 So, we ended our previous lecture by defining what is called a polyhedron. I will just revise that for you quickly. So, a polyhedron P is a polyhedron is if it is the intersection of finitely many half spaces, alright. So, if you can find finitely many half spaces, so that such that their intersection defines is equal to P, then that then P is a polyhedron. So, what this means is there exist some say m of these m of m vectors a1 till am in Rn, let us assume P is a subset of Rn. There exist these m vectors in Rn and also m scalars like this v1 till vm such that P can be expressed as the common region that belongs to all of these half spaces defined by these vectors and this constant. So, P is x such that ai transpose x is less than equal to bi for all i from 1 till m. This in we as I said we can consolidate these a's by writing them in the in this way you write say for a1 transpose as the first row of a matrix a2 transpose as a second row and so on all the way down till am transpose that gives you a matrix a, you put the b's together in a vector called b and then that gives you that P is actually nothing but the region x defined regions defined by x such that a x is less than equal to b. So, it is the region that is defined by finitely many linear inequalities. Now, this is this sort of way of defining a polyhedron where we first say it is the intersection of half spaces and then you go about finding the half spaces and give you that gives you this sort of representation. Now, polyhedron may not have may have multiple representation. So, on the face of it what looks like a poly, what looks like it is not a polyhedron could also end up being a polyhedron. For example, here if you look at this, if you look at this kind of representation, this is saying that the polyhedron is all the x's that satisfies some linear inequalities. But then you can also have a set like this. Let me show you for example, let us look at this set in R2. So, this is plus 1, this is minus 1, 0, minus 1 plus 1 and I am looking at this b. So, this region, what is this region? Well, this is one way of representing this region, let us call this set P, this region let us call this P, one way of representing it is to say well, this is x in R2, sorry R2 such that if I look at the absolute value of the first coordinate that is less than equal to 1 and the absolute value of the second coordinate that is also less than equal to 1. So, the absolute value of the first coordinate and the absolute value of the second coordinate are both less than equal to 1. That is actually this region. Both coordinates have any point in this region you take a point say here for example or here wherever you want, if you look at its coordinates in absolute value they are less than equal to 1, that is actually this region. But if you look at these inequalities, the absolute value of x1 less than equal to 1 and the absolute value of x2 less than equal to 1, these are not linear inequalities. They are not of this kind because the absolute value function is not a linear function. The absolute value function is a the absolute value function. How does the absolute value function? What does the absolute value function look like? It looks like this. So, this is absolute value of P and here is T, this is 0, this is going to plus infinity, this is minus infinity. So, the absolute value function is not a linear function. So, if you look at these inequalities, I have written P as an as the common region of two nonlinear inequalities. So, but that does not mean P is not a polyhedral. The point is P is still the intersection of finitely many half spaces and what are those half spaces? Those half spaces can be seen here itself. So, you can take this as one of three half spaces, you can take this as another half space, take this as another half space and then finally this as another half space. The common region between all of them is actually our set P. No, no, no, A transpose X less than equal to B. So, definition of a half space is that it is A transpose X less than equal to B. So, what we have is that this P is, if you look at it in one way, it is defined using as the common region as a common region of nonlinear inequalities. Whereas if you look at it in another way, it is actually the intersection of finitely many half spaces. So, because it is an intersection of finitely many half spaces, it is a polyhedral. So, the point is this here is our definition of a polyhedral and from here all we can say is that there exists some way of representing it as an intersection of finitely many half spaces. That means there exists A1, A2 up till AM and B1, B2 up till BM such that their intersection gives you P. Now, they may not be the one that you have chosen at the moment. Here you have chosen some inequalities, they are nonlinear inequalities, that does not mean there are no linear inequalities that will describe the same reason. So, in particular you can see observe here that P is also the intersection of these inequalities. So, the absolute value of X1 is less than equal to 1 if and only if X1 is less than equal to 1 and also greater than equal to minus, less than equal to 1 and greater than equal to minus 1 likewise for X2. So, these are actually the half spaces that I have just highlighted here. So, these are these, this is, this here is one of your inequalities, this is another inequality, this is another inequality and this is another inequality, total of 4 of them can put them together in a systematic form X in R2, X1 less than equal to 1, X2 less than equal to 1, X1 minus X1 less than equal, less than equal to 1 and minus X2 less than equal to 1. So, those define R4 half spaces, is this clear? So, a polyhedral is the intersection of finitely many half spaces. Now, this as I said and of course you objects this what this example showed you was that objects that on the face of it may not be written as an intersection of linear inequalities can still be turn out to be polyhedral. Now, another thing to point out here is the importance of finite. If I do not allow, if I allow for infinitely many half spaces, then many, many sets end up being polyhedral, which are not really what we want them to be. So, for example, just take the circle, there is circle with this, the region inside, I take a tangent here, I can, this circle is actually the intersection of all of these regions. This region defined by this tangent, this region defined by this tangent, etc, this region defined by this tangent, etc, etc. I draw a tangent at every point, I will get infinitely many such tangent, infinitely many such half spaces, their common region is actually the circle. The circle is not a polyhedral because it is not an intersection of, it cannot be written as the intersection of finitely many half spaces. So now, this is one kind of issue related to representation where you have a polyhedron that can, you can have a set which is represented by nonlinear inequalities and can still be a polyhedron. The other type of issue of representation, which I want is what is extremely important for optimization, which is what I want to talk about right now. So, this is, for this we need a couple of concepts. So, let me introduce these concepts for you. First is what is called