 In this video, I wanna give you another example or two of an equivalence relationship. How do we show one? How do we show something's equivalent relationship to be specific? So let's, for our seven consideration, let's consider N by N matrices. So let's say we have two matrices A comma B, which are N by N. We say that the two matrices are similar. So A is similar to B, and we'll denote this as A twiddle B. We say that A is similar to B if there exists a non-singular matrix P, which is also N by N. In this context, non-singular matrix means it has a matrix inverse. We say that A is equivalent to B is similar to B if there's some non-singular matrix such that A equals P, B, P inverse. So there's some factorization of A that involves the matrix B in a manner of speaking. And so we claim that this is an equivalence relationship. Now, when we talk about matrices, we should specify what are the appropriate scalars for that matrix. I'm not specifying that here because it honestly doesn't matter which fields you choose for your matrices. For the sake of argument, I should say for the sake of example, you can just pretend they're real numbers and we'll make much consequence here whatsoever. We just have to understand what matrix multiplication means in this context. So how do we show that this is an equivalence relationship? Well, to show something's equivalent relationship, the proof always has the following template. We prove the three properties, the three axioms, the reflexive axiom, the symmetric axiom, and the transitivity. And so we use the definition of the relationships to show reflexivity, just take an arbitrary element and show why it's equivalent to itself. So note, for example, note that A is equal to INAIN inverse here where IN is equal to the IN by IN identity matrix. So this is the matrix with ones along the diagonal zero everywhere else. The identity matrix has the property that if you take any matrix A and times it by the identity, this is equal to A. So this is the identity matrix. And so note that we have this factorization and I should mention the identity matrix is non-singular because the identity matrix, its inverse is itself. This is a non-singular matrix. So this then tells us that A is similar to A. So every matrix is similar to itself. The reflexive property typically falls out pretty quickly. And we will show the reflexive property here because we had an identity matrix. How do we show symmetry, right? For symmetry, what we have to do is we have to start with an assumption. Assume that A is similar to B. That's how you always start a symmetry property proof. Start with one assumption about the relationship. Then you unravel the definition. So there exists some non-singular matrix P such that A equals P, B, P inverse. That's what it means for the two things to be related. Now you wanna start unraveling this equation here, right? So then if you're not sure what to do then you can always work this thing backwards, right? We wanna conclude with something like therefore B is related to A. B is similar to A. And that of course means that B equals Q, A, Q inverse. So we have to find some non-singular matrix so that you can factor B using A in that non-singular matrix. So coming back here, we might think of how does one accomplish that. Notice that if we take P, B, P inverse, we can times both sides of the matrix by P inverse. I should say this is gonna be equal to A, of course. I'm gonna multiply on the left-hand side by P inverse. And I'm gonna write on the, I'm gonna multiply on the right-hand side by P. Do be aware that matrix multiplication is non-commutative. So the order in which you multiply does matter. If you wanna cancel the P on the left-hand side, you have to multiply by P inverse. And if you wanna get rid of the P inverse on the right-hand side, you have to multiply by P on the right. So those things would cancel out in that manner. So notice what we have here is that P inverse times P, B, P inverse times P, that's gonna equal, every, that looks kinda sloppy. Oh well, we know what it says. That's gonna equal P inverse A, P. And so then continuing down this vein, what we see here is that B is equal to P inverse A, P. Now that's not exactly what we need, right? We need a Q, A, Q inverse. So we need to have the inverse on the right-hand side. How do we do that? Well, note that P is equal to P inverse, inverse. And so thus we see that our matrix Q is actually just gonna be P inverse. So we get B equals P inverse A, P inverse inverse. And that actually finishes up the proof we wanted. This proves the symmetric property. That oh, if A is similar to B using the matrix Q, or matrix P, excuse me, then B is similar to A using the inverse. So we are able to prove the symmetric property because we have inverse matrices. Non-singular matrices are invertible and they have an inverse. So transitivity is the last one we wanna show. To show transitivity, we're gonna assume, we assume that A is similar to B and that B is similar to C. And