 Welcome back everyone. We're still gonna be in section 6.1 in our textbook linear algebra done openly about eigenvalues and eigenvectors. We've seen already what an eigenvalue is, an eigenvector is, so if you missed that, check out the previous video about that one. And then also in another previous video, we've learned how to check to see if an eigen, or if a vector is an eigenvector or not. In this video, I wanna actually talk about how does one identify if a scalar is an eigenvalue, right? Because when you are checking if something's an eigenvector, you can just multiply the matrix and the vector together. And when you factor the product, the eigenvalue kind of pops out. But in this situation, if we had a matrix A, take it to be 1652, this was of course the same matrix in the last video right there. Could we check if seven is an eigenvalue of that? Well, the issue is I don't know what vector to check out, because if you're an eigenvalue, that means there's some vector X, so that multiplying X by A is the same thing as multiplying X by seven. But are we just gonna try every vector in the world? Well, if we're looking for vectors in R2, there's a lot of options there. We might spend a long time, and I wanna keep this lecture video kind of short. So how can we do that? Well, in order to test if we have an eigenvalue or not, let me kind of give you some explanation how the plan of attack is. If you have an eigenvalue, that means you satisfy this characteristic equation right here, AX equals lambda X. Well, this is an equation after all, and if you start manipulating that equation, you could, for example, you could subtract lambda X from both sides of the equation having the effect that moves it to the right-hand side. So then you get AX minus lambda X equals zero. And of course, multiplying by lambda is the same thing as multiplying by the diagonal matrix, lambda I, where capital I is the identity matrix. And so this is the matrix, lambda I right here. This would be the matrix where you have lambdas along the diagonal and zeros everywhere else, okay? And so scaling vectors, the same thing as multiplying by a scalar matrix. This right here is what we call a scalar matrix. And so then the reason I wanna introduce this scalar matrix is that you'll notice that both terms here are divisible by X. So if you factor it out, we end up with this expression right here, A minus lambda I times X equals zero. So what we're gonna see here is that lambda, lambda is an eigenvalue, it's an eigenvalue of A if and only if A minus lambda I is actually singular, right? So the issue going on here is that this vector X right here, this X would satisfy this equation exactly when X is inside the null space of A minus lambda I, all right? Now, the null space of A minus lambda I right here, this will always contain zero, but the zero vector, but as we defined earlier, an eigenvector has to be non-zero. So is there any non-zero vector inside the null space? A non-trivial null space would only happen when your matrix is singular. So we're trying to check is A minus seven I singular or not. So that's where we're gonna continue here. We're gonna take A minus seven I and we have to see if this matrix is singular. And there's a couple of ways you could check for singularity. You could take the determinant of that thing, that's an option. I'm gonna try just row reducing it to see what's going on there. So the matrix A was given above one, six, five and two and we're gonna subtract from it the scalar matrix sevens along the diagonal, you subtract seven from each of these. And so when you do that, you end up with negative six, six, five and negative five, right? Like so. And so now with this matrix in hand, we could try to solve the homogeneous system of equations. So we wanna solve A minus seven seven I X equals zero. And so we know that this comes down to solving the augmented system A minus seven seven I augment zero, like so. Now of course, when you're augmenting zero, that doesn't really do much. I mean, if we look at this thing, we would actually take negative six, six, five, negative five, augment zero, zero. And if you exclude the last column, it's not such a big deal. Cause for a homogeneous system as we reduce this it's never gonna change anything. So things to notice very quickly is that the first rows, everything's divisible by six. So I'm gonna divide everything in the first row by negative six and the second row, everything's divisible by five. So I'm actually gonna divide everything by five just for convenience there. And actually, I think, yeah, we'll do that. And so if we do that, this will row reduce the first one becomes one, negative one, zero. The second row becomes one, negative one, zero. And so you can see quite clear here, like, oh, the two rows are identical. Just take row two minus row one. And then this thing will row reduce to be one, negative one, zero, and then a row of zero, zero, zero, zero. Like so. If we retranslate this augmented matrix as a system of equations, or really in this case just a single equation, right? We could determine that x one minus x two equals zero. That is to say x one equals x two. And so we can get a vector x that'll look like one one, right? So we get this vector right here. This vector right here, x spans the null space of a minus seven i. And I claim this is an eigenvector right here. And so if we were to check it, so this is just like we did previously, right? If we take the matrix a, then make sure you take the original matrix a, not this, not a minus seven i. So if we take a times x, which remember a was one, six, five, and two. And if you times it by one and one, notice what you get here is you're gonna get one plus six of the first bit. You get five plus two, which is the second bit. And so this ends up with seven seven, which when you factor out the seven, you get one one right here and Bob's your uncle. We found that one one is an eigenvector, but we also confirmed the fact that we wanted to show earlier is that seven was a eigenvalue of this. So what we've seen here right here in this calculation is that we could see how to check if you're, we're able to check to see if we had an eigenvalue or not. So coming back up to this location right here, we were prepared to say that yes, seven is an eigenvalue at that location. And that's because the homogeneous system, A minus seven IX equals zero has a non-trivial solution. We saw that by seeing the free variable right here in the system. And then as we continue to solve the system, we can actually find a specific eigenvector for that eigenvalue. And so that's kind of impressive. We can find our, that is we can check, is this number and eigenvalue or not by solving this homogeneous system of equations, the AX minus lambda, sorry, the A minus lambda I, X equals zero. All right, so stay tuned for the next video to kind of see why, well, because we don't just wanna guess every single eigenvalue in the world, right? So at some point we gotta figure out how to know, how do you know to guess seven? And we'll talk about that forthcoming. See you next time.