 In this video, I want to introduce the idea of the orthogonal group, which is an important group of real matrices, and in order to do that, I want to remind the viewer of some important properties from linear algebra. So if we recall from linear algebra, the so-called transposition operator, the transpose map, it's typically denoted by some type of capital T, sometimes lowercase t, or some symbol that kind of resembles a t in some regard. And so if you have a matrix A, A transpose is you're going to switch all of the rows to columns and columns to rows. So if the typical element of your matrix is Aij, so this element Aij is in the i-th row j-th column, the transpose will interchange the roles of rows and columns. So the number Aij will then move, so the number in the i-th row j-th column will move to the j-th row i-th column, so those things go things get mixed around. Again, you've probably seen something like this before. You can click the link on the screen right now to see some examples of this if you need a little bit of a refresher. Some properties of the transpose operator I want to make mention here, that if you take the sum of two matrices and you take its transpose, this is equal to A transpose plus B transpose. So if you come down here for a second and you take the transpose operator, if you take the group R to the m by n, excuse me, with respect to matrix addition, so this is an abelian group, the transpose will convert this to the matrix group, I should say the vector space, Rn by m with respect to addition. What this says here is that because A plus B transpose is equal to A transpose plus B transpose, this tells us that this is an additive homomorphism. So it's a homomorphism with respect to matrix addition. Now when you talk about a set of matrices with respect to addition, it's not really good form to call it a matrix group, even though it's a group, matrix groups typically refer to matrix multiplication, so it might be better to call it a vector space, which it is. So the transpose with respect to matrix addition is homomorphic. The transpose also preserves scalar multiplication if you take R times, so R is just some scalar, so we can pretend it's like a real or complex number or whatever the scalar of the field is. You get R times A here, transpose, you can take the scalar out, R times A transpose. Now I should mention here that we talk about the transpose, although analogies of the transpose exist for basically any field, we probably are talking about the real vector spaces, real matrices in this context. For complex matrices, we never talk about the transpose, we talk about something else called the Hermitian transpose, excuse me, the conjugate transpose, that's usually what's called in generality. And then for other vector spaces, again, you can talk about transposes, but it doesn't have the same geometric benefit that does for the real numbers. So we are talking about the real numbers in this video as we lead to the orthogonal group. So if you put properties one and two together, this tells you that this map is actually better than an additive homomorphism, it's actually called a linear transformation because it preserves addition and scalar multiplication. So that's a property of this transpose. Some other things we see here, if you take the double transpose, you get back to the original thing, so the transpose is transpose is the matrix. So this says that the transpose as an operator has order two, something we also say about the determinant. The determinant is not affected by the transposes, and you can kind of get this from the usual Laplace cofactor expansions that I could cofactor expand any row or column and it would give me the same determinant. Well, if you switch the rows and columns, since you can still expand over any row or column, that doesn't affect the determinant. And the last thing I want to mention here is that the transpose satisfies the so-called shoe sock principle, that if you take A times B transpose, this gives you B transpose times A transpose. So with respect to matrix multiplication, this is not a homomorphism, but this is what sometimes referred to as an anti-homomorphism. It's homomorphic except it switches the order of the product. And if the operation's non-commutative like matrix multiplication is, then that actually gives you something different than a homomorphism. So the shoe sock principle we've mentioned many times in this lecture series, that is just sort of a mnemonic device to remember what's called a anti-homomorphism. It puts it in the wrong order. The inversion map of a group is, in fact, an example of an anti-homomorphism. So with the transpose now out of the way, that is, we've reviewed it. Now we're ready to define what we call the orthogonal group. The orthogonal group is typically denoted as O of N, but sometimes it's denoted as O sub N of R. This is to emphasize that the scalars in play are real numbers. And like I said, the orthogonal group can be defined using the transpose operator, which has analogies to other rings of scalars. So we can talk about orthogonal groups for other scalars other than the real numbers. But in the classical sense, the orthogonal group is always referring to the orthogonal group of real numbers. So oftentimes just O of N is mentioned. We don't specify the real numbers because it's inferred. Because if we're talking about the