 So let's try to find the eigenvectors for this 2 by 2 matrix without using determinants. This all started when we considered v to be any non-zero vector. If vav was a dependent set, then we had an eigenvector. But if vav was independent, but vava squared v was dependent, we could factor a polynomial expression in a, and then one of the factors would give us an eigenvalue and, at the same time, compute an eigenvector. So what can we do with that? Well, let's start off by picking any vector v. And we'll choose our vector v equal to 1, 0. And we find av, which is our matrix a times v, and a squared v. And the easiest way to find that is that's av times a again. Now, since these are vectors in R2, this set of three vectors must be dependent. And so let's take a look at that. Because they're dependent, there must be non-zero values x1, x2, x3, where the linear combination of vav and a squared v is equal to 0. And we can find a solution by row reducing the matrix of column vectors. So we'll take our matrix of column vectors and put it into row echelon form. Well, it's already in row echelon form, but we might want to simplify it a little bit. We can factor a 2 out of our second line. And we can parameterize our solutions. And since we want a specific solution, let's pick a value for t. So let's pick t equal to 1. And so one of our solutions is going to be. And that gives us our coefficients x1, x2, x3. And so we know that 2v minus 3av plus a squared v is equal to the 0 vector. And I'll rearrange that a little bit. So we know that a squared v minus 3av plus 2v is the 0 vector. Well, let's take a look at that. Everything has this right factor of v. So we'll rewrite that as a squared minus 3a plus 2i times our vector. And this matrix expression factors. Now let's find that a minus i applied to v. So we know what a is. We know what i is. So a minus i applied to v will be. And we have a minus 2i applied to 7, 2 gives us the 0 vector. And so that means 7, 2 is an eigenvector for lambda equal to 2. Now there may be other eigenvectors. But that's OK. We can rearrange the factors. So again, we have a squared minus 3a plus 2i. Well, this time let's factor it as a minus i times a minus 2i. And again, we'll find a minus 2i applied to our vector v. So that's a minus 2i applied to v. And that's going to be 6, 2. And so 6, 2 is an eigenvector for lambda equal to 1.