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 |