 Since our eigenvalues come from solving an equation, it's possible that some of those solutions might be complex numbers. In particular, since the characteristic polynomial of an n by n matrix is an nth degree polynomial, it's possible that an eigenvalue will be a complex number. We can still find the associated eigenvectors using the arithmetic of complex numbers. For example, suppose I want to find the eigenvalues and the eigenvectors of the matrix 1, 1, negative 1, 1. So we'll find our characteristic polynomial, which will be 1 minus lambda squared plus 1, and that gives rise to a corresponding characteristic equation, lambda squared minus 2 lambda plus 2 equals 0. As a general rule, if you have a kind and nice teacher, they will give you characteristic equations that are easily factorable. I am not a kind and nice teacher, but that's okay. You know how to solve the quadratic equation because you have the quadratic formula. So we'll solve the characteristic equation using the quadratic formula and get solutions 1 plus or minus i. And we can find our eigenvectors in the usual fashion. The only real difference here is that we're dealing with complex coefficients. So for lambda equal 1 plus i, the eigenvectors will be found by row reducing our matrix, and we get solution negative i times x1 plus x2 equals 0, and that gives us parameterized solutions x2 equals is, x1 equals s. And if we let s equal 1, we get the eigenvector 1i. For the eigenvalue 1 minus i, the eigenvectors will be found by row reducing the matrix, and we'll have our solution negative i x1 plus x2 equals 0, and our solutions parameterized will be x2 equals is, x1 equals negative s. This time we'll let s equal negative 1, and that'll give us the eigenvector 1 negative i. And something that's worth noting here is that our eigenvalues are complex conjugates, and our eigenvectors are also complex conjugates. And you might ask yourself, self, is this a coincidence? And it turns out that it isn't. In particular, suppose a is an n by n matrix with real entries. If lambda is a complex eigenvalue with associated eigenvector v lambda, then the conjugate of lambda is also an eigenvalue, and its associated eigenvector will be the conjugate of our original eigenvector.