In this segment we will take an example of
how we can find eigenvalues and eigenvectors
of a square matrix so let s suppose somebody
says find eigenvalues and eigenvectors of
this matrix: [A]=1.5, 0, 1,-0.5, 0.5, -0.5,
-0.5, -0.5, and 0, 0 so in order to find the
eigenvalues of this particular matrix what
we are going to do is we are going to find
the determinate of this matrix where we take
the [A] matrix and from that we subtract lambda
times the identity matrix and we'll put that
determinate equal to 0.
So what that means is that what I m going
to get is the determinate of this matrix so
it will be the [A] matrix but the only thing
that should be different would be that our
minus lambda in each of the diagonals added
to it so we get 1.5-lambda, 0, 1, -0.5, 0.5-lambda,
-0.5 and then -0.5, 0, and lambda and the
determinate of this has to be shown to be
equal to zero.
Now I can find the determinate of this particular
matrix right here by using several methods
such as using the forward elimination steps
or gaussian elimination.
I can use the co-factor method and that s
what I m going to use here so if I use the
co-factor method let me for example take this
as my (?? 01:53) which I will use for expansion
for the cofactor method so what that means
is I take the first row, first column so I
get (1.5-lambda) and since its first row first
column it s one plus one for the first row,
first column that s two so it s even so it
ll be positive in the front here and then
it will be the determinate of the matrix which
is left over once I get rid of the first row
and the first column because first row, first
column so I get rid of the first column and
the first row so I m left with this two by
two matrix right here and the determinate
of two by two matrix this one here would be
simply multiplying this by this and subtracting
what I get from multiplying this by this so
that would be [(0.5-lambda)(-lambda)-(0)(-0.5)]
so that takes care of that but then I ll have
the zero so I ll get zero from there so I
don t need to show that and from there this
is the first row, third column so first row
1, third column 3 1+3=4 it s even so I ll
have plus here plus whatever this quantity
here which is one and then the determinate
of whatever is left over if I take my first
row and third column out that will be just
this two by two matrix and the determinate
of this two by two matrix right here is simply
this multiplied by this minus this multiplied
by this so that will be [(-0.5)(0)-(0.5-lambda)(-0.5)]
and that will be equal to 0.
So if I do the expansion of all this this
is what I m going to get I m going to get
lambda^3+2lambda^2-1.25lambda+0.25=0 that
s what I m going to get as the cubic equation
and you can very well see that there is a
lambda here, a lambda here, and lambda here
so I expect that I would get a cubic polynomial
for determinate and that would be equal to
zero.
Ok, so we have this cubic polynomial equation
lambda^3+2lambda^2-1.25lambda+0.25=0.
And we will see that lambda=1 is a root here
because I put 1 here I get -13+2(1)2-1.25(1)
+0.25=0 and that gets me 0=0 so it looks like
that just by observation I am able to say
that.
So what that means (lambda-1) is a factor
of lambda^3+2lambda^2-1.25lambda+0.25 and
what I m going to do is if (lambda-1) is a
factor of this you can always do long division
like this 
and let s suppose we do it we get lambda^2
here so that's lambda^3+lambda^2 there s minus
here so I get lambda^2 plus here so that cancels
lambda^2-1.25*lambda+0.25 and then this will
go by multiplying the lambda plus lambda so
I get lambda^2-lambda here and then I subtract
I get this cancels I get -0.25*lambda+0.25
and then I say -0.25 I ll get -0.25*lambda
here +0.25 here and that gives me no remainder
so that is the quotient when I divide this
particular cubic polynomial by lambda-1 as
the factor so what that basically tells me
is that my -lambda^3+ 2*lambda^2-1.25*lambda+0.25=(lambda-1)(-lambda^2+
lambda-0.25).
And this one here will give you (lambda-1)
and I can write this as (-lambda+0.5)(lambda-0.5)
so 
if I do that I will get this second polynomials
you can always find out the roots of this
you can always find the zeros of this second
of polynomial by using quadratic equation
and that s what you will get as the roots
-0.5 and -0.5 so 0.5 and 0.5 that s what you
will get as the roots and that s how you will
be able to do it so here by putting this equal
to zero you ll get lambda=1 as one of the
zeroes and you get 0.5 as another zero and
0.5 as another zero of this polynomial or
if you put it equal to zero you get three
roots of the cubic equation and that s what
you re getting here and what we are finding
out here is that you get three eigenvalues
1, 0.5, and 0.5 but these two eigenvalues
are not distinct they are the same and we
have to figure out how we can find the eigenvectors
corresponding to the time when we have when
we don t have distinct eigenvalues.
So let s look at how we can find the eigenvectors
corresponding to lambda=0.5 which is a repeated
eigenvalue.
So in this case what I want to find out is
that if I have [A-lambda*I][X]=[0] so all
I need to find is [X] corresponding to lambda=0.5
so in that I have [A-lambda*I] is as follows:
1.5-lambda, 0, 1, -0.5, 0.5-lambda, -0.5,
-0.5, 0, -lambda times the eigenvector let's
suppose x1, x2, x3 equal to 0, 0, 0.
So when I'm going to put lambda=0.5, what
do I get from here?
So far as the [A-�I] is concerned, it will
be I ll get 1 from here, 0, 1 and then I will
get -0.5, and I ll get 0 here, and -0.5 here,
and get -0.5 here, 0 here, and -0.5 here times
x1, x2, x3 equal to 0, 0, 0 and what you are
finding out here is that it looks like that
hey this equation here looks the same as this
first equation here looks the same as this
second equation here because all it is that
the first row is getting multiplied by -0.5
and the same thing is with the third row,
third row same as the second row, so you basically
have only one equation, one linearly independent
row, in this whole matrix right here because
this is dependent on this and this is the
same as this so that's also dependent on this
so there is only one independent row here
and what is the solution so if I choose x1=-a,
I can choose x3=a so a can be arbitrary real
number because if I choose a and a I ll get
zero and if I choose a and a I ll get zero
again if I choose a and a for x1 and x3 I
will get zero in respect to what I choose
for x2 because its 0, 0, 0 here no matter
what I choose for x2 so I can choose x2 to
be some other real number b what that basically
tells you is that -a, b, a is the eigenspace
corresponding to lambda=0.5 so what about
the eigenvectors so if we want to find the
eigenvectors corresponding to lambda=0.5 what
I will do is I will have to see what this
means so you got a, b, and a and I can write
this as a times -1, 0, 1 plus b times 0, 1,
0 so I get a times -1, 0, 1 plus b times 0,
1, 0 if I add these two if I add this linear
combination of vectors this the results which
I will get, and if we go from that perspective
these are the two eigenvectors so those are
my two eigenvectors which are corresponding
to lambda=0.5 and similarly lambda=1 so similarly
there is another eigenvalue lambda=1 similarly
you can find that the eigenvector corresponding
to lambda=1 would be as follows would be 1,
-0.5 and -0.5 that s what you are going to
get as the eigenvector corresponding to lambda=1
if you follow the same procedure so what we
are finding out here is that this is your
third eigenvector so this is the third eigenvector
you have a three by three matrix it had three
eigenvalues 1, 0.5, and 0.5 two of the eigenvalues
were not distinct so we had to find eigenspace
first and then corresponding to eigenspace
you had two eigenvectors and those are your
two eigenvectors and this is your third eigenvector
corresponding to the third eigenvalue which
is 1 so those are the three eigenvectors which
we have this one, this one, and this one corresponding
to the three eigenvalues.
And that's the end of this segment.
