 We are again in section 6.2, the characteristic polynomial of the textbook linear algebra done openly, and what I want to do is take a look at I want to find the eigenvalues of a square matrix. It's two by two, so it's not so bad And we're gonna find the eigenvalues using the characteristic polynomial Remember the characteristic polynomial is the determinant of the matrix a minus lambda i or if you prefer you could take the determinant of Lambda i minus a doesn't make much of a difference there just it's just a matter of Preference at that situation and so we're going to calculate the determinant of a minus lambda i So we're just going to subtract the lambdas from the diagonal injuries and we compute that as it's just a two by two determinant We'll take The product minus the product there the diagonals and so you get two minus lambda Multiply that by negative six minus lambda and then we're going to subtract from that three times three and So we are going to have to foil These first things out first outside inside last and so if you look at all those possible combinations You end up with negative 12 minus 2 lambda plus 6 lambda Plus lambda squared and then you also get that minus 9 there 3 times 3 Some combining some like terms we get a lambda squared in terms of the lambdas We're going to get a 4 lambda and then in the end we end up with negative 21 as our Characteristic polynomial now the main reason we're looking for the characteristic polynomial is we desire to factor it if we factor the Characteristic polynomial the eigenvalues are the roots of that polynomial So we want to look for factors of our constant term because notice the leading coefficient there is a 1 We want to look for factors of the constant term negative 21 Which add up to be positive 4 and we can accomplish this by doing let's say 7 minus 3 7 minus 3 is 4 and then it multiplies gives negative 21 and so then this would factor as lambda plus 7 and Lambda minus 3 and so if we set the characteristics polynomial equal to 0 We can set each of these factors equal to 0 lambda plus 7 equals 0 would imply that lambda equals negative 7 and then lambda minus 3 Equals 0 would imply that lambda equals 3 and so there we have it We have the the the two eigenvalues for this matrix They both have algebraic multiplicity of 1 and which which forces the geometric multiplicity to likewise be 1 And so if we wanted to calculate the eigenvectors associated to these eigenvalues We could then calculate their eigenspaces using the techniques. We've talked about before All right, and so for two by two matrices. This is typically how this calculation is going to go Stay tuned for the next video. We're going to show you where there is a potential anomaly that can happen That kind of justifies why we've cared about complex numbers so far this semester. I'll see you then