so what does that mean? So there exist matrices P and Q such that A equals P, B, P inverse. And we also have that B equals Q, C, Q inverse. We wanna show that A is similar to C. So there has to be some non-singular matrix which when multiplied by C appropriately gives us A. Now in this example, you'll notice here, it's like, okay, we use the identity matrix for the reflexive property. We use matrix inverses for the symmetric property. How do we do the transitive property? We have to put these together somehow. Well, B equals B, right? We could substitute this inside here. And so then we could see that thus we're gonna get that A equals P times Q, C, Q inverse, P inverse. And so by the associative property, this will equal PQ times C times Q inverse, P inverse. And so this seems like we're going in the right direction, right? If we could, maybe our non-singular matrix that relates C to A is QP or PQ, excuse me. But what about this one right here? Now the thing with inverse operations here that if you put your socks on, then your shoes, then you take your shoes off, then your socks, right? The order in which you undo the operations is backwards. So the order of operations matter when it comes to matrix multiplication, multiplication is non-commutative. And so when you take the inverse, you actually have to go backwards. So note here that, excuse me, PQ inverse is equal to Q inverse, P inverse. This is called the shoe sock property because if you reverse the order, that is if you do the process backwards, you put your socks on, then your shoes, you have to take your shoes off, then your socks to invert it. And so then because of this observation, we see that this is equal to PQ times C times PQ inverse. And so therefore we see that A is related to C. And so this shows that the, that this similarity condition is transitive. This shows us that we have an equivalence relationship because it's reflexive, symmetric, and transitive. And we've proved that it's a requirements relationship. Similarity is an important equivalence relationship on matrices. For example, if two matrices are equivalent, they'll have this thin characteristic polynomial. They'll have the same determinant, they'll have the same trace, they'll have the same eigenvalues. Those eigenvalues will come up with the same multiplicities. There's a lot of important stuff that happens. In terms of choosing a representative, there's actually a couple options. The Jordan canonical form is a good representative for similar matrices. There's the rational canonical form. And if you study more linear algebra, you'll see this is a very important equivalence relationship. I think I'm gonna squeeze in one extra relationship inside of this video here. It's not really a continuation of the other one, but just a very quick one, right? If you have two real value functions, F and G, these are real value functions from R to R, and we'll actually say they're differentiable. We can say that two functions are equivalent if and only if they have the same derivative. And this is gonna be an equivalence relationship, right? Why is it reflective? Reflexive, I should say. The reflexive property holds because of the following that F prime is gonna equal F prime, which implies that, I guess I should say it this way, right? So if F is equal to itself, right, this implies they'll have the same derivative, which then implies that F is related to F. That's pretty easy. Symmetric property. If, so let's say if we have F is related to G, I guess we have a different symbol here like this. We're saying, oops. If F is related to G, that implies that F prime equals G prime, which implies that G prime equals F prime, which implies that G is related to F. There you go. And then for transitivity, equally as simple, you get that, all right? So if, if F is related to G, and G is related to H, that implies that F prime is equal to G prime. But then the second relationship says that G prime is equal to H prime. And then that implies, and I should be more careful here, that implies, so G prime equals H prime. By transitivity of equality, you get that F prime equals H prime, which then implies that F is related to H. Very, very basic arguments. This is the typical argument one does here. And so you just have to show all three properties to show that you have an equivalence relationship. Now in calculus, if two things have the same derivative, you can also argue it at the following statement. They only differ by a constant. And honestly, when you start doing antiderivatives, you often then make that specification. Here's an antiderivative plus an arbitrary constant. And if you're a poor calculus student, then you probably are like, oh yeah, you always set C equal to zero by default. So you're choosing a representative of an equivalence class, which in calculus, you typically don't want to do that. You get in trouble for getting your plus C because you're choosing a representative instead of choosing the class of equivalent functions there.