 Now, there is the couple of points I want to make and then I will state one proposition. So, so consider, did somebody have a question, was there a question? So I'm defining an orthonormal square matrix, yeah go ahead, okay, so let's consider an orthonormal square matrix. This is a matrix such that it has orthogonal and unit norm columns, that's the definition of an orthonormal matrix and it has the property that another way to define it is a matrix Q satisfies Q transpose Q equals Q Q transpose equals the identity matrix. So by the way, in general, it's always true that for square matrices, if AB equals the identity matrix, then BA is also equal to the identity matrix, okay, this is easy to see. If BA is not equal to the identity matrix, then BA minus I is non-zero, which implies now if I pre-multiply by A, then ABA minus A is not equal to zero, but AB is equal to the identity matrix, so which implies A minus A is not equal to zero, which is a contradiction, okay. So for square matrices, it's always true that if AB equals I, then BA equals I, therefore, if I say Q transpose Q equals I, then Q Q transpose must also be equal to the identity matrix. So, there is another definition for an orthonormal matrix, which is the definition I'm going to use for stating this proposition, which is that a matrix A has to be equal, yeah? Sir, in above proof, BA, we are taking not is equal to I, yeah, so why I am writing ABA is equal to A? AB is equal to I, right? If AB equals I, that's what we are trying to show here, that if AB is I, then BA is not is must also be equal to I, so AB is equal to the identity matrix, identity times A is always equal to A, whatever A is, yes sir, thank you, yeah, I have a doubt, so on the top you have written Q transpose Q equals Q Q transpose equal to identity matrix, but the dimension of Q transpose Q will not be equal to the dimension of Q Q transpose, right? Square. Oh, okay, it's for n cross n, okay fine, okay, if it is, suppose if it's in rectangular matrix, then how will this? Then it's no longer true that if Q transpose Q is I, then Q Q transpose must be equal to I, okay, but either of them will hold true, right, like at least, I mean, if one doesn't hold true, then other will definitely hold true, right, in case it's an orthogonal matrix. So if the columns are orthonormal, then Q transpose Q will contain as its entries inner products between columns, okay, and so that will be equal to the identity matrix, but Q Q transpose need not be the identity matrix, okay, similarly, if the rows are orthonormal, then Q Q transpose will be the identity matrix, but Q transpose Q need not be the identity matrix, okay, fine, okay, sir, yeah. So this is said to be an orthonormal matrix if it preserves the inner product. So what I mean by that is, if I take the inner product between ax and ay, that is equal to the inner product between x and y, for every x and y belonging to r to the n, okay, so that's also another definition of an orthonormal matrix, okay. So here is the proposition, so the following statements are equivalent, A is, so here A is a matrix in r to the n cross n, B is, A preserves the length, by that I mean, A v for every v in r to the n, so recall, okay, that's the definition of the, in fact, for the moment I won't write the subscript 2, because this result is actually more general, it doesn't require the inner product to be defined the way we have been defining it until now, we've seen only one example of an inner product, there are other possible definitions for the inner product, but right now we are working with the dot product as the notion of the inner product, but any other valid notion of the inner product can be used to define a norm defined like this, okay, and that is called a norm that is induced by an inner product, and this is true for any definition, any valid definition of an inner product, okay. Can you explain the previous definition, A x, A y equal to x, y, what is A x, this is a column? Yeah, x is a vector, so A x is a vector, so it's looking at the inner product between A x and A y. Okay, so what is x and y, it is numbers or vectors? It's right here, x, y belongs to r to the n, so if it's a column 1 and column 2, it will be 1 and 2. What do you mean, x is a vector in r to the n, y is a vector in r to the n, x and y are vectors, x is a vector and y is a vector, so if you compute the inner product between x and y, whatever number you get, this is the same number as what you would get if you computed the inner product between A x and A y, and that's true for any pair of vectors I choose. Okay, I think this is important that everybody understands the definition, otherwise this entire proposition makes no sense. So do you have other questions about the definition of an orthonormal matrix? Please ask, the definition is that it should preserve the inner product, that is you take two vectors in r to the n, if you find the inner product, that is x, y, and