an extreme point. So, consider a some convex set. Now, X is said to be an extreme point of S if the following is true, following implication is true. That means the implication is that X can be written as a linear combination of a convex combination to be precise. So, of two other points in S. So, X is equal to lambda X 1 plus 1 minus lambda X 2 where lambda is strictly greater than 0 and X 1 not equal to S 2 belong to S. So, if X can be written in this way, then it should imply if X, let me remove this, X can be written as in this way where X 1 X 2 belong to S, then this should imply that X 1 equals X 2. So, if X can be written as the convex combination of two points using a strictly positive weight for each of them. So, X can be written as some lambda which is X 1 plus 1 minus lambda X 2 where lambda is strictly positive. Then it has to be that X 1 equals X 2 and when X 1 is equal to X 2, they are actually both equal to X itself. So, an extreme point is a point that cannot be written as a as a strict convex combination of two distinct points in the set. So, an extreme point is one for which this implication holds. So, whenever you try to write it as a convex combination like this of two different points, it has to be that those two points are actually the same and it is equal to the point itself. As a convex combination of two points and the two points have to actually coincide. So, what does an extreme point look like? Let us take for example, just I was just writing drawing a circle takes a circle is a convex set. What are the extreme points of this set? Every point on the boundary is an extreme point because the circle does not have you take any point on this on the boundary. If you try to if you are able to write it as a convex combination of two other points, then the circle would end up having would end up having a flat flat surfaces. So, this sort of point cannot be written as a convex combination of two other points in the set. Every point in the interior can of course be written as a convex combination of two other points. There are always two other points around it whose midpoint will be this point. So, every point on the boundary of the circle is actually an extreme point of the circle. So, the extreme points actually are more interesting when you are talking of polyhedron. So, if I draw a polyhedron like this, this is an intersection of half spaces. So, the half space is being this, this half space, this half space, this half space, this half space and likewise this half space. So, now tell me what are the extreme points of this polyhedron? Once again a point here in the interior cannot be an extreme point. Oh yes, of course, sorry, I said convex combination but did not make that clear. Yes, lambda strictly between 0 and 1. Yes. So, we will come back to the polyhedron here. So, this sort of point is not an extreme point. What about, what are the other points which cannot be extreme point? Yeah. So, if I take some point here in the middle of an edge here, this sort of point is also not an extreme point. So, I can write it as a convex combination of two other points like this. So, eventually if you think about it, what remain as extreme points are actually only these points. This one, this one, this one, this one and this one. These are the extreme points of this polyhedron. Let me draw for you another kind of polyhedron. So, this is an unbounded polyhedron. So, what are the extreme points of this set? Let us say this goes on forever. What are the extreme points of this? It has only two extreme points which is this one and this one. These are the extreme points of this polyhedron. Now, this figure is actually very interesting. If you think about it, this polyhedron is unbounded. So, what does that mean? There is a sort of a direction here along which if you keep going, this sort of direction. Just imagine you started off from here and you keep kept proceeding along this direction. You keep going, you will remain in the set. There are directions like this for a polyhedron. Once it is unbounded, verify that it is unbounded means gives you that there will be some direction along which you can travel such that you will never leave the set. So, and what is amazing is that these directions are such that they do not it does not matter where you started from means. So, if there is a direction like this, let us say a direction like this along which if you can. So, if you started from a point here and went in this direction forever, you will remain in the set. But I can go along that direction starting from some other point also in the set and I will still remain forever in the set. So, for instance, I can start from this point here and still go in this direction and remain in the set. So, there are directions like this. Once a polyhedron is unbounded, there are directions such that if those directions are such that you can start from any point in the set, keep walking along that direction and you will never leave the polyhedron. These directions are what are called extreme rays of the polyhedron. Sorry, these are these directions are what are called rays of a polyhedron. Sorry, not extreme rays, they are called rays of a polyhedron or also called recession directions. So, ray. So, this actually this particular thing that I just mentioned, this does not really require the set to be a polyhedron. So, long as it is a convex set, if there is a direction along which you can keep going starting from some point, you can go along that direction starting from any other point and you will still remain in the set. This is the shape of, this is the nature of the shape of convexity. So, a ray is defined like this. So, let us consider a convex set. Let us take that convex set. So, consider a convex vector Rn is called a ray if the following implication is true. So, x belongs to the set S implies that x plus mu times D belongs to the set S for all mu greater than equal to 0. So, if it is unbounded, then if it is bounded, then the only such vector will be D equal to 0. So, once x is in S, this x plus mu D is in S for all mu greater than equal to 0. So, x plus mu D is, so the way you should look at this is that here is your point x, here is the direction D, x plus mu D as mu get is this since as you go from for mu greater than larger and larger mu, you keep going along this dotted line that I just further and further from x along the direction D. So, for all mu greater than equal to 0, this point belongs to this x plus mu D belongs to S. So, this such a vector is what is called a ray. So, another name for it is called D is also called also called a recession direction. And the set of all recession direction of a set S is often denoted by S infinity, all D such that D is a recession direction of S. This actually is turns out to be a cone and it is called a recession cone of S. It is very easy to check that this is going to be a cone and it is called a recession cone of S. So, if the set is bounded then the only vector that satisfy that would satisfy this sort of condition is the vector 0. So, in fact, it is this is this statement is also true that S infinity this cone is as is equal to just the 0 vector over the singleton set with only the 0 vector if and only if S is bounded.