CHAPTER 10.07: EIGENVALUES AND
EIGENVECTORS: How do I find eigenvectors of a square matrix? Example 2 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 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-λ, 0, 1, -0.5, 0.5-λ, -0.5 and then -0.5, 0, and –λ 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 row 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-λ) 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-λ)(-λ)-(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-λ)(-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
–λ3+2λ2-1.25λ+0.25=0 that’s what I’m going to get as the cubic
equation and you can see very well 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 –λ3+2λ2-1.25λ+0.25=0. And we will see that λ=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 (λ-1) is a factor of –λ3+2λ2-1.25λ+0.25
and what I’m going to do is if λ -1 is a factor of this you can always
do long division like this and let’s suppose we do it we get –λ2 here so
that’s –λ3+λ2 there’s minus here so I get λ2 plus here so that
cancels λ2-1.25 λ+0.25 and then this will go by multiplying the
λ plus λ so I get λ2-λ here and then I subtract I get
this cancels I get -0.25λ+0.25 and then I say -0.25 I’ll get -0.25
λ here +0.25 here and that gives me no remainder so that is the quotient
when I divide this particular cubic polynomial by λ-1 as the factor so
what that basically tells me is that my -λ3+ 2λ2-1.25λ+0.25=(
λ-1)(- λ2+ λ-0.25). And this one here will give you (λ-1)
and I can write this as (-λ+0.5)(λ-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 λ=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 λ=0.5
which is a repeated eigenvalue. So in this case what I want to find out is
that if I have [A-λI][X]=[0] so all I need to
find is [X] corresponding to
λ=0.5 so in that I have [A-λI] is as
follows: 1.5-λ, 0, 1, -0.5, 0.5-λ, -0.5, -0.5, 0, -λ times the
eigenvector let’s suppose x1¬, x2, x3 equal to 0, 0, 0. So when I’m going to
put lambda is equal to 0.5, what do I get from here so far as the [A-λI] 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 an 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 λ=0.5 so what about the
eigenvectors so if we want to find the eigenvectors corresponding to
λ=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 λ=0.5 and similarly λ=1 So
similarly there is another eigenvalue λ=1 similarly you can find that
the eigenvector corresponding to λ=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 λ=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. |