you definitely try to complete this. So suppose W1, W2 and W3 are subspaces of a vector space V right. Suppose I have given you this the following are true W1 plus W2 is equal to W1 plus W3. Second W1 intersection W2 is equal to W1 intersection W3 and third W2 is contained in W3. You have to show that W3 is equal to W2. I will just drop you some hints about how you can proceed I am not going to complete this. So what is the first way to attack such problems? A lot of work has already been done. If you have to show two subspaces are equal what is the technique that we should be using? Sorry both are contained in each other. Yeah already we have seen that we are required to show these two are equal and one side of the inclusion is already shown. So if we can show that W3 is also contained in W2 will be done. So as a first step I mean blind guess forget about any special techniques or tricks or anything. First step should be first logical step should be choose any arbitrary object in W3 and try to show that it must also be contained in W2 and you will be done. So what you do is choose W3 in W3. Now by that diagram that I had drawn what can you immediately say about something that belongs to W3? Where should it also be contained? Sorry yeah. So W3 again I am going to write that statement clearly which I have explained already why. So clearly W3 belongs to what? W3 plus W1. But what does this assertion tell me? This must therefore mean that W3 belongs to W2 plus W1. What does that mean? Think about it. That entire diagram the reason why I like this problem is because that diagram that I have just illustrated. If you have the picture in mind it helps you along with the proof because otherwise you might get stuck. What does this mean? Think about the definitions we have given okay. Not either this is a plus this is a sum. So where should it belong? Yeah I heard somebody say something. Why? This is a bigger set right. So just because something belongs to a bigger set should not be that it belongs to a smaller set. You have to still do some work to get there. What does this mean? This simply means that W3 can be written as a sum of two objects from W2 and W1. And what would that mean? So this means that W3 can be written as some W2 plus W1. Think about what the assertions are here right. What can you say next? What is this W2? Where is it coming from? Sorry. Where is this W2 coming from? W2 of course I mean I have not written it but if it comes to from W2 where must it also belong to? W3 also because it has been given. So therefore you take it to the other side. What you have is an object that belongs to W3. If it belongs to W3 what about this fellow? So I leave that to you as an exercise to complete okay. You just have to use these relations. Please complete this. It is very important that you do. It is an important exercise. It will help you solidify your understanding of all the discussions and about relations between these subspaces that we have had up until this point. There are several other interesting problems in the problem sheet but this you should try to complete on your own. Now we are going to make another interesting claim about the span. So suppose I give you a set of vectors which in and of itself is not a subspace. Now think about this problem constructively. You want to build the smallest subspace which is the least like the redundancy free subspace which contains at least all of the vectors that are part of that set. So what is the smallest possible subspace that contains every object in a given set? You follow? If I give you a set of vectors that in and of itself clearly need not be a subspace but you might be interested in building a subspace because you know subspaces have beautiful properties. They help us in posing problems in linear algebra in a nice sophisticated fashion. So given a set of vectors we might be interested in building a subspace out of it. How do we go about doing that? What is the smallest subspace? I can of course say oh hang on. This is part of some vector space. So let me consider the whole vector space. Obviously that whole vector space is going to contain every vector you have picked out in your set. That is a given. But I do not want so many redundancies. It might often be the case that you are cooking up a vector space out of a given set of vectors which contains lot more than what you want. You want just enough to contain every vector that is in your given set but you do not want anything more. You want to sharpen this vector space. So it turns out that here is where we have this result that the span of S is the smallest subspace containing every vector in S. So again we are going to please do not look at these proofs as some sort of exercise in futility as in they are to be forgotten. During the course of these proofs we are discussing several techniques and as you will see if you glance through your problem sheet you will be required to prove a lot of things. The way to write those proofs, the way to argue, the way to reason, the way to proceed with them the hints are laid out in these proofs that we are doing here. They may not be replicable exactly but the line of reasoning should follow along very similar lines. So, this is what we are going to now try and prove. How do we do this? Any ideas? So, yeah we, that is not very handy here. What we are going to do is we are interested in looking at this subspace which contains all the vectors in S and just about enough. So, let us take a look at all those subspaces, all possible subspaces that contain every member of S. Suppose, W i or rather this is a set B a set of all subspaces containing S. So, now observe this argument. We are interested in what exactly? The intersections, correct. So, we are interested in this. So, this is i belongs to some index set which we define as some W star. Clearly, I have left that as an exercise for you, but again it is a trivial step. As I have said finite or I mean sorry countable intersections will also lead to a subspace. So, this is a subspace. See at every step it is important to check that whatever claims we are making we have proved them or at least given you enough justification to convince you that it is so. So, the objects we are dealing with here each of these is a subspace, but the fact that this is a subspace follows from the fact that which I have not completed completely proved that countable intersections are also