CHAPTER 10.06: EIGENVALUES AND EIGENVECTORS: How do I find eigenvectors of a square matrix? Example

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 λ1=3.421 and λ2=0.3288 so these are two eigenvalues which we found in the previous segment let’s 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 λ=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]-λ[I])[X]=0 so if λ is the eigenvalue of [A] then if we subtract λ 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 λ 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 λ=3.421 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 λ1=3.421 and you can do the same thing corresponding to λ2=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.