then you take A x and A y, those are also two vectors in r to the n, and you find there inner product, and those two numbers must be equal for all pairs of vectors in r to the n, not for just one particular pair. Sir, actually in the inner product line, a matrix into a vector is actually a transformation, right? Yes. So what is the meaning, I mean, inner product is actually the length, I mean, the component of x along y or something in that sense, right? I mean, inner product between x and y means the component of x along y. Correct. So it's preserving the relative orientations of x and y, that is what it means. Okay. In whatever frame we transform this vector. Yes. Yes. So if, for example, x and y were orthogonal, they were perpendicular to each other, but collinear with each other, then even after transformation by A, they will remain collinear with each other. That's what it means. Okay. Okay. And mathematically, this is the precise meaning of a matrix being an orthonormal matrix. It preserves all inner products, not just when x and y are orthogonal, or not just when they are collinear. Thank you, sir. Hello, sir. Just a question, like it is, when a multiplied a vector with A, it's not changing the length. So is it, it will rotate the vector? Yeah. That's a good way to think about it. It will rotate the vector, but it will not change its length. So in some say, so another way to put it is that every rotation matrix is actually an orthonormal matrix. And every orthonormal matrix can be thought of as a rotation matrix. Okay, sir. Sir, is this sometimes called orthogonal matrix as well? Yeah. So that's actually some, that's actually a good point. Depending on the textbook, some textbooks refer to this, this kind of matrix as an orthogonal matrix. Some textbooks refer to it as an orthonormal matrix. I prefer the terminology of orthonormal matrix because it also clearly tells you that every column is unit norm. An orthogonal matrix, if you really wanted a more general definition, it is a matrix where the columns need not be unit norm, but they are still orthogonal to each other. Okay, that could be a more general definition of an orthogonal matrix. But yeah, just to be clear, I want to call it an orthonormal matrix in this course and it's a matrix whose columns are unit norm and are orthogonal to each other. These are two equivalent definitions, but for the purposes of showing this proposition, I will go with this definition, meaning that it's an orthonormal matrix if it preserves in a product. So the third property is that A transpose is equal to A inverse, A is an invertible matrix, which means that A transpose A equals A transpose which equals the identity matrix. And the fourth property is that the rows of A form an orthonormal basis for r to the n. And the last one is that the columns also form an orthonormal basis for r to the n. Okay, so how do you prove such a proposition? So it says that these statements are equivalent, meaning that if I tell you that A is orthonormal, it's the same as me telling you that A preserves length. So if you have five statements like this, the way to show that all of these are equivalent is to say, for example, you take the first two statements, you should show that A implies B and B implies A. That means A and B are equivalent. Then you take the third statement C, then what you have to show is that either A or B, one of those two statements implies C and C implies either A or B, one of those two statements. Then it means that A, B and C are equivalent. Then similarly you take D and you show that D implies one of these three statements, A, B or C. And you should show also that A, B or C, one of these three statements implies D and so on. If you show all of that, then you've shown that these statements are all equivalent. Of course, there are many ways to do it. For example, you show A implies B and B implies A, then you show B implies C and C implies B, then you show C implies D and D implies C and D implies E and E implies D. So that also means that these statements are all equivalent. So let's see how to show this. So first we'll tackle A implies B. So we want to show that if A is an orthonormal matrix, then it preserves the length, meaning that for any vector, if I look at the length of V, it's equal to the length of the transformed version of V, which is A V. So and vice versa. If A preserves the length of every vector in R to the N, then A must be an orthonormal matrix. So to show first A implies B, an orthonormal matrix implies that it preserves the length, then what I need to do is I just need to consider the inner product between AX and AX. And this, the inner product between AX and AX, because A is an orthonormal matrix, by definition it preserves inner product. So basically this is equal to the inner product between X and X. So that means that if I... So this is nothing but AX squared and this is nothing but X squared, which means that