 Let's take a closer look at the problem of finding eigenvalues and eigenvectors. The important thing to observe is that once you've found the eigenvalues, it's straightforward to find the eigenvectors. But how do you find the eigenvalues? We can solve a system of linear equations. And remember that the eigenproblem is solving the equation transformation matrix times column vector equals scalar multiple of column vector. Rewriting this matrix equation as a system of equations in standard form gives us, and we can extract the corresponding coefficient matrix, which we'll write as a minus lambda times the identity matrix. So that means when we try to reframe the eigenvector eigenvalue problem as a system of equations, our coefficient matrix can be expressed as A minus lambda times I, where I is the identity matrix of the appropriate size. And here's where the determinant is useful. If the determinant of this coefficient matrix is non-zero, there will be a unique solution to the system of equations. In this particular case, that unique solution is vector x equal to the zero vector. But we've specifically excluded the zero vector as a possibility for an eigenvector. So if the determinant is non-zero, the only solution we get is the zero vector, which is not an eigenvector. And that means for any possibility of a solution, we have to have the determinant equal to zero. And this gives us something called the characteristic equation. For example, suppose we want to find the characteristic equation in all eigenvalues for this matrix. The eigenvalues will be the values of lambda that make our determinant A minus lambda I equal to zero. So we'll find the determinant, and the determinant equal to zero is the characteristic equation. This equation has solutions lambda equals one, seven, and negative two. These will be the eigenvalues. If lambda equals one, remember that the system of equations we're trying to solve corresponds to matrix applied to vector x1, x2, x3 equals one times x1, x2, x3. And so this gives us a system of equations. We can then re-reduce that system of equations and get our parameterized solutions. Since we only need to find one of the eigenvectors, we'll let S be one. And that tells us that eigenvalue one has eigenvector one, zero, zero. Similarly, if I take the eigenvalue lambda equals negative two, our system of equations is going to be matrix applied to x1, x2, x3 equals two times x1, x2, x3. And we'll reduce our coefficient matrix and get us our vectors in parameterized form, x1, x2, x3 equals S times one, negative three, three. And so again, taking S equal to one gives us the eigenvector one, negative three, three. And finally, for our eigenvalue seven, our system of equations will be matrix equals seven times vector, re-reducing that system of equations gives us our parameterized solutions. And again, we'll let S equal one to get one of the many eigenvectors, 14, 30, 24. And so now we have our characteristic equation, our eigenvalues, and all of our eigenvectors.