subspaces. So, now what does my problem reduce to? My problem essentially now reduces to showing that the span of S must be equal to this because this why my very understanding and definition is indeed. If you look at the intersection of all those sets, all those subspaces sorry which contain the set S, then you will eventually end up with the best refinement possible. So, all that I now need to show is that this W star is indeed equal to the span of S. Now, I am back in business with the usual tricks that I use which is to try and show that one is contained in the other and the other way round. So, how do I argue about that? So, again I will use that oft used term. Obviously, what is obvious? What do you think of span of S? You choose all the alphas to be 0 except 1 alpha i. It will imply that every member of the set S belongs to span of S. So, obviously, span of S contains S. Therefore, span of S is one such candidate for a W i and we know that span of S is a subspace right. So, span of S is definitely one such candidate for W i's is it not? Obviously, span of S is for some i is or can be written as span of S is equal to W i. What does that mean? When I take this intersection here, the object I end up on the right hand side is at least contained in each of them. So, therefore, if I am taking now the intersection over all W i's and span of S happens to be sitting as one of these fellows here, then this fellow must be contained in span of S because this fellow is contained in each W i you see that is the definition of intersection. When I am saying this fellow belongs to the intersection of this that means this fellow belongs to W i for every i and span of S happens to be one such W i for some i right because span of S contains every member in the set S. So, therefore, span of S definitely contains this. So, therefore, W star is contained in span of S right any doubts? Now all that I need to show is that this is also contained in this. So, I pick out any arbitrary object from here and I have to show that this must belong to this. How do I do that? So, I have to show that span of S is contained in W i star W star how is it how is it true or why should it be true? Any arbitrary object in span of S can be written as what linear combinations of fellows in right, but what does that mean? I mean this object you agree that it contains every object in S right. So, what does it mean? What can I say about this? So, consider S belonging to span of S. So, S is written as summation alpha i S i over i where S i belongs to S. What does this object belong to? Each of these S i is belong to every W i. So, each of the this element must belong to every W i know. You see the point? It is very funny sounding, but it is true right. S i comes from the set S each of the W i is contain every element in S their subspaces. Therefore, a linear combination of fellows from a subspace must also belong to that subspace. So, this must belong to W i for all i right. See once you follow the definition things become very simple. At first glance this might seem like a huge deal right oh how am I going to even see this, but once you find your steps and define your objects in the right manner things become very obvious. It is not a big proof right, but it is already done you see it belongs to W i for every i that means S belongs to the intersection of all the W i's i belonging to this index set which is nothing, but W star. So, I started with an object in span of S and I end up showing that it belongs to W star which means that span of S is contained in W star right. So, based on this here and this here it stands to reason that span of S is exactly the W i star that W star that I am looking for which is the smallest subspace that contains every element in the set S. So, that is why span is a special set ok. The moment I give you a set of vectors you cook up the span of that set of vectors and that is the smallest subspace that can accommodate all those vectors that I have given you in the set and yet also have the structure of a subspace right which this this. So, S i belongs to S, but S i is contained in S means it is contained in every W i because what is the definition of W i. So, W i is a collection of all those subspaces which contain S that means every object in S is contained in every W i. So, that means these S i's which are coming from S must be contained in every W i. If they are contained in every W i they must be part of the intersections of all those W i's that means they must be part of W star. So, every S i is part of W star, W star is a subspace. So, therefore a linear combinations of all those S i's which are coming from W star now as we have seen must also be a part of W star, but that is nothing, but S the linear combinations of those S i's. Therefore, S must also belong to W star, but S originally started with we originally started with the claim that S belongs to the span of S and now we cannot help but agree that S must also belong to W star. Therefore, any object you pick out from span of S must also belong to W star. Therefore, span of S must be contained inside W star. Earlier we have seen W star must be contained in span of S two subspaces contain in each other means they must be equal, right. So, we have already seen this W 1 union W 2 need not necessarily be a vector space, but it is definitely a set of vectors and the moment there is a set of vectors you can cook up a span of that set, but that will end up becoming a vector space a subspace, right. So, I can safely say that let us again start with that same thing W 1 W 2 these are subspaces, right and in general W 1 union W 2 the span of this, this is the vector space no doubt the moment you put this span it becomes a vector space whatever sitting inside it whether it is a vector space or a set does not matter the moment you are taking a span of something you can also take a span of a vector space does not matter, but that in itself is a vector space the moment you are acting letting the span act on it, right. So, this is our vector space the claim that I am going to make is any guesses W 1 plus W 2 beautiful relation is not it we have seen this come like a full circle now we started with intersections being subspaces then we saw sums being subspaces then we saw unions not being subspaces unless some special conditions were met then we saw this object span which was cooked up at least constructively theoretically using these intersections countable intersections and so on and now we see, but even if this object fails to be a subspace you take its span make it a subspace and that happens to be nothing but this nothing more or less than the sum of these subspaces, right. So, if you can show this in some sense we have kind of completed a study of these objects like sums intersections span and unions and the interplay between these relations and operations that you perform on subspaces, right, alright. Again standard technique for showing this equality what do you think is the standard way of showing the equality of this again containment yeah you will get bored of this by the time we are done with this course but that is the idea, right, okay. So, consider W belonging to W 1 plus W 2, okay implies W can be written as small W 1 plus small W 2 with W 1 belonging to W 1 and small W 2 belonging to W 2, right. It is already done one side is it see what can you say about W 1 it is definitely an object in W 1 union W 2 it is either in W 1 or W 2 it is precisely in W 1 same with W 2. So, this W 1 plus W 2 yeah what can we say definitely need not belong as we have seen to W 1 union W 2. However, if you allow for the span now it becomes a vector space. So, W 1 belongs to W 1 union W 2 W 2 belongs to W 1 union W 2. So, what can I say this is just 1 times W 1 plus 1 times W 2 if you think of W 1 union W 2 as a set then this comes from that set this also comes from that set yeah. So, this belongs to span of W 1 W 2 more formally speaking this is the set but is in this contained in what of course, W 1 union W 2 yeah the set W 1 W 2 is contained in the set W 1 union W 2. So, therefore, whenever something belongs to this smaller set obviously belongs to the bigger super set do not call this a vector space it is not it is not necessarily a vector space, but it is a set for sure. So, set theoretically if you see that a vector belongs to the span of a subset then it obviously belongs to the span of a super set right. So, therefore, you start with something implies that W belongs to this. So, therefore, W 1 plus W 2 is contained in span of W 1 union W 2 one side is done for the other side always remember definitions are your friends not your enemies. So, if you remember what definitions are therefore, just use the definitions that is all that you need. So, what can you say this is a set clearly at least a set unless there is this one of them contained in the other it is not a subspace it is a set. So, the moment I say that this W is in the span of this what does the span tell me. So, it is basically a linear combination of fellows in this set that is sitting inside this span yeah. So, W can be written as again recall just a while back I had written this set as a disjoint union of three sets I had partitioned this. So, I will just say this is a summation of say alpha I W I where I belongs to the index set for S 1 plus beta I just let me write this down then I will explain what this is. Let us call it beta j W hat j where j belongs to the index set for S 2 plus summation gamma k W till day k, k belongs to the index set of the set three with set S 1 being W 1 deletion W 2, S set 2 being W 1 intersection W 2 and S 3 being W 2 deletion W 1. So, what have I done because I know that at least this is a set if not a subspace. So, let us treat it as a set. So, any object that is sitting inside this set belongs to either of those three parts set S 1 or set S 2 or set S 3, but cannot belong to any two of them simultaneously. Three partitions of the set three disjoint partitions of the set clear. So, now if it yes yeah yeah. So, these index sets are not necessarily finite I mean fine linear combinations we take only finite fellows. So, if it is infinite then there are problems with other things like you know we do not take infinite even when we say linear combinations if you recall we only restricted it to finite sets. So, the set might be infinite, but when we say that we are taking objects like these those are basically finite because otherwise you have issues such as convergence and other things right. If you have infinite objects you are you are not always sure that if you are taking an infinite sum even a very you know crude thought process will lead you to believe that unless you have notions of convergence and other things you cannot talk about infinite sums. So, we only restrict linear combinations and all of these things to finite sets therein. So, everything that you can express is over a finite sum. So, finite number you have plucked out from this S 1 finite number you have plucked out from S 2 and a finite number you plucked out from S 3 because overall this W this fellow can be represented as a finite linear combination of fellows in W 1 union W 2. So, if it is a finite number then there is a finite number coming from this set finite number coming from S 2 finite number coming from S 3. So, then this carries forward right. So, what can we say about this now I am saying it is already done I mean there is this like two different ways in which you can approach this you can either put your brackets like these and say what about this fellow S 1 and S 2 that would be the part that comes from W 1 and this is the part that comes from W 2. And clearly then this fellow W has been written as a sum of two vectors one of which comes from W 1 the other comes from W 2 or it is a matter of pace really if you wish to combine this and this together then the thing inside the pink braces is something that comes from W 2 and the thing that is left behind after that is something that comes from W 1 again you know that is also an object that belongs to W 1 plus W 2 the sum of the two subspaces. So, therefore, you start with an object that is in the span of the union of the two subspaces and you end up showing that it cannot help but be also in the sum of those two subspaces. So, therefore, this containment relation that the union of W 1 and W 2 the span there of being contained in W 1 plus W 2 leads you to conclude that the two subspaces that is these two are equal. Please complete that a few lines remain to be written just formally to complete this and then draw a line and say QED. So, that is that is an important relation alright. So, we will discuss about other things in the next lecture such as linear independence and other such related notions but all in good time. So, for today let us call it to close now. Thank you.