 Welcome back to the lecture series based upon the textbook linear algebra done openly. As always, I am your professor today, Dr. Andrew Misaline. In this section, we are gonna talk about, while we're in section 6.3 and we're talking about this, seemingly made up word diagonalization and when a matrix is diagonalizable, right? I would caution you before you start trying to play those tiles down in the game, scrabble, you verify whether your dictionary includes this word or not, but it's a perfectly acceptable word in linear algebra here. A matrix, you say, is diagonalizable when it's similar to a diagonal matrix. We'll talk about that a little bit more, but the idea of diagonalization is very closely related to the existence of a so-called eigen basis. Is there a basis for a vector space that can be comprised only of eigen vectors for a specific matrix A? So we have some matrix A, which is in by N and can we construct a set of linear independent eigen vectors? So we're gonna be interested in what conditions can we guarantee independence between eigen vectors? And there is a special case for which independence is guaranteed. If we have, say, R eigen vectors, X1, X2, X3, up to XR, these are eigen vectors of the matrix A and these correspond to distinct eigen values, lambda one, lambda two, up to lambda R. So what we're saying here is that A times the vector X1 is just equal to lambda one X and also A times X2 is equal to lambda two X2. So that's, again, we're an eigen vector, eigen value there. AX3 equals lambda three X3. You get the idea here, continue on to the end of the list. So we have eigen vectors and these are associated to different eigen values. So lambda one is not equal to lambda two, which is not equal to lambda three, proceed until we get to the end of the list. Under this circumstance, we actually can guarantee that eigen vectors associated to different eigen values are gonna be linearly independent. And how do we do that? The proof here uses a mathematical concept known as mathematical induction. And the idea is basically the following. You start off with a so-called base case. Is it ever true? Is there a simple case where it is true? And for this situation, a base case, well, we could say that, well, if you have a single eigen vector, if you have a single eigen vector, this set is linearly independent because after all, a set of one vectors is independent if and only if it's non-zero and eigen vectors are not allowed to be zero. So we do get that it is independent sometime. And admittedly, I might even start off with you could do the empty set, right? If you have no eigen vectors, r equals zero in that case, that also would act as a base case right here, but it feels a little bit more natural. Well, maybe for those somewhat very early to linear algebra, it might feel a little bit more natural to start with r equals one. So you have this base case. And so the next thing we're gonna do is we're gonna be like, okay, if it's true for r equals one, then we can proceed to show it's true for r equals two and then proceed to show it's true for r equals three. Then we do that and that and now we just kind of pretend you down this progression until it's like, okay, so let's assume it is true up to just some specific value k, right? So we've proven it's true for one, for r equals two, three, four, five, up to some value r equals k. So let's consider then the next iteration then. Consider the case where we have x one, x two, up to x k plus one, many eigenvectors. So what happens to that case? Well, if we're interested if something is linearly independent or not, we look at the homogeneous system. So we wanna consider c one, x one, plus c two, x two, and we take this sum all the way up to c k plus one, x k plus one, and suppose this is equal to zero. So we wanna consider this equation right here. If the set of vectors is linearly dependent, then there should be only the trivial solution to this homogeneous system of equations. Well, taking this equation, let's first multiply it by the matrix A. So we multiply everything on the left-hand side, c one, x one, all the way up to c k one, x one, k k, sorry, x k one right there, and then on the right-hand side, we times a zero right there. Well, the right-hand side's pretty easy to deal with, right? We're gonna end up with a zero right there, but also on the left, we can distribute this A onto each of the pieces in this term and so you're gonna get A times x where all of those x's are different eigenvectors. And so when you times the eigenvector by the matrix, it's the same thing by times in that scalar, c one, lambda one, x one, and then you get c two, lambda two, x two, right? And this pattern continues up until we get to c k plus one, lambda, need a little bit more space there. Whoops, lambda died. Lambda k plus one, x k plus one, and this is equal to zero. So make notice of this equation right here. We're gonna come back to it in just a second. Also, so if you take the original equation right here and we times everything by A, bunch of these landas pop up everywhere. Also, what we could do is we could take the original equation and we can multiply everything by lambda k plus one. Oh, that's poor pinmanship, k plus one. We're gonna times that by all of these ones, c x one, all the way up to c k one, x k plus one, lambda k plus one times the zero vector. Well, on the left-hand side, we can distribute the lambdas. So we get these lambdas and distribute all the pieces. And so we're gonna get c one, lambda k plus one, x one. We're gonna end up with a c two, lambda k plus one, x two. And this will then continue on until you get to the end, c k plus one, lambda k plus one, x k plus one, like so. And this equals the zero vector right there. And so we're gonna put a green star next to this one. And so I want you to compare these two star equations right here. And in fact, let's subtract them. Let's take yellow star minus green star. What's gonna happen in that situation? Well, you can combine like terms, you can combine the x ones together. And when you combine the x ones together, you're gonna get c one times lambda one minus lambda k plus one times x one. And then you're gonna do the next one. You're gonna get c two times lambda two minus lambda k plus one, like so. You're gonna get x two right here. And then you continue on down this pattern until you end up with some c k. And then you're gonna get lambda k minus lambda k plus one, x k plus one, and then the last term, you're gonna get a c k plus one times lambda k plus one minus lambda k plus one times x k plus one. And this should all equal zero. All right, so now we've reached the point where we're gonna actually sort of benefit from the ingenuity of how we constructed this argument here. You'll notice this very last term, we have this k or lambda k, k plus one minus lambda k plus one. That's the same number, that's gonna cancel out. So this whole last term disappears in the sum. And so looking at the left-hand side, we now have a multiple of x plus one plus a multiple of x plus two all the way up to a multiple of x plus k, that equals zero. Now, by the assumptions we've said already, it's like, oh, we've already shown that x one up to x k, that set is linearly independent. We've already established that. Because remember we went from one to two to three to four to five, we did up to k. So this is an independent set. So the only way that this set can combine to be zero, this would imply that each of these coefficients, c i times lambda one, or lambda i should say minus lambda k plus one, this equals zero. The only way an independent set can combine to give you zero is if all the coefficients were zero themselves. Now by the zero product property, if this thing is zero, if this product is zero, this would imply that either c i equals zero or this would imply that lambda i equals lambda i plus one. That's because lambda i minus lambda k plus one would equal zero, you move to the other side. And this possibility is not possible, remember? Because we are assuming that all of these eigenvalues are distinct. So since they're distinct, that factor's not zero. So it must have been all of the original factors were zero as well. So come back up to this equation right here. We had this dependence relationship. We had this dependence relationship on our vectors right here. It had to be that c one, c two, c k are all zero. So we get a zero, zero all the way up to this. So all of the coefficients are zero and this would then tell us that c k plus one times x k plus one is likewise zero. Because everything else went to zero. But how can these all be zero as well? Well, this would happen only if they get using that sort of zero product property here. We're getting that either c k plus one equals zero or x k plus one equals zero. Which the second possibility is not possible because eigenvectors can't equal zero. So we've now established that every coefficient in this linear combination must have been zero. And this establishes the fact that, okay, if x one up to x k is linear independent, that will imply that x one up to x k plus one is gonna be independent as well. And so as this pattern continues like a string of infinite dominoes, as one domino falls and knocks over the next domino, the next domino, the next domino. And we can see that all these, if we have a set of eigenvectors associated to distinct eigenvalues, that set of vectors is necessarily going to be linear independent. And we're gonna use this in our forthcoming discussion about eigenvectors and eigenbases more specifically.