 So, what is being said here? See anything that belongs to the kernel of this, what can you say about objects that belong to the kernel of this? What is so special about objects that belong to the kernel of n? What is n? Remember what n was? n was a minus lambda i. So, what can you say about any object that belongs to