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.