 So, in the previous lecture we have just seen some properties and probably descriptions of subspaces and we have seen that there is a convenient way of checking whether something is a subspace, right. Given that it is sitting inside a vector space, so the vector space would be an extension of that subspace, given that it fits in with certain rules obviously, that is the reason why the extension is a vector space. The subset that we have chosen has to meet this property that you pick out any two arbitrary elements from the subset say v1 and v2 and take alpha v1 plus v2, right. And if that also belongs to the aforesaid set subset, then the subset under consideration is a subspace, but the very important point which I am not sure if it got highlighted enough is that even before you check alpha v1 plus v2, you should always check that it is a subset after all, which is to say that if I give you this v with this structure over f and if I give you this w, right, the first check would obviously be this inclusion. The fact that this w is after all the subset of this, otherwise none of it makes any sense whatsoever that is a very crucial check. So that you of course check for say some alpha w1 plus w2, whether it belongs to w for all alpha in f and w1, w2 in w, alright. So today we shall see some interesting properties of subspaces, some further interesting properties and also we will see that we can cook up several new subspaces given some subspaces by performing certain operations that are set theoretic in nature, sort of, alright. So we look at some of those newly cooked up subsets and also certain things that are non-examples of subspaces. So you do certain operations on these subsets, sometimes you land up with a subspace, sometimes you do not land up with a subspace and in fact it can be shown that in certain cases you will not land up with a subspace. So the proof of the contrary, alright. But for any of that let us define a very, very important subspace, we have already seen two important subspaces with respect to matrices, the image and the kernel. As it turns out the matrices are nothing but some special cases of what we call linear operators or linear transformations for rectangular matrices, operators is for square matrices. We will see that later when we discuss more about linear transformations, more on that we will follow but in a more abstract sense given that you have a set S which is contained inside a vector space V, alright. So suppose you have the set S contained inside this vector space V, okay. Under this notation sometimes it is also called SP of S or also called the span of S, okay. So as of now this S is just a set, alright. This S is merely a set, nothing more. V is a vector space but S is just a set but what we are going to now define this span of S will turn out to be a subspace of V. So how do we define this? Okay, these are just notational purposes so I am just going to keep them aside but here is how it is defined. This is a collection of all such S belonging to V such that S is written as summation of i going from 1 through N, alpha i S i with S i coming from the set S for all i, alright. And of course we will only take finite numbers so this N is not infinite it is some finite collection, okay. So we are talking about spans of finite sets but you can also extend this. The set by the way can be an infinite set, okay. So let us try and get an idea about what this looks like in the spaces that we are familiar with and then we will try and investigate certain more properties of this. So for instance consider the two-dimensional the R2 for example the Euclidean space, yes. See oh so sorry sorry that is S i for all, yeah so with S i belonging to S for all i, yeah that is correct. So that means each of these S i's they come from S, yeah so S1, S2, S3, S4 suppose you have till S500. So for all i's from 1 to 500, S1 through S500 they all come from the set S. So basically combining elements in the set S, okay maybe I should also write alpha i as part of the field, alright. So we will just look at some examples, let me draw them with different colors. Let me take first a single vector say V, okay that is the set S just a single vector, okay. Then let us say this is my set S, okay and then let us say this is my set S. So obviously when I say vectors here I mean points on R2, okay. So what do you think is the span of this first one the one I have drawn in green, no we do not yet know what dimensions are technically speaking. So what should it look like straight line along which direction? So you mean straight line along this, so just this and along this, great. So everything extending in both directions along this. What about this set of points? Is this by the way a subspace in itself, this set S, what do you think? If you take any two arbitrary vectors from here which is essentially saying any two arbitrary points here, can you check whether alpha V1 plus V2 also belongs to this set will it? You understand what I am saying? So I am saying in one case this is the set S, in the other case this is the set S, in the other case this is the set S, so the collection of all these points this snake like object here. Is this a subspace? But it is a set right and for talking about spans we do not require this to be a subspace. So it is fine it can be any arbitrary set, so these are all sets that is true. Is this vector taken alone a subspace? It is not. What about this set by the way coming back to this, is this a subspace? How can you say it is not a subspace, at least geometrically what is the intuition? I mean only thing I am telling you is it is not a straight line. So why do you think it is not a subspace? Argue geometrically let us say on this figure, R2 we understand well enough right, okay and what