orthogonal group, it's got to be real numbers. Now the orthogonal group gets its name because it'll be the subset of the general linear group of orthogonal matrices. What's an orthogonal matrix again? Well, depending on your linear algebra background, there's a couple different ways we could have defined orthogonal matrix. How we're going to define it for this video is the following. We're going to use transposition to do this. So we define a matrix to be orthogonal if the transpose of the matrix is equal to the inverse of the matrix. So if the transpose is the inverse, then we call that an orthogonal matrix. Well, if the transpose is the inverse, that means the matrix has an inverse. It's a non-singular matrix. So every matrix inside the orthogonal group is necessarily non-singular. This is going to be a subgroup of the general linear group. All right, so that's something that should go without saying. Now I'll leave it up as an exercise to the viewer here to prove that this set of matrices is in fact a subgroup, right? It's closed under multiplication. It contains the identity matrix, and it also contains inverses. I'll let you double check that. The properties from the previous slide are exactly the properties one needs to prove that property, to prove that theorem, that statement right there. But like I said, this might not be the way that orthogonal matrices were defined in your linear algebra class. So we could define them as the transpose is the inverse. Another way of defining it is the following. You could say that q transpose, q transpose is equal to q transpose. So excuse me, q transpose is equal to q transpose q, which is equal to the identity, right? Now I do mention both of these because matrix multiplication is non-commutative. So just because one order goes one way doesn't mean the other order will be the same. So sometimes people take this as the definition. So if you multiply a matrix by its transpose and either direction get the identity. Now it's very easy to go from here back to here, right? Because we've said before, if you walk like an inverse and you quack like an inverse, you have to be an inverse. The q transpose is acting like the inverse right here, right? Because if I take q times q inverse, that should equal the identity matrix. So the transpose is acting like an inverse so you get this direction. And of course, if your transpose is your inverse, then since this statement holds, then this statement holds as well. So you can very quickly go back and forth between these two statements here. That the transpose being the inverse is equivalent to this identity right here. Now another statement, and this is often how I think orthogonal matrices are defined in a linear algebra setting in the following way. And this will suggest that the transpose is the same as the inverse. And this will suggest the name behind the orthogonal matrices. A square matrix u, square, of course, meaning it's n by n, the same number of rows as column, square matrix is orthogonal. So we're defining it here by orthogonal, we mean its transpose as its inverse. A matrix is orthogonal if and only if its column vectors form an orthonormal set. What does orthonormal mean again? So if you think of the column vectors here as vectors, of course, because we're calling them column vectors, orthonormal means two things. It first means it's an orthogonal set. So if we take the dot product of these different vectors, their dot product will equal zero. That's a zero scalar, excuse me. And then normal here means they're each unit vector. So the length of the vector is one. So each of the column vectors has length one. And if you take the dot product of any two vectors, you always get back to zero. And so that's why we call these things orthogonal matrices because the set of vectors is orthogonal, is orthonormal. Now some people are like, well, why don't we call them orthonormal vectors or matrices? I don't know. Their color on orthonormal matrix has an orthonormal set of column vectors. I should also mention that this will be equivalent to showing that the row vectors form an orthonormal set. I won't prove that one. I'll prove it for column vectors, but it's very easy to go back and forth between them because, again, the transpose and the inverse are acting the same way. So what's the argument there? So this is an infinite statement, so we have two directions to go. So first, suppose that U is an orthogonal matrix. That means if it's an orthogonal matrix, its transpose is equal to its inverse. OK, now consider the column vectors of U. We'll call the first column U1, the second one U2, all the way up to Un. Now, because the transpose, I guess I should be calling it U not Q, but whatever, because the transpose is its inverse, we have that U transpose U is going to equal the identity, right? We get this thing right here. But if we look at it term by term, how do you compute this thing? Well, the general elements of U will look like Uij. U is the number in the i-th row, j-th column. So when you multiply these things together, U transpose U, well, this just actually makes a matrix of dot products. When you take U transpose U, this just gives you a matrix where an arbitrary element in this matrix is going to look like you take the i-th row, transpose the i-th, or the j-th column. That's what it is. And the critical thing to see here is that Ui transpose