AX equals X. So this means that... So this is what we wanted to show that A preserves length. So A implies B. Similarly, B implies A. What I need to do is I need to show that if A preserves the length, then A is an orthonormal matrix. That means that all inner products are unchanged by multiplication by A. So what I'll do is I'll consider... Sir, could you explain how inner product of AX with itself is equals to inner product of X with itself? That's from the definition. So what I want to show here is that if A is orthonormal, then it preserves length. So if A is orthonormal by definition, because it's an orthonormal matrix, it preserves the inner product. The inner product between any two vectors. In fact, Y need not be equal to X. So if A is an orthonormal matrix, then AX, AY is equal to the inner product between X and Y. This is true for all pairs of vectors X and Y. And so all I'm doing is to take a special case of this, where Y equals X. So the inner product between AX and AX is equal to the inner product between X and X. So you can see that although the statements look quite different from each other, saying that A is an orthonormal matrix, meaning that it preserves inner products and saying that A preserves length, they seem different from each other in terms of statements, but in fact, it follows trivially from the definition. Is it clear? Yes, sir. Thank you, sir. So if I take the inner product between AX plus Y and AX plus Y, this because A preserves length. So now I want to show that B implies A. So A preserves length. So this is nothing but the length of A times X plus Y square. And so this must be equal to the length of X plus Y square. But if I expand this out, by using the linearity of inner products, then this thing can be written like this. It's the inner product between AX and AX plus two times the inner product between AX and AY plus the inner product between AY and AY. And the right-hand side, if I expand this out, I get the inner product of X with itself plus two times the inner product between X and Y plus the inner product between Y and Y. Now once again I use the property that A preserves length. So this quantity inner product between AX and AX, this is equal to the inner product between X and X. And this term is equal to the inner product between Y and Y. So this XX will cancel with this XX. This YY cancels with this YY. And the two and two can cancel. So you are left with AX, AY equals XY. So it preserves inner products and so then it is orthonormal. Then we consider C, the statement C. So the moment you are considering inner product of AX, AX is X, X. So that moment only you are considering it to be orthogonal, right? That is what they are showing here actually. Yeah, but in the second last step you already considered AX, AX is X, X. So what we are trying to show here, I think it's important to pay attention a little bit to what we want to show. So what we want to show is that B implies A. What is that? If A preserves the length, that is if norm of AV equals norm of V for every V in R to the N, then A preserves the inner product. That is the inner product between AX and AY is equal to the inner product between X and Y for all XY in R to the N. That is what we want to show. And that's what we are doing here. It's not a difficult proof. It's very simple. But it's good to keep in mind exactly what it is that we are showing. So in writing the first step, I'm using the fact that B is true. I'm saying if B is true, then I want to show that A will preserve an inner product between any pair of vectors. So this is true because A preserves the length. This is also true because A preserves the length, AX, AX equals X, X. That's also true because A preserves the length. And similarly, AY, AY equals YY is because A preserves the length. And a consequence of this is that I'm getting AX, AY inner product is equal to the inner product between X and Y, which means that A is orthonormal. Sir. Yeah. Sir, so you gave the definition that an orthonormal, it's an A is an orthonormal matrix if it preserves the inner product. But that statement does not speak about the converse. That is, if it preserves the inner product, then it should be orthonormal. That part is not implied by the definition, right? Yeah. So this is also a very good point. In mathematics, a definition is always an if and only if statement. Okay. So we write it like this. A matrix is said to be an orthonormal matrix if it preserves the inner product. But when we are saying that it is said to be or it is defined as an orthonormal matrix, we mean that this is an if and only if condition. Okay. Okay. Okay. By definition, an orthonormal matrix is one which preserves the inner product. All definitions in mathematics are like this. We define something to be in a particular way. It means that A is equivalent to B. If I define A to be equal to B or A to B, or if I say a matrix A is defined to have a property X, if it satisfies