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Current time:0:00Total duration:5:39

Example solving for the eigenvalues of a 2x2 matrix

Video transcript

In the last video we were able to show that any lambda that satisfies this equation for some non-zero vectors, V, then the determinant of lambda times the identity matrix minus A, must be equal to 0. Or if we could rewrite this as saying lambda is an eigenvalue of A if and only if-- I'll write it as if-- the determinant of lambda times the identity matrix minus A is equal to 0. Now, let's see if we can actually use this in any kind of concrete way to figure out eigenvalues. So let's do a simple 2 by 2, let's do an R2. Let's say that A is equal to the matrix 1, 2, and 4, 3. And I want to find the eigenvalues of A. So if lambda is an eigenvalue of A, then this right here tells us that the determinant of lambda times the identity matrix, so it's going to be the identity matrix in R2. So lambda times 1, 0, 0, 1, minus A, 1, 2, 4, 3, is going to be equal to 0. Well what does this equal to? This right here is the determinant. Lambda times this is just lambda times all of these terms. So it's lambda times 1 is lambda, lambda times 0 is 0, lambda times 0 is 0, lambda times 1 is lambda. And from that we'll subtract A. So you get 1, 2, 4, 3, and this has got to equal 0. And then this matrix, or this difference of matrices, this is just to keep the determinant. This is the determinant of. This first term's going to be lambda minus 1. The second term is 0 minus 2, so it's just minus 2. The third term is 0 minus 4, so it's just minus 4. And then the fourth term is lambda minus 3, just like that. So kind of a shortcut to see what happened. The terms along the diagonal, well everything became a negative, right? We negated everything. And then the terms around the diagonal, we've got a lambda out front. That was essentially the byproduct of this expression right there. So what's the determinant of this 2 by 2 matrix? Well the determinant of this is just this times that, minus this times that. So it's lambda minus 1, times lambda minus 3, minus these two guys multiplied by each other. So minus 2 times minus 4 is plus eight, minus 8. This is the determinant of this matrix right here or this matrix right here, which simplified to that matrix. And this has got to be equal to 0. And the whole reason why that's got to be equal to 0 is because we saw earlier, this matrix has a non-trivial null space. And because it has a non-trivial null space, it can't be invertible and its determinant has to be equal to 0. So now we have an interesting polynomial equation right here. We can multiply it out. We get what? Let's multiply it out. We get lambda squared, right, minus 3 lambda, minus lambda, plus 3, minus 8, is equal to 0. Or lambda squared, minus 4 lambda, minus 5, is equal to 0. And just in case you want to know some terminology, this expression right here is known as the characteristic polynomial. Just a little terminology, polynomial. But if we want to find the eigenvalues for A, we just have to solve this right here. This is just a basic quadratic problem. And this is actually factorable. Let's see, two numbers and you take the product is minus 5, when you add them you get minus 4. It's minus 5 and plus 1, so you get lambda minus 5, times lambda plus 1, is equal to 0, right? Minus 5 times 1 is minus 5, and then minus 5 lambda plus 1 lambda is equal to minus 4 lambda. So the two solutions of our characteristic equation being set to 0, our characteristic polynomial, are lambda is equal to 5 or lambda is equal to minus 1. So just like that, using the information that we proved to ourselves in the last video, we're able to figure out that the two eigenvalues of A are lambda equals 5 and lambda equals negative 1. Now that only just solves part of the problem, right? We know we're looking for eigenvalues and eigenvectors, right? We know that this equation can be satisfied with the lambdas equaling 5 or minus 1. So we know the eigenvalues, but we've yet to determine the actual eigenvectors. So that's what we're going to do in the next video.