happens? So that the point does not belong to this set anymore, so obviously you forget about linear combination just the scalar multiplication does not keep it, I mean closure under scalar multiplication is not valid right. What about this object is this a subspace? It is not a subspace, but what do you think are the spans of these two fellow zone will look like, the whole R2 is it, is it true? Just this thin looking object here, all points here is just a small set right and yet I mean some of you are claiming here that this span of this set, this snake like serpentine line is apparently this whole R2, is it true? What is it essentially saying? What does that entail? It means that you are claiming any arbitrary point that I pick out here, yeah I should be able to write down as a linear combination of certain points that I pick out here on this, okay. We will investigate that in further detail, but as it turns out the answer is correct. About this is the same conclusion true that the span of this set is the entire R2, is that clear to everybody, is it obvious to everybody? If it is not obvious to even a single of you please raise your hand, because these insights are important, at least geometrically in R2 we should be able to assess what this looks like, alright. What about say something like this, so if you understood that then you should be able to answer this, so let us say this is x, this is y, this is z, so this is R3. Let us say on the xy plane I draw some closed curve like this and I take all the points inside this and call this S, what is the span of this set, the entire xy plane not R3, not the entirety of R3, okay good. So it means you are getting a hang of it, but okay we will let it rest there, so it is good that we have a geometric understanding at least so far as we can stretch our imagination as to what this span entails, okay. So having understood what this object called a span of a set is, crucial thing I reiterate again remember this is not required to be a subspace, any set given any set you can always define a span of that set and the span then becomes a subspace that is something we have to prove now, okay. So the claim is that this is a subspace, of course geometrically with those examples you have already seen, right. So of course the fact that the span of S belongs to V does not need to be proved, it is obvious from the definition because each of these objects is coming out of this, right, each of these objects is coming out of the set S which is in turn contained inside the vector space V, yeah. So that fact that this whatever you cook up inside this span must be contained inside V goes without saying, so the first check is done. So then it makes sense to investigate whether if you pick up two objects from the span say V1 and V2 and then try out this check of alpha V1 plus V2 whether that is also contained inside the so called span, it makes sense, right. So consider V1, V2 belonging to span of S, it means that we can write V1 is equal to summation say alpha i S i where of course S i belongs to S and V2 is equal to summation beta i S i, S i belong to S and of course alpha i beta i they come from the field. What do we need to check? We need to check that what happens to alpha V1 plus V2, okay. Look at alpha V1 plus V2, what can we say about it? We have alpha V1 plus V2 given by what exactly? Alpha alpha i plus beta i times S i, the sum thereof, isn't it? What do we know about this number? Of course alpha is part of the field, so because of the closure under the field addition and multiplication this is also part of the field. So let us call this summation alpha hat i S i and what is my conclusion then? Anything that can be represented in this form must therefore belong to the span, right. So this naturally belongs to the span of this set S. So whenever I have picked out V1 and V2 from the span of S, I immediately see that alpha V1 plus V2 no matter what alpha I pick, no matter what V1 or V2 I might have picked up, these were all arbitrary, right. This object must also belong to the span of S, hence prove that span of S is a subspace, okay. So it is a very important subspace as we shall see later, okay. We will push for very important result related to the span but before that we need to understand a few more constructions with these subspaces, okay and how to cook up newer subspaces from given existing subspaces, okay. So the next important thing is intersections of subspaces, okay. So we start with the assumption that suppose W1 and W2 which both, so these are subspaces, not just subsets but subspaces, subspaces of V. I am not even mentioning the field it is understood from the context that over some common field so that they are subspaces of V, each of them. The claim is that W1 intersection W2 which obviously is contained in V is also a subspace, okay. It is just a bit of semantics once you understand what intersection implies and you know how to write proofs little bit, you can just push this through but that is what we are here to learn so let us try and give you a sketch of this. How do we say that this will be a subspace? Again you have to consider two arbitrary objects that belong to the intersections of these two fellows and take alpha times the first plus the second and show that that also must belong to this, right. So we have to argue. So suppose let us call it a proof. Suppose let us say V1, V2 belong to W1 intersection W2 that immediately means that V1 belongs to W1 and the intersection means it is a logical and right and V1 belongs to W2 similarly V2 belongs to W1 and V2 belongs to W2, right. What can we say about? So now again look at alpha V1 plus V2 where of course alpha belongs to the field. What can we say about this object? See W1 and W2 are subspaces themselves because of