 So, let us consider v is equal to real matrices right n cross n real matrices and look at you know what this is this is called the trace sum of the diagonal entries of a matrix that is a trace of a matrix. So, you have a and b two matrices right they get mapped to trace of a b transposed. This is also an example of an inner product over the space of matrices. If you change it now we will introduce complex again you can take the pair of matrices a b complex matrices and they can get mapped to the trace of a b Hermitian. When I say Hermitian it is basically the transpose of the conjugate. So, b Hermitian is defined as take the conjugation of b and then it is transposed right that is also an example of an inner product. You can go ahead and check that there are square integrable functions. So, this that is also a vector space by the way square integrable functions. So, square integrable functions are those such that if you integrate them f t squared dt is going to be bounded. So, this is a vector space of square integrable functions. So, you take l squared and l squared to let us say r. So, these are square integrable real functions and you can integrate them from minus infinity to infinity f t g t dt ok f and g getting mapped to this is also an inner product alright. Let us take the case of continuous functions over some interval in r. So, real continuous functions over some interval i where i is some interval over the real line alright. So, that is also a vector space and you can just see that if you take f and g and you let it be mapped let us say some lambda some generalization you take the integration over the interval i you have lambda t f t g t dt for lambda t which is positive over i ok that is also an inner product. So, you see all these are exotic vector spaces and yet they all share this common property that they have an inner product. So, inner products need not necessarily look just like those dot products that you have seen over Euclidean spaces rather anything that meets those four properties which we have defined or identified with inner products qualifies as a legitimate inner product over the corresponding vector space right. So, now we will define inner product spaces and it is quite straightforward any vector space v over r or c having an inner product say this defined on it is an inner product space denoted by v along with just this inner product right. So, we put those two things together and that becomes an inner product space right. So, you have seen what essentially definition of an inner product we have hopefully grasped some sense about its properties what it entails and all and I would be check for whether a given function is an inner product. We have also now seen that any vector space armed with this notion of an inner product of course for that the vector space has to be defined only over r or c that is a prerequisite, but once it is armed with an inner product it passes off by the moniker of an inner product space right and we have seen some examples along the way, but now we will see certain consequences of a space being an inner product space it satisfies certain properties there is certain things about the topology of an inner product space that intrigues us it allows us to do much of the mathematics in very much the same way that we did for Euclidean spaces you see because it can allow us to associate this idea of certain things like angles that we inherited from Euclidean spaces right. So, that is what we shall now look at. So, this definition is clear right I am going to erase this definition now. So, here is a proposition that I have let V with this inner product be an inner product space right. So, this is a function that we are now going to define from V to the field given by ok. So, what does it do? It goes ahead and picks out objects arbitrary objects inside the inner product space because it is an inner product space. So, this inner product is well and truly defined because it is going to be positive definite therefore, I can go ahead and take the square root without worrying about whether it is a real field or the complex field this is always going to be real and this object is what I am going to define as this function ok. So, suppose this is the inner product space and this is the function. So, what do we have then? Then firstly we have right. So, this is the property of homogeneity sorry yeah thank you I miss that. So, this is the property of homogeneity. What other property? Of course, we have already given that as a property of the inner product itself. So, no surprises that it is going to be a property of the this function as well what is it? Positive definiteness right has to be. So, again that is positive definiteness. So, even though I am saying these are parts of the proposition you should check that there is really nothing much to prove in this homogeneity and this positive definiteness. Why is the homogeneity true? Well again by virtue of how it is defined here right you are only going to take the positive square root. So, you take alpha, alpha v and alpha v right it is just going to pull out that alpha and alpha squared if it is complex then it is alpha and alpha conjugate which is still the square of the moduli of alpha when you take its square root you just get the moduli of alpha out yeah. So, the homogeneity whether it is real or complex field does not matter. So, these two are really nothing much to prove. The first non-trivial property that this function exhibits is the following can you guess what this is Cauchy Schwarz inequality. So, actually there is the name a third name also often associated the name of Binet. So, Cauchy Binet Schwarz inequality C B S or Cauchy Schwarz whichever way you like it right. So, I still write it as C S inequality for all v 1, v 2 belonging to v this is celebrated Cauchy Schwarz inequality in Euclidean spaces this turns out to be the fact that cos theta is less than 1 and you are done with the proof right. But we will prove this of course for general vector spaces and it is very very powerful because now after we have done this proof I would urge you to go back and look at those different inner products that we have defined just a while back on some of those exotic vector spaces and apply the Cauchy Schwarz inequality corresponding to the inner products that have been defined on those spaces. And then you will begin to see that all of them do not look like that Euclidean space and those inequalities may be quite a handful to prove particularly the ones with traces or the integrations of those functions and so on and so forth. And yet because of this Cauchy Schwarz inequality they are as true as the inequality that you manifested in terms of dot products when you study Euclidean spaces right. The fourth important property this is also another inequality can you guess what inequality triangle inequality. So, we have v 1 plus v 2 yeah what is this the triangle inequality basically says the sum of two sides of a triangle is always going to be greater than the third side right. They can be only equal if it is a degenerate triangle when they are all along the same line right. So, this is the triangle inequality not sure if you will get that to the end of the triangle inequality today but at least I will try to prove in whatever time is remaining try to show you the proof of the Cauchy Schwarz inequality in today's lecture and close it there right. So, these are the four properties the first two I put it to you are straightforward for you to verify follow from the definition itself. So, the Cauchy Schwarz inequalities what we shall now try to prove all right. So, consider this particular object ok what can we say about this object of course it is going to be positive not negative at least yeah and I am totally agnostic to the choice of field that is whether it is real or complex all you are going to use in this proof is just the properties of inner products. So, what is this as per our definition of this function that we have not named yet, but now we can actually call it the norm. So, this is the square of the norm right. So, what is this we know how to define this in terms of an inner product right. So, this means that the inner product of v 1 minus v 1 v 2 times v 2 upon norm of v 2 squared and v 1 minus again the same object v 1 v 2 v 2 upon. Now, you might think that oh hang on suddenly he is brought in this humongous looking expression from out of nowhere again of course, because I know how the proof works. So, I just a up front I proposed it, but there is also some intuition behind this if you think of and nowhere better than to start our intuitions from Euclidean spaces that we understand. So, what is this when you take the inner product of a vector with another the dot product of a vector with another in the Euclidean space what is it that you are exactly doing you are looking at you are projecting that one vector on the second vector right and then if you are dividing it by the length of the second vector then what you have is essentially the length of the first vector or the projection of the length of the first vector on the second vector yeah and then if you are multiplying it by v 2 upon the norm of v 2 it is like the unit vector. So, you are basically taking that length along the direction of v 2 yeah and you are subtracting it from the original vector. So, just think geometrically in Euclidean spaces at least what sort of an object this vector represents exactly. So, that is all that the argument starts from. So, so long as there is this orthogonal component some nonzero orthogonal component we should be able to show the Cauchy Schwarz inequality if it is if it vanishes altogether that is those two vectors are aligned to begin with then you have equality right that is the intuition. So, this huge looking expression is not just pulled out from thin air but also there is a good amount of intuition behind this. So, now let us go ahead and carry out the algebra with this. So, what is this the first term with the first term here that is just v 1 with respect to v 1 when we take the first with the second remember there is a sesquilinearity to be considered because it is not necessarily a real inner product space. So, we have to accommodate that possibility right. So, this object is actually a scalar either a real or a complex number right. So, we have to accommodate that possibility if we do that then what happens to this minus v 1 v 2, but this object will come out, but because it comes out and it is a scaling factor of the second argument. So, what do I have to do? Conjugate of course, there is nothing to conjugate in this the conjugate will only show its effect on this right. So, this is v 1 v 2 the conjugate thereof divided by v 2 squared yeah. So, this with this comes here this with this comes here now I need this with this of course I do not need to conjugate anymore why because this is of course the coefficient of the scaling factor of the first argument. So, first argument it does not matter it is completely linear in the first argument right. So, I have minus v 1 v 2 upon v 2 squared the norm squared that is and this is v 2 v 1 v 2 v 1, but this v 2 v 1 is bugging me. So, I am just going to flip the order and take the conjugate. So, that these two terms end up looking similar you agree. So, I might ask you to do this in one more step, but I will just use the fact that I can delete chalk marks easily agreed last plus of course, this fellow is unaffected, but this fellow will come with its conjugate, but what happens when you multiply a fellow with its conjugate it is a square of the moduli of that fellow. So, what you have essentially is v 1 v 2 the moduli the square there of and v 2 the square and square. So, it is the fourth power, but on top here you have. So, this is what v 2's norm squared which is greater than or equal to 0 yeah, but what are these after all again this is the same as this is it not because it is just x and x conjugate x and x conjugate and this is also x and x conjugate which is a moduli of x squared right. So, I am going to erase this first line and continue here. So, you agree that this term and these two these are exactly the same terms only that one of them will cancel out with the other leaving only just one of those negative terms behind terms with the negative sign behind. So, in one short I am now going to write this as v 1 squared because that is what this is right v 1 squared and this and this will cancel out. So, let me just cancel this out with this. So, that I will be left with just one of those terms minus from whence I can infer upon taking appropriate square root right. So, irrespective of whether this is the real field or the complex field the Cauchy Schwarz inequality is true and we use nothing other than just definitions that we have introduced earlier in the lecture today. Any doubts on this? Otherwise we will put an end to this discussion here today. Sorry in the second term of the. Here, here, this one, this, this one, yeah, this one. So, what I am doing is I am taking this with this and then I am taking this with this. So, when I am taking this this is the first argument, this is the second argument and the second argument is carrying out I mean carrying forward this scaling factor. So, that when it gets pulled out its conjugate has to be pulled out right because that is exactly what denies it from being bilinear. If it is over the complex field we have to accommodate the possibility that it is not bilinear, but say square linear. In the second argument it is not completely linear and it is this scaling factor. So, when this gets pulled out it does not come out by itself it comes out with its conjugate. So, that is what brings out this conjugate here right. Ok, thank you.