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 lambda is. So
you re looking at a non-zero column vector
which you multiply by [A] turns out to be
some scalar lambda times the column vector
[X]. So what we want to do we want to be able
to find out hey what is this lambda? How do
I find the eigenvalues of this square matrix?
So what that means is that [A][X]- "lambda"[X]=[0]
that s a zero-vector right there so I get
[A][X]- "lambda"[I][X]=[0] what that means
is that ([A]- "lambda"[I])[X]=[0] and the
reason why I multiply this "lambda" 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 lambda from
[A] which is not allowed because [A] is a
n by n matrix and lambda 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
lambda 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]- "lambda"[I]), not just [A],
but ([A]- "lambda"[I]) will have to be singular
so the det([A]- "lambda"[I]) will have to
be zero. What that this matrix right here
([A]-"lambda"[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 lambda so if you are able
to solve this set of equations right here
det([A]-"lambda"[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]- "lambda"[I]) it s going to turn
out to be of this form "lambda"^n+C1"lambda"^(n-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.