the properties of subspaces. What can we say about this object? V1 belongs to W1, V2 belongs to W1 therefore alpha V1 plus V2 must belong to W1. By the same token V1 belongs to W2, V2 belongs to W2 therefore alpha V1 plus V2 must also belong to W2, yeah. This must also by definition of subspaces. So I am not even writing that by definition belongs to W1 and W2 implying that alpha V1 plus V2 also belongs to W1 intersection W2 which is all I needed to show. So therefore intersections of subspaces are also subspaces. In fact we can extend this further and make a claim I leave that to you as an exercise to try, show that countable intersections of subspaces are also subspaces. In short, yeah, this belongs to some index set WI. So that is a subspace when each WI is a subspace. Countable means this need not be a finite set but it is obviously some index set. So it is not 1.534, it is 1, 2, 3, 4 those kind of numbers that can come, right. So you can use something like what is the technique that you know most often that comes in handy in such proofs. You know it to be true for a certain number, you use induction, right. You use let us say it is true for intersection of k and then there is nothing preventing you from extending it to k plus 1 because once you have gotten it for k objects that becomes one subspace. Now you take then it is at every step you are only considering two subspaces at one go. You take the intersection of the first k and take the k plus first. So it is again boils down to two sets, yeah. So that is about the intersections of subspaces, okay. Immediately the moment you have intersections it kind of begs the question as to what happens when you take the unions of subspaces because you know union and intersection operations kind of come in hand in hand together. You are you talk to think about it like that. So we ask the question as to what happens if you take unions of subspaces, okay. So suppose you have the claim I am going to make W1, W2 are both subspaces, okay. Here is the claim. W1 union, W2 is a subspace if and only if either W1 is contained in W2 or W2 is contained in W1. Otherwise, in general the union of two subspaces fails to be a subspace, alright. An interesting result but if you think about it for a while it is not so surprising. See for example just in R2, in R2. So this is the x direction, this is the y direction. Every vector is given of this form xy. So you take the span of let us let me use the color. Let us call this set S1 and let us call this set S2, okay. By the intuitive understanding that you might have already obtained about what a span is consider the span of S1 union the span of S2, alright. These are subspaces you agree? Of course by definition a span is a subspace we have just proved that. What is the span of S1 going to look like geometrically? The entire x axis, span of S2 is the entire y axis, alright. What about objects in S1 union S I mean span of S1 union span of S2? What sort of objects can you pick? You can either pick objects from here or you can pick objects from here. So let us pick two non-trivial objects. Let us pick this point from here and this point from here. The moment you add them, yeah, by the parallelogram law of vector addition this is what you end up getting. Do you not? But this point or this vector is neither a part of the span of S1 nor a part of the span of S2, right. So it should really come as no surprise that the unions of subspaces do not always result in a subspace. At least intuitively you should already be figuring this out. But of course there still remains the task of proving this and also this containment condition. But really that containment condition is just quite straightforward. I mean if one set is part of another set then what is the union of the two sets? It is nothing but the bigger set, right. So just consider the opposite when W1 is contained in W2 then the union of W1 and W2 is just W2. Similarly when W2 is contained in W1 the union of W1 and W2 is W1 and of course we are saying already upfront that W1 and W2 are subspaces. So then the union will obviously be a subspace. Nothing fancy being said there. So one part is obvious. If W1 is contained in W2 then implies W1 union W2 is equal to W2 is a subspace again by definition. Similarly if W2 is contained in W1 again we have W2 is contained in W1 union W2 is equal to W1 is a, you can fill out the gap, okay. So that part is done. Now comes the crucial bit where we will disprove, where we will say that under certain situations when these conditions are not met indeed it will always fail to be a, I mean to say this is precisely the condition. See that is what if and only if does. You might think oh this is an over kill. I mean this is really a trivial statement. One is contained in the other what is there to check. But it turns out that that is exactly what you need. You are not really overreaching or you know going for an over kill. The moment this fails you are 100% sure that it is not going to result in a subspace. How are we so sure? So let us say suppose W1 is not contained in W2 and W2 is not contained in W2 and W1 is not contained in W1. I hate to draw pictures but sometimes the pictures do help. I mean pictures of sets and objects like these. In this case if you think of two sets like the typical Venn diagrams that you are probably used to seeing. If you take the unions of two sets you can essentially look at that set as combination of three sets. One part which is essentially the intersection of the two. The other part which is the first set deletion the second set. The other part the third part which is the second set deletion the first. What happens then? It creates a so called partition of the overall set that is the union of the two sets. Partition means the union leads