 Now I'm just moving on to this kind of presentation. Some of these equations might be a bit long for the blackboard So let's have a go now. We've seen some Eigen values and they were distinct and different from each other. We had a polynomial We solved it and we had lambda sub 1 and lambda sub 2 But what if they repeat they both real but they repeat look at that lambda sub lambda square plus 6 times lambda That's 9 equals 0 that will result in lambda sub 1 equals lambda sub 2 equals negative 3 If it was a third-order polynomial with a 3 by 3 a matrix size for matrix A I might have gotten a yet another value for them for lambda sub 3 there In any way, we're dealing here with real repeated values Eigen values Now depending on the order of the polynomial will have this lambda minus lambda sub 1 to the power m Yeah, we have it squared now. They are two possibilities here From this eigenvalue eigenvector, I might get many eigenvalues for lambda sub 1 From this eigenvalue lambda sub 1. I should say I can get many eigenvectors And actually get m and that should be less than n and it's the size of the matrix a matrix a remember And if I can get many eigenvectors or more than one eigenvector for that repeat eigenvalue lambda sub 1 Then I'll have a solution set that looks like that Now I must say at the end there that should have read c sub m and k sub m But you get the point you'll just carry on they'll have the different constants and different eigenvectors as fast case of 1 case of 2 Exetus concern the other The other possibility is a bit more difficult if I only get a single eigenvector for the eigenvalue Lambda sub 1 then something else is going to happen in this case We'll have this set of solutions It'll be x of 1 x of 2 until we get to x of m whatever the value of m was was going to be and look how it's constructed They with x sub m will have the eigenvector k sub m 1 Times t to the power m minus 1 over m minus 1 factorial times e to the power lambda upon t and we carry on carry on Carry on with that till the end we'll have One over one there as far as the t to the power m minus m and over zero factorial Which is this one over one right at the end So if you look back at x sub 1 and x sub 2 at the top of there and you put in one and then two Into this x sub m equation you'll get those two values You'll get those values for x sub 1 and x sub 2 now We usually don't write the eigenvectors as k sub 2 1 k sub 2 2 Etc it's written as it is in the bottom two equations The x sub 2 equals k and then p or if it was an x sub 3 It will be k p and q and look the deteriorating values of this of the powers of t So that was x sub 3 so we're going to start with t squared and then with a p There's only a t to the power 1 and then with a q there's t to power 0 And you'll notice at the bottom as well It'll be 2 factorial which is 2 then 1 factorial which is 1 and 0 factorial which is this one So that's how we construct it if we can only get one vector for that repeat eigenvalue y sub 1 Difficult to understand but once you've done an example just look back at this video. It'll make a lot of sense Let's just see what happens now if I have M equals 2 so I have x sub 2 there That's going to be the k the t and then the p as well But remember I can always write x prime the matrix x prime as a times x So let's just expand both sides the x prime and the a x x sub 2 prime from the top The first term k t e to the power lambda sub 1 t you'll have to use the power rule and you can see the result there And if I take my x sub 2 and I multiply it by the matrix a I'm going to get a a k t e to the power Lambda sub 1 t and a p e to the power lambda sub 1 t So now I've got these two equations and I can equate them to each other Bring everything to the left-hand side. So it's going to equal the 0 vector on the right-hand side column vector And I can also group terms as you'll see in the last line So all I've done is I've written stuff in a certain order that makes sense And you can pause the video and think back to linear algebra as to why it makes sense in this order and look at this If you look at that first line think about it It can only equal the 0 vector if a k minus lambda sub 1 k and the a p minus lambda sub 1 p minus k both Equals 0 they've got a equal 0 for this to equal 0 on the right-hand side the 0 vector I should say on the right-hand side and The only way that that can be is I can take k out as a common factor Remember I've got to put the identity matrix i in there otherwise. It's a matrix minus a scalar So it's a minus lambda sub 1 i times k equals 0 And for the second factor in the parentheses there I just bring the k over to the other side take p out as a common factor and If I had more Values to work with if my value for m was larger say x sub 3 I Would have had to queue in the p-day and so it would just carry on and with this information. We can do some examples