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. Lets 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 were 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 Im 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 thats 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 were getting is a second order polynomial equation we got a second order polynomial here and equal to zero. So its 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 (3.753.092)/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.