Y, it means that X and Y are equivalent to each other. Okay. Now, let's do the next step. Next. If I consider the inner product between AX and Y, I will make a statement. This is equal to the inner product between X and A transpose Y. This can be seen by just writing out the expansion of what this will be. So, basically, this AXY is nothing but summation i equals 1 to n, sigma j equals 1 to n, Aij, Xi and Yj. And if you expand this out in terms of the entries of A, you will find that this is also exactly the summation. So, I should say. Okay. So, now what I can do is instead of Y, I will replace Y with AY. Then what happens is that I get, so I should write that out so we don't have confusion later. Then what I get is the inner product between AX and AY. Now, because, so suppose A is an orthonormal matrix, then it preserves inner products. So, that means that this is equal to XY. And this is equal to the inner product between X and A transpose A times Y. And this is true for every X, Y. So, if A is orthonormal, then we have that this is equal to this for every XY in R to the N. So, then that means that what I can do is I can take example vectors for X and Y, X equal to EI, the ith column of the identity matrix and Y equal to EJ, where EI equals the ith column of I N cross N. Then what I have is, if I consider the left hand side, this EI EJ. So, EI EJ is equal to, I can write it as delta IJ, which is this equal to 1 if I equals J and 0 otherwise. Then the right hand side is the inner product between EI and A transpose A EJ, which basically if you expand this out, you see that you will see that this exactly pulls out the IJ ith entry of A transpose A. So, that means that the IJ ith entry of A transpose A is delta IJ, which means that A transpose A is the identity matrix. So, then it means that A is A transpose is the inverse of A, A implies C. So, we have that. Now, conversely, if A transpose A is equal to the identity matrix, then if I consider the inner product between X and A transpose AY, this is equal to, this is the identity matrix. So, this is always equal to X and Y, the inner product, because A transpose A does not change Y at all. But then the left hand side, this, I can move this A transpose over here and I can, so more generally, so let me just write it like this, X A transpose times AY. I will consider this to be some vector, A transpose times something is equal to the inner product between AX and AY. And so, thus, AX AY equals XY for all XY in R to the N and does C implies A. And the last step is just that, if A transpose equals I, it means that the rows of A are orthonormal. So, we have seen that, so these two are equivalent statements, because this is just computing the inner product between rows of A. And if this is equal to I, it means if I take any pair of, any distinct pair of rows, they are orthogonal to each other. And if I take any row, it has unit inner product with itself. So, the rows of A are orthonormal, they are both equivalent statements. And so, C implies D and D implies C. And similarly, A transpose A equals I is the same as saying that the columns of A are orthonormal. Yes. Sir, in order to show that inner product of AX and Y is same as inner product of X and A transpose Y, can we use this formulation that inner product of X comma Y is same as A transpose Y? Yeah, it's the same point. Yeah, it's exactly the same point. Yeah, it's exactly the same point. This is it. This is the point. Because, you know, this summation sometimes does not click how to write in form of this or two summations. But directly doing the transpose, we can easily separate it out. Yeah. So, but that will be valid. Yeah. So, you can do it that way also. Okay. Okay. Thank you. So, the thing is that, you know, it follows from the definition of the inner product that we said. So, you have to take the transpose of the first vector and then multiply it with the second vector. So, if you think of it that way, it follows immediately. That's also a valid way to write it out. Okay. Any other questions? Sir, is that true for all A? No, right. Of course, this is always true. AX comma Y inner product is always equal to X comma A transpose Y. What the previous student just said is that if I think of X, Y to be equal to Y transpose X, if I think of it this way, which is how I defined the usual inner product earlier, then if I were to take AX and Y, this is equal to Y transpose AX, which is also equal to A transpose Y whole transpose times X. Yeah. And which I can write as X transpose times A transpose Y. No, I don't need to do that. To prove the equivalence of five statements, we are actually going through ten proofs, but actually six proofs would suffice, right? Because A implies B, B implies C, C implies D, D implies E, then E implies A, then all of them will imply each other. Yeah. So, that's also a valid way to show it. There are many ways to write out this proof. I've shown you one way here.