Uj. If you take a vector, transpose another vector, this is just what the dot product does in the first place. It's just the dot product. So when you look at this matrix, you just get a bunch of numbers, which are going to be dot products. But this thing is going to equal the identity matrix. So if you're the identity matrix, you get ones along the diagonal and zeros everywhere else. So if you're in the diagonal, it means the row and column index are the same. So Ui.Uj here, you're going to get a one if you're on the diagonal. That happens exactly when you have i equal to j. And if you're off the main diagonal, you're going to get a zero, which happens exactly when i doesn't equal j. This number right here, this expression, zero when they're different, one when they're the same. This is often abbreviated in mathematical literature as delta sub i sub j. And this is commonly referred to as Kronecker's delta. So whatever you see in a mathematical text, Kronecker's delta, that means it's one when i and j are the same, and it's zero when i and j are different. So because U transpose U is equal to the identity, that means that the entries of this matrix are going to be Kronecker's delta. So you can get ones along the diagonals and zeros everywhere else. But that's exactly what it means to be orthonormal, right? If i and j are different, that means these are different vectors, different column vectors, their dot product is zero, so they're orthogonal. But if we get i Ui.Ui, that is i is the same, you're going to get that Ui.Ui equals one. Well, notice that the vector dot itself, this is just the length of the vector squared. So taking the square root, you get the length of the vectors one. So it is, in fact, an orthonormal set. And this whole argument was an if and only of statement. Every step was reversible, so that we actually proved both statements at the same time. So why do we care about the orthogonal group whatsoever? Well, geometrically speaking, the orthogonal group, ON, consists of all rotations and reflections of Rn. So essentially, the orthogonal group consists of the so-called isometries. Can I spell this word? Isometries, there we go. Try that again, isometries. They're isometries of the vector space Rn. That is, we're going to talk about those transformations of the vector space, of the geometry of Euclidean space Rn. So that distance angles don't change, right? Isometry, this word here means same measure. So distances between points don't move when you act by an orthogonal matrix. Angles don't move. Dot products don't change when you do, when you multiply by an orthogonal matrix. And it's this group of symmetries. We'll talk more about this in the next lecture, why we care so much about the orthogonal group. Now, let me do give you an example of such a thing, because these things are not necessarily easy to produce on their own. Let's take you to be this three by three matrix, which we're going to have three square root of 11, over square root of 11, excuse me, negative one over square root of six, negative one over the square root of 66. And then you get a bunch of other things similar to it. One over the square root of 11, one over the square root of 11, two over the square root of six, one over the square root of six, negative four over the square root of 66, seven over the square root of 66. You may wonder why all the square roots here. Well, remember, the column vectors need to form an orthonormal set. So in particular, they have to be unit vectors. So when you look at the columns right here, you'll notice here that when you take its length, you're going to get the square root of nine plus one, plus one, which is 11, over 11, which are going to get one in the end. So that square root of 11 has to be in the bottom to compensate for the fact that the sum of the squares here, nine plus one plus one is 11. And this one, you get the sum of squares, one plus four plus one is six, and then one plus 16 plus 49 is 66, all right? So, and you can check also that term by term, these are going to be orthogonal with each other, right? When you take the dot product, it's sort of ignoring the bottoms because we can factor out scalars. Essentially, you're going to get negative three plus two plus one, which is zero. You're going to get negative three minus four plus seven, which is zero. You're going to get one minus eight plus seven, which is zero right there. So we can check very quickly that the column vectors form this orthonormal set, but it also should be true that when we multiply these things out, we should get the identity, right? You transpose you should be the identity. So if I just take the matrix and transpose it, you get the following, and you look at all the possible dot products right here, I'll let you pause the screen and double check all of these. But if you take, for example, the first row times the first column, you're going to get nine over 11 plus one over 11 plus one over 11. So you get nine plus one plus one, which is 11 over 11, that's going to give you a one. Okay? If you take the first column, a first row times the second column, you're going to get negative three over the square root of 66 plus two over the square root of 66 plus one over the square root of 66. So that's negative three plus two plus one over the square root of 66. That's what I was talking about earlier about the columns being