In this segment we'll talk about "How can
we find eigenvalues of a square matrix" through
an example. Let s suppose somebody says hey
find eigenvalues of this square matrix right
here gives you 3, -1.5, -0.75 and 0.75. So
this is the matrix that is given to you and
you are asked to find the eigenvalues of this
matrix then we know that hey det([A]-"lambda"[I])=0
where [A] is this square matrix right here.
So that what we re seeing is the determinate
of 3, -1.5, -0.75, plus 0.75 minus "lambda"i
which is 1, 0, 0, 1 is equal to zero. If we
do the matrix subtraction I m going to get
something like this (3-lambda) because lambda
is right here,(-1.5) because (-1.5- lambda)0
is that quantity itself,(-0.75) and (0.75-lambda).
So that s what we get when we subtract the
two matrices so we have to find the determinate
of this matrix and put that equal to zero.
The determinate of a two by two matrix is
simple, you simply multiply this by this and
subtract what you get from the product of
this and this. So that turns out to be then
(3-lambda)(0.75-lambda)-(-1.5)(-0.75)=0. And
if we expand this, this is what we are going
to get, we are going to get lambda^2-3.75*lambda+1.125=0.
As you can see that we have a two by two matrix
of which we are trying to find eigenvalues
and what we re getting is a second order polynomial
equation we got a second order polynomial
here and equal to zero. So it s a quadratic
equation so everybody knows how to find the
solution of a quadratic equation so it is
lambda=[-(-3.75) "plus or minus" [(-3.75)2-4(1)(1.125)]]/[2(1)]
so 
based on this we get (3.75 'plus or minus"
3.092)/2 so will have two roots we will get
3.421 and we get 0.3288 so those are the two
eigenvalues here so lambda1=3.421, lambda2=0.3288
as the two eigenvalues of this particular
square matrix. And that is the end of this
segment.
