In this segment, we will talk about how do
you find eigenvectors of a square matrix.
So let s suppose somebody says hey find eigenvectors
of this matrix right here: 3, -1.5, -0.75,
plus 0.75.
So the first thing I m trying to understand
is that eigenvectors correspond to eigenvalues
so this particular 2 by 2 matrix is going
to have two eigenvalues and corresponding
to each eigenvalue we will have an eigenvector
so in the previous segment we have already
found out that the two eigenvalues corresponding
to this particular square matrix is lambda1=3.421
and lambda2=0.3288 so these are two eigenvalues
which we found in the previous segment lets
go and see how we can find the eigenvector
corresponding to this eigenvalue and this
one you can then find by yourself so what
we are saying here is that hey let then this
column vector x1, x2 be the eigenvector corresponding
to lambda=3.421 so we re saying hey let this
two rows and one column vector, or this column
vector, be the eigenvector corresponding to
the first eigenvalue.
How do we go about finding that?
We already know that from the definition of
eigenvalues is that hey ([A]-lambda[I])[X]=0
so if lambda is the eigenvalue of [A] then
if we subtract lambda times the identity matrix
from [A] and we multiply the eigenvector it
has to give us zero.
So if we do that what we have now is: 3, -1.5,
-0.75, 0.75 minus lambda which is 3.421 then
1, 0, 0, 1 which is the identity matrix times
[X] which is x1, x2 is equal to zero, zero.
So all I m doing is substituting the value
of [A] matrix substituting the value of lambda
and then putting the identity matrix right
here and x1, x2 is the assumed eigenvector
and it has to satisfy this particular set
of equations.
From there this is what we get, we get: -0.421,
-1.5, -0.75, -2.671 times x1,x2 equal to zero,
zero and we already know that hey x1,x2=0
is a solution to this particular set of equations
because if I put x1=0, x2=0 I ll get the left
hand side same as the right hand side but
what we are looking for is a non-trivial solution
so how do we look for the non-trivial solution
is that hey lets choose x1=s if x1=s then
we get from the first equation -0.421s-1.5x2=0
that s what we get from the first equation
right here and from here we get x2= -0.2808s
so if that is the case then we know that the
eigenvector corresponding to lambda=3.421,
so this was our eigenvalue, so corresponding
to this eigenvalue 3.421 the eigenvector is
x1,x2 and we know that hey if I assume x1
to be s some number s x2 will be -0.2808s
so this is the eigenvector corresponding to
this eigenvector right here but we said s
right there we can take s to be outside of
the scaler and get 1, -0.2808 so we can consider
this to be the eigenvector corresponding with
that eigenvalue but any if we multiply x1,
x2 by any scaler that also is an eigenvector
corresponding to that eigenvalue but we can
also convert this into a unit vector if you
want to that might make it unique in some
ways you still have two possibilities in here
because a unit vector is based on the magnitude
of the vector but all those things can be
done later so we can at least say that hey
1, -0.2808 is an eigenvector corresponding
to lambda1=3.421 and you can do the same thing
corresponding to lambda2=0.3288 which was
the other eigenvalue which we had corresponding
with this the eigenvector turns out to be
this to be 1, 1.781 you can do that as part
of your homework that hey this is the eigenvector
corresponding to this eigenvalue corresponding
to this eigenvalue following the same procedure
this will turn out to be one of the eigenvectors
corresponding to this eigenvalue.
And that is the end of this segment.
