CHAPTER 10.13: EIGENVALUES AND EIGENVECTORS: Theorems of eigenvalues and eigenvectors Part 6 of 6     In this segment we are talking about some theorems corresponding to eigenvalues and eigenvectors and one of the theorems says that if λ1-λn are eigenvalues of nxn [A] matrix so your squared matrix which is nxn then you can find then you can find det(A)= λ1 * λ2 *…λ3 absolute value so in order to find the magnitude of the det(A) all you have to do is you take all the eigenvalues find the product of those eigenvalues and you get the det(A) the magnitude of that which can also be written as this: I is equal from 1 to n, lambda i. But you also now lets us make sense out of one of the theorems that if one of the eigenvalues is zero then the [A] matrix is not invertible its singular it cannot inverse cannot be found which makes sense because if one of the eigenvalues is zero then the product of all these is going to be zero in the Spector of what the other eigenvalues are then the det the magnitude of the det(A) is going to be zero which means the det(A) is going to be zero which basically means that the [A] matrix is singular so that’s just another way of proving that particular theorem but let’s take an example and see what we mean by that so if somebody says im going to give you an [A] matrix [A]=2, -3.5, 6, 3.5, 5, 2, 8,1, 8.5 and they are also telling that im giving you the eigenvalues of this square matrix  λ1= -1.547, λ2= 12.33, λ3= 4.711 can you find the magnitude of the determinant  and you got to say yes I can find the magnitude of the determinant  because I know that the magnitude of the det(A) is nothing but det(A)=λ1 * λ2 * λ3=(-1.54)(12.33)(4.711) absolute value and this number turns out to be =89.88so that’s the magnitude of the det(A) which basically means that the det(A) is either -89.88 or +89.88 but you are able to find the magnitude of the det(A) which might be the most important thing to find in most cases and that’s the end of this segment