orthogonal with each other. And if you go through all the possibilities, you're going to get ones along the diagonal, check that, and you can get zeros everywhere else, above and below the diagonal. So this is an example of a orthogonal matrix. All right? And so I mentioned earlier, write that orthogonal matrices correspond to these isometries. Sometimes the word, instead of isometry, they use the word rigid motion. That is, we're talking about how we can transform the plane, but it's rigid that as things aren't flexible, so the little distances between vectors doesn't change over time. Since multiplication, well, here's a theorem here. So theorem 1214 here tells us that if you have an orthogonal matrix, then if you take the dot product of two vectors, but you also look at the images of those vectors under the linear transformation multiplication by U, it doesn't change the dot product, right? So UX dot UY is equal to UX, or UX dot Y, excuse me. So the dot product between UX and UI is the same as the dot product between X and Y. And as a consequence, distance in the Euclidean space is defined by the dot product. If the dot product is preserved, sometimes it's called the inner product. If the dot product is preserved, that means the distances between vectors will be preserved. Angles in Euclidean space is defined using the dot product. If the dot product is preserved, that means angles is gonna be preserved. So multiplying by an orthogonal matrix is an example of this isometry, this rigid motion. Now, so again, since multiplication by orthogonal matrices preserves dot products, like I said, it's gonna preserve lengths, distances, angles, orthogonality, that is when vectors are perpendicular to each other. So as a consequence of this theorem, U here, we're gonna see that the length of UX is the same thing as the length of X. The length of the vector does not change when you hit it with an orthogonal matrix. In particular, orthogonal matrices are exactly those linear transformations of RN that preserve lengths, the so-called rigid motions. Now, I should emphasize that not, orthogonal matrices are not the only isometries of the plane, but they're the only isometries which are linear transformations in a vector space sense. So what's the argument here? It actually is, it goes really straightforward here. So if you take the left-hand side of this identity, UX dot UY, we'll just take that. Well, the dot product is the same thing as taking the transpose with respect to matrix multiplication. You get UX transpose times UY, for which the transpose, like we saw earlier, it's an anti-homomorphism, the Schusack principle. You get X transpose, U transpose, and then UY. But redoing parentheses, you get U transpose U. Aha, that's the identity since it's an orthogonal matrix. So it just disappears. You get U transpose X, which is the same thing as the dot product. So orthogonal matrices are like perfectly designed matrices. You see the condition right here. These are the perfectly designed matrices that'll preserve inner products and then everything that we derive from an inner product. And so closing this video, we've talked a lot about the orthogonal group, which is an important, it's one of the classical matrix groups. Related to the orthogonal group is the so-called special orthogonal group. Like I mentioned in the previous video, special means the determinant here is one. So the special orthogonal group denoted S O of N, or sometimes called S O N of R to emphasize the scalars are real. But if we don't specify other scalars, it should be inferred that the scalars are real. That's in the classical sense, the special orthogonal group only is over real coefficients. So the special orthogonal group is the subset of the orthogonal matrices whose determinant is equal to one. In particular, S O of N is the intersection of O N with SLN of R. So it's the intersection of orthogonal group with the real special linear group. As such, the special orthogonal group is a subgroup of the orthogonal group. It'll also be a normal subgroup of the orthogonal group because the special linear group is in fact normal inside of the general linear group. Now I cannot say that the special orthogonal group is normal inside of the general linear group. No, no, no, but it will be normal inside of the orthogonal group. And also, right, you can prove that the determinant of an orthogonal matrix, so Q is orthogonal, you can prove that orthogonal matrix is determinant is always plus or minus one. If it's a plus one, that indicates that the matrix is essentially a rotational matrix. And if it's negative one, that essentially means that it's a reflection matrix. And so when you take the special orthogonal group, you're just taking those with determinant one, you throw out the negative ones. There's just as many determinants being one as there are negative one inside the orthogonal group. And so you can actually make the argument that the index of the special orthogonal group is two because half of the orthogonal matrices have determinant one, that is the rotation matrices, and the other half have determinant of negative one, the reflection matrices. And so therefore the special orthogonal group is essentially the group of rotations of the vector space Rn. And since it's a subgroup of normal two, that gives more reason why it's normal because index two subgroups are always normal.