 In this video, we're going to present the solution to question number 12 for the practice exam number three for math 2270 We are given a three by three matrix a and we're told that Lambda equals two is an eigenvalue of a with this information in this matrix We need to compute a basis for the eigen space of a corresponding to lambda equals two now the The the eigen space of a corresponding to lambda equals two that means the null space of a minus two times the identity or if you prefer you could take the null space of Two times the identity minus a it doesn't matter if you take a minus lambda the identity or Lambda identity minus a because these matrices only differ by a negative one which will not affect the eigenvalues the characteristic polynomial The eigen space any of it so you can do either one. It doesn't really matter So if I take two I minus a what that means is I'm going to take two times the identity So you get two zero zero zero two zero zero two and you subtract from it a Four four negative two one four negative one Three six negative one what you get is the following you're gonna get two minus four Which is a negative two you're gonna get zero minus four which is a negative four you're gonna get zero plus two For the first row for the second row you get zero minus one you get two minus four and you're gonna get zero plus one For the last row you're going to get zero minus three You're gonna get zero minus six and then you're gonna get two plus one which is three like so In which case now we need to row reduce this matrix We wouldn't we need to find a basis for the null space of this matrix right here now because Lambda equals two is an eigenvalue We can guarantee that there's gonna be a row of zeros in this situation And how we want to proceed, you know, there's a lot of things we could do The first thing I'm gonna do is I'm gonna grab the first row and Move it to the top because I want a one there. I'm also gonna times it by negative one I'm gonna times the first row by negative one there. So what that then gives us is A new matrix we're gonna get one two negative one We're gonna get negative two negative four and two and we're gonna get negative three negative six and three like so And so now they're having a one in my pivot position. I can notice below The second and third rows are the scalar multiples of the first row. I'm gonna take row two here I just take row two because the second row is just the first row times negative two So I'm gonna take row two. I'm gonna add to it two times for a one that's gonna cancel out to give me a row of zeros I'm gonna do the same thing for row three, but this time I'm gonna take three times row one For which that's just gonna cancel out these rows and then we find the ref of this matrix to i minus a will be one Two negative one zero zero zero zero zero zero like so For which case We could revert this matrix back to the corresponding Homogeneous system of equations and solve for it That is you could solve for the dependent variables with respect to the free variables There's gonna be two free variables here x two and x three are gonna be free variables x one will then to be dependent upon that For which you could rewrite this as simply just x one is equal to negative two x two plus x three Right because this equation was x one plus two x two Minus x three is equal to zero you could go from there and start unpeeling things from there But when you have the ref you need to find a basis for the null space You can rip it apart really quickly from what you see right here So what we in fact see is that the null space of two i minus a is going to equal the span Of the following two vectors You're going to get two vectors because they coincide with the free variables And so filling out the chart here We have a free variable in the second position and in the third position So we're just kind of like a blank right here For the x one which we'll come back to in just a second So you get one zero and zero one And then filling in that blank We're just going to look at this row right here because this corresponds with the first pivot and we just switch the signs So we're going to get a negative two and a positive one So a basis for the eigen space here is going to be negative two one zero one zero one We need to find a basis for two i minus a Right and then once to do that you find the ref which we did right here And then you can read the basis of the null space from the ref