 So, let us start looking at the notion of a metric space. So, X is a non-empty set and D is a function on X cross X to non-negative real numbers. So, X, Y goes to D, X, Y. But the following properties, one, the distance, we call it as D as the distance is always bigger than or equal to 0, equal to 0 if and only if X is equal to Y. Distance between X and Y is same as the distance between Y and X. In the third property, the distance between X, Y is less than the distance between X, Z plus distance between Z, Y. For all X, Y and Z belonging to, so this is symmetric and this is triangle inequality. So, essentially this is modeled on, so let us look at examples. The motivating example was that X is the real line and D X Y is mod of X minus Y. So, the idea was that this notion of distance gives you the idea of saying something is close to two points or close to each other and gave us many concepts on real line. So, let us look at some more. For example, let X be any non-empty set and define D on X cross X. So, let us define D X Y is equal to, it is to be 0 anyway when X is equal to Y and let us put it equal to 1 if X is not equal to Y. So, distance between any two points is defined to be 0 if of course X is equal to Y and two distinct points the distance is defined to be equal to 1. So, easy to check that obviously it is symmetric non-negative 0 if and only if X is equal to Y and D X Y is less than or equal to the triangle inequality is again obvious because the right hand side could be either 0 or bigger than 1. So, obviously it is a metric. So, any set has got this kind of a metric called discrete metric. So, this is called a discrete metric on any set. Not very interesting, but in some way let us look at some more. We had let us take X is equal to R 2 and let us take a point X Y belonging to R 2. In the real line we had the notion of absolute value of a number and that was the distance of the point from 0. So, in R 2 let us introduce that. So, the distance of the point X Y from 0. So, let us so that is X square plus Y square called the norm of the vector with components X Y. We would like to check that this has properties similar to that of absolute value function in R. So, meaning what? 1, the norm of we will be using underline X for the vector with components X and Y is bigger than or equal to 0 and it is equal to 0 if and only if the vector X is the 0 vector. Let us look at the second one that says that if I multiply a vector by alpha then this is mod alpha times absolute value. Absolute value of A B is equal to absolute value A into absolute value B and the third property namely X plus Y is less than plus the triangle inequality property for all X Y belonging to R 2. One way of looking at that would be that actually if you look at a triangle in R 2 with sides X and Y then the hypotenuse the length is less than or equal to or triangle inequality any two sides the sum of any two sides is bigger than. Let us try to prove it analytically. I would like to prove this that for any two vectors this is so. Let us try to introduce something which is interesting in its own respect. So, what is called the dot product in R 2? Given X 1, Y 1 and X 2, Y 2 belonging to R 2 one defines the dot product. So, of this X 1, Y 1 product X 2, Y 2 to be equal to sigma X 1, X 2 plus Y 1. In fact, it will be a good idea to prove it for all not only R 2. Let us prove it for R n. But anyway, we will do it a bit later probably. So, this is called the dot product in R 2. What is the, why we want to define that? So, dot product, if I look at dot product of X dot X itself that is X 1, Y 1 dot product with X 1, Y 1 that is sigma X i square i equal to 1 to 2 and that is norm of X square. So, that is the most interesting thing that happens. It relates the dot product with the notion of magnitude or absolute value of a vector. I have written X i. So, let me, because I have used a different notation here. So, let me write that because components are, so it is X 1 square plus Y 1 square. So, that is equal to norm X square, where X has components X 1 and Y 1. So, this dot product is related with the distance in this following way. It also has interesting properties. So, let me write some properties of dot product. Let me write the dot product as a function. So, here the function R 2 cross R 2 to R. So, how is it defined? It is a dot product. I am just using a different notation, so that things become very clear. So, for a vector X and a vector Y, it is the dot product of X with Y for every X, Y belonging to R 2. For any two vectors X and Y, the dot product is this notation I am using. So, with this notation, it is called the inner product. Dot product is also called inner product. It is the product of a vector with itself. So, why we are doing this? The reason is the following. So, you can treat it as a function of two variables X and Y. X belongs to R 2, Y belonging to R 2. So, the properties R 1 is always bigger than R equal to 0, equal to 0 if and only if X is equal to 0. That is obvious that we have already seen that. Or anyway, because it is sigma X i square, so that is 0 means each component must be equal to 0. Second, let us take two vectors X and Y. So, let us take vectors X, Y and Z belonging to R 2 and let us look at the vectors X plus Y, Z. So, what is that equal to? So, that will be components of this will be sigma X i plus Y i and comma that will be Z i. So, product distributes over addition. So, this is same as X comma Z plus Y comma Z. So, saying in the first component, if I add and take the dot product is same as the sum of the dot products and third property. In fact, I can write here alpha X Z is same as alpha times X Z. The multiplying scalar comes out in the dot product because sigma X i, so alpha will come out from the sum. Fourth property, X Y is same as Y X. It is symmetric. Dot product of X with Y is same as dot product of Y with X and fourth and fifth. Similarly, this combined with the second combined with this one will give you that X comma Y plus Z is same as X plus Y, sorry X comma Y plus X comma Z. So, essentially what we are saying is positive definite. The first property, it is linear in the first variable as well as in the second variable and it is symmetric. Linearity because of alpha also comes out. So, it is a linear thing. We will come to the generalizations a bit later. So, let us say it has obviously these properties. So, proofs are obvious or easy to verify just ordinary algebra. And of course, we had that fact namely X dot X is equal to norm of X square. What we want to do is prove a property of this which is very important called Cauchy which was inequality. So, what does it say? It says if I take vectors X and Y belonging to R2 and I look at the dot product of XY, that is a real number. Take its absolute value that is always less than or equal to norm of X square into norm of X. There are many ways of proving it. Let us look at a way which we analyze it to everything. So, let us prove of this. Look at X plus Y, X plus Y, the vector X plus Y, X plus Y and look at the norm square of that. That is always bigger than or equal to 0. Let us write this in terms of the dot product. So, this is X plus Y, X plus Y, norm square is equal to the dot product of the vector with itself. So, let us expand it using linearity. So, that is X comma X plus Y comma X plus X comma Y plus Y comma Y. So, that is equal to norm of X square 2 times. That will not lead it to bigger than or equal to 0. Let us just keep it as 1. This will not give us the required thing. Let me just look at it. So, one way of this would not lead to the proof, but just keep it as it is. There is one obvious way of looking at it, but that requires the knowledge of saying that the dot product A and B is norm A, norm B into cos theta. But how does that formula come? So, I am just debating about that. This is another way. Let us just look at this proof. One can use calculus also. We will do the calculus proof later. So, we want to prove this inequality that the dot product less than equal to norm U norm V. Obviously, if either U or V is 0, then both sides are 0. So, no problem. So, let us assume both of them are not 0 and divide U by norm of U. So, define a new vector called U dash and V dash. U dash is U divided by norm U because it is not 0. I can divide by it. So, what will be the norm of U dash? That is one norm. Norm of V dash also is equal to 1. So, let us look at this term U dash minus the dot product V dash with itself. So, that is a vector we should look at. So, are you able to follow from here or shall I write it up? So, let me write probably. So, we had what we had written x and y. So, let us keep it x and y. x and y are vectors. So, we want to prove norm of is less than or equal to norm x. So, let us. So, if x is equal to 0 or y is equal to 0, then left hand side is equal to 0 equal to right hand side. So, no problem. So, let x not equal to 0, y not equal to 0. So, this is a key thing. So, let us write x minus x, y, y and product with itself. So, x minus x, y. So, what is this quantity? On one hand, this is equal to norm of x minus x, y times y square, norm of that square. Same vector with dot product with itself, which is bigger than or equal to 0. On the other hand, let us expand this. What is it equal to? So, this is x comma x, first term minus x comma x, y, y. So, it will be two times. x, x, first term and then x with this one, x with itself and then x comma minus, so that we have taken. So, this one. So, that minus sign. So, minus x, y, y, x plus algebra. So, last term will be minus plus. So, fourth term will be minus plus. So, x, y, y comma x, y. We will have four terms. First term, x with itself, that is, x comma x dot product. Second term will be taken x with the second term here. So, that will be x comma this with a negative sign, negative alpha comes out and x will be minus of that thing with respect to, is that okay? So, minus x, y, y that is x and this thing. So, that is equal to norm of x square minus two times x. So, x, y. So, let me write x, y and y. So, this term and this term together. So, minus two plus. So, this is norm of x, y. So, this is a scalar comes out. So, norm of x square minus two times x, y. And this is another scalar. So, that is into x, y. One scalar comes out and the same thing plus x, y square norm of y square. The scalar comes out x, y square and norm of y square. This is also same quantity. So, let us write x square minus twice of that same quantity. So, let us write if norm x is equal to 1 is equal to norm y. Then what do we get? In case, so that is what we are looking at 1 minus. So, this will be x, y. So, x, y, x, y square. So, you can write square plus. Yeah, that is all. No? Let us go back and see what we are writing is. So, I am looking at this vector x minus dot product of x with y times y. This is a scalar. This is a scalar times y. So, I am looking at a x minus a scalar times y norm of that. So, that is bigger than or equal to 0. So, that will be equal to dot product of this vector with itself. Now, open this. This is this minus this plus dot product of this minus. This is a linearity. That will give you four terms. So, what will be the first term? Dot product of x with itself. So, that is the first term here. Second term will be dot product of x with negative of this thing. So, that negative sign I have taken it out. So, dot product of x with that vector. Other will be the switch to over negative sign. So, two cross terms and then the last one will be the dot product of this vector with itself. So, this vector with itself, four terms. Now, let us simplify. The first one is norm of x square. This term is same as this because of symmetry and this scalar comes out. So, it is two times norm x dot product of x with y and y. So, that is these two terms are combined together to give you this two times that. And the last one is the dot norm of dot product of x, y, y. Let us simplify that a bit further. So, this scalar comes out as square norm of y square. So, this is this term. Now, let us assume norm x is equal to 1 and norm y is equal to 1. So, what do you get? So, this quantity is 1. This is 2 times and this is 1 here. So, it is only minus 1 times this. So, what does and this is bigger than or equal to 0. So, that is what we said. This is bigger than or equal to 0. So, implies that norm of x, y square is less than or equal to 1. And what is 1? You can write as norm of x norm of y because norm x we assume to be 1. So, what we are saying is this proof works very well when norm x is equal to 1, norm y is equal to 1. If not, then what you will do? We have already seen if not, you can divide x by norm x, y by norm y. So, if x, y not equal to 0, the same thing will give you norm x over norm x, y over norm y. Now, both are of norm 1. So, if I take dot product, this will be less than norm x over is less than or equal to 1. That means, dot product of x, y is less than or equal to norm of x into norm of x. That is all. So, basically the idea is take that linear combination namely and expand it to one side. It is bigger than or equal to 0, other side by using the property of the dot product. So, this proves Cauchy-Schwarz inequality. So, what I wanted to do was using that. So, this is not required in Cauchy-Schwarz inequality. So, let me remove from here. So, let us remove it from here. Cauchy-Schwarz we have proved. So, let us prove now what is the property that norm of x plus y is less than or equal to norm x plus norm y for every x, y belonging to y dot. So, proof that is what the other thing we should be looking at. So, we are trying to prove that the norm of x plus y is less than or equal to norm x plus norm y. So, let us compute it. Square is the dot product with itself and that is equal to this. We are computing the norm of x plus y. So, write it as a dot product, expand. So, you get this. Now, the idea is this is less than or equal to norm of x square plus norm of y square mod of 2 times inner product. So, that is less than or equal to 2 times norm x norm y by Cauchy-Schwarz inequality. So, that is a crucial thing here that Cauchy-Schwarz inequality gives you and this is nothing but norm of x plus norm of y whole square. So, that implies that norm of x plus norm of y is less than norm x plus. So, finally, what we are saying is that norm or the magnitude in R2 also has the property similar to that of absolute value function.