CHAPTER 10.05: EIGENVALUES AND
EIGENVECTORS: How do I find eigenvalues of a square matrix? Example 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]-λ[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 λi
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-λ) because λ is right here,(-1.5) because (-1.5- λ)0 is that quantity
itself,(-0.75) and (0.75-λ). 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-λ)(0.75-λ)-(-1.5)(-0.75)=0.
And if we expand this, this is what we are going to get, we are going to get
λ2-3.75λ+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
λ=[-(-3.75)±√[(-3.75)2-4(1)(1.125)]]/[2(1)] so based on this we
get (3.75±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 λ1=3.421, λ2=0.3288 as the
two eigenvalues of this particular square matrix. And that is the end of this
segment. |