to the exact set and their intersections are empty sets. So disjoint. So you create basically disjoint sets. In other words if you have sets like so suppose this is S1, this is S2 then this part is something you can call as S1 union S2 and this part is what you call S1 deletion S2. Sorry this is intersection. Yeah thanks you are right. And this part is basically S2 deletion S1. Now the moment I am saying that neither set is completely contained in the other what can I immediately infer? With this picture in mind what can I immediately infer? I mean look at these objects S1 deletion S2. If this was an empty set what would happen? S1 would be completely contained in S2. Similarly if S2 deletion S1 were an empty set then this would be completely S2 would be completely contained in S1. Now extending the same idea here the moment we are saying that W1 is not completely contained in W2 and W2 is also not completely contained in W1 we might immediately say that this implies W1 deletion W2 is not the empty set and W2 deletion W1 is also not an empty set. It is clear? Why this must be so? Right? Okay good. So all right once we have that what can we say that W1 union W2 is nothing but the set W1 deletion W2 union W1 intersection W2 union W2 deletion W1. Okay? Where these two are at least non-empty sets. Okay? So pick V1 belonging to W1 deletion W2 and V2 belonging to W2 deletion W1. All right? Look at W1 deletion W2 deletion W1. V1 plus V2 say it is defined as some vector V. All right? Can V belong to W1 or W2? I have to show that V can neither belong to W1 nor belong to W2 in which case V will fail to belong to W1 union W2. Look there is no doubt about the fact that both of these V1 and V2 they are coming from this set. Now if this is to be a subspace then this object this fellow here which I have now defined as this V must either come from W1 or from W2 only then it will meet the condition. Of course that is not enough but if one case fails then failure is assured at least right. If I just show this for this choice it is not enough to prove it is a subspace but failure to meet is enough to prove that it is not a subspace you agree. If I have to show it is a subspace then I have to show for every possible value of alpha. I have just chosen one alpha alpha is equal to one here but if it fails to hold true for alpha is equal to one no point in looking for anything else. I have already found a flaw in this reasoning right that it needs to be a subspace. So suppose on the contrary V belongs to W1 observe that minus V1 belongs to W1 why because W1 is a subspace. So whenever V1 belongs to W1 the additive inverse of V1 which is minus V1 denoted by minus V1 must also belong to W1. Now suppose you are assuming that V belongs to W1 and minus V1 belongs to W1 what does this immediately lead us to? V minus V1 belongs to W1 because there is no doubt about the fact that W1 is a subspace that is our choice but what is V minus V1? What is V minus V1? That is V2. So V2 belongs to W1 is it possible? Why have we chosen V2 to come from? V2 comes from W2 deletion W1. So it comes from that part of W2 which is not common or overlapping with W1 right. So which means that this is an absurd proposition, reductio ad absurdum we have reduced it to absurdity right. So this is a contradiction of course the same thing holds the other way round as well where you are assuming that this belongs to W2 and then you take minus V2 which also belongs to W2. So you take V minus V2 that must also belong to W2 but that is only V1 which if it belongs to W2 it contradicts our very choice of V1. So therefore I again leave it to you to complete this part you have to just replicate this part once again and then finally conclude that since V which is basically a combination of two fellows from W1 union W2 can neither belong to W1 not to W2 and hence does not belong to union of W1 and W2. So therefore this set W1 union W2 now it is indeed a set and not a subspace is not closed under this operation and therefore it fails to be a subspace unless these two properties are true. So this is exactly the property that you need right it is not like you are making some leeway or some relaxation this is if and only if clear it has to belong to the other part no if you subtract it where must it belong to it must belong to this V2 but that choice the first stop with the choice of V1 and V2 that we made they were not allowed to belong to that intersection that is why I showed you the W1 union W2 can also be written as the union of three disjoint because W1 and W2 may not be disjoint but then I partitioned it into three sets which were disjoint and then I said that whether you pick an object from this set on the left hand side which is the union of W1 and W2 or you pick an object from the right hand side which is the union of those three sets and that choice then I made from this specific description of those three sets and I said that V1 must be in W1 deletion W2 V2 must be in W2 deletion W1 where they are sitting in W1 and W2 may not be very obvious but the moment I am saying that V1 is allowed to be in W1 but not in specifically not in W2 and then later on I come back to the conclusion that oh hang on it must also belong to W2 that leads to a contradiction right that is the reason why I wrote all that up and said that look you can also look at this union of these two sets as the union of these three sets and then I am looking specifically at those two which are clearly disjoint because W1 deletion W2 and W2 deletion W1 are definitely disjoint there is nothing common between those two right yeah right so we have this and the important claim which we shall see in the next module is you might have already guessed that this object W1 plus W2 is a subspace okay so we will prove that next