CHAPTER 10.04: EIGENVALUES AND EIGENVECTORS: How do I find eigenvalues of a square matrix?

In this segment we will talk about the basis of how do we find eigenvalues of a square matrix. So if somebody is telling you that hey I want you to find the eigenvalue of a matrix [A] then we only know that it has to satisfy this condition that there has to be some eigenvector some vector , column vector, x which is non-zero keep in mind because if it is zero it’s going to be satisfied for no matter what the value of λ is. So you’re looking at a non-zero column vector which you multiply by [A] turns out to be some scalar λ times the column vector so what we want to do we want to be able to find out hey what is this λ how do I find the eigenvalues of this square matrix. So what that means is that [A][X]- λ[X]=[0] that’s a zero-vector right there so I get [A][X]- λ[I][X]=[0] what that means is that ([A]- λ[I])[X]=[0] and the reason why I multiply this λ by [I] is because if I multiply [I] by [X] I get [X] itself so there is nothing wrong with doing that but the reason why I introduce this identity matrix there is because so that when I’m doing the subtraction here I am doing the subtraction between the matrices of the same size that allows me to do that if I didn’t do that I would be subtracting λ from [A] which is not allowed because [A] is a n by n matrix and λ is just a scalar so then if you look at this particular set of equations and somebody says hey I gave you a set of equations for which the right side is zero one of the solutions is zero itself.

So x=0 is a solution to this set of equation but what we started with was that hey x has to be non-zero so we can call this λ to be an eigenvalue so the only way it is possible that you wouldn’t get so what that means is that this particular set of equations does not have a unique solution because if it had a unique solution x=0 is one of the solutions zero-vector is one of the solutions that would be a unique solution so if we are seeing that this one does not give a unique solution the only way that is possible is if this whole quotient matrix which is ([A]- λ[I]), not just [A], but ([A]- λ[I]) will have to be singular so the det([A]- λ[I])=0 will have to be zero. What that this matrix right here ([A]- λ[I]) has zero determinate, means that it is singular, means that it doesn’t have an inverse and that’s how we are going to be able to find λ so if you are able to solve this set of equations right here det([A]- λ[I])=0 then you will have a solution what is going to happen is that if [A] is an n by n matrix so we’re talking about the order of the matrix is going to be n, what’s going to happen is that when you find the determinate of this matrix right here ([A]- λ[I]) it’s going to turn out to be of this form λn+C1λn-1+…+Cn=0 that’s what you’re going to get. The form of the expansion of the determinate of this matrix here is going to turn out to be of this form so you’re going to have a matrix you’re going to have a polynomial equation the polynomial order of this polynomial equation will be n so since we have a left order polynomial equal to 0 you’re going to get n roots of that polynomial and those will be the eigenvalues of the [A] matrix. And that’s the end of this segment.