 The fact that eigenvalues produce eigenvectors allows us to consider what we might call an eigenspace. Before we do that, we'll have to take a look at some important ideas. First, remember that our characteristic equation is going to be the value of the determinant of a minus lambda times i, and this will always be a polynomial in lambda. You should take a moment to understand why this is the case, and that means that solving the characteristic equation requires finding the roots of a polynomial. But one of the things you know is that it's possible for a value to be a repeated root of the characteristic equation. And this leads to what is known as the algebraic multiplicity of an eigenvalue, and the algebraic multiplicity of an eigenvalue lambda is the multiplicity of the root lambda in the characteristic equation. For example, let's say we want to find the eigenvalues and the algebraic multiplicities for the matrix 5, 1, negative 4, 1. So setting up our characteristic equation, which will be the determinant of a minus lambda i equal to 0, is going to give us the characteristic equation lambda squared minus 6 lambda plus 9 equals 0. And we'll solve this. We'll use the quadratic formula and get our solutions 3 or 3. So remember the algebraic multiplicity of an eigenvalue is the multiplicity of the root lambda in the characteristic equation. And in this case, lambda equals 3 has multiplicity 2. It appears twice as a solution, and so lambda equals 3 is an eigenvalue with algebraic multiplicity 2. What happens next? After we've found our eigenvalue, we row-reduce a minus lambda i to find the eigenvectors. And when we do that, it's possible that a given value of lambda may produce more than one eigenvector. And this leads to the notion of geometric multiplicity. The geometric multiplicity of an eigenvalue lambda is the number of linearly independent eigenvectors it produces. So for example, let's consider the geometric multiplicities of the eigenvalues of our matrix. So we found lambda equals 3 is the only eigenvalue. We can then use this to find the eigenvectors. So solving for our eigenvectors is going to give us in parameterized form the eigenvector s times negative 1, 2. Or if we let s equals 1, we'll have the single eigenvector negative 1, 2. And again, the geometric multiplicity of lambda is the number of linearly independent eigenvectors it produces. This eigenvalue lambda equals 3 gives us 1 eigenvector, so its geometric multiplicity is going to be 1. Let's take a look at another example and we'll find the algebraic and geometric multiplicities of the eigenvalues of this matrix. We'll set up and solve our characteristic equation and the solutions are going to be lambda equals 4 twice and lambda equals 3 once. And so that means lambda equals 4 is an eigenvalue with algebraic multiplicity 2. To find the geometric multiplicity, we need to find the eigenvectors for lambda equals 4. So setting up our system of equations and reducing our coefficient matrix, we find that the eigenvectors have parameterized solution s100 plus t010. And that means we have two linearly independent eigenvectors. So lambda equals 4 has geometric multiplicity 2 with eigenvectors 100 and 010. Similarly, for lambda equals 3, we'll set up our equation, row reduce the coefficient matrix, and find that the eigenvector is s times negative 1, negative 2, 1. And this is the only eigenvector. So lambda equals 3 is an eigenvalue with algebraic multiplicity 1 and geometric multiplicity 1 with the eigenvector negative 1, negative 2, 1. And we close with the following idea. Given any eigenvalue lambda, the set of eigenvectors for lambda forms a vector space. And this is something that you should prove. This allows us to define an eigenspace. Suppose A has eigenvalues lambda with geometric multiplicity n and associated eigenvectors v1, v2, and so on. Then the set of linear combinations of vectors in our set of eigenvectors forms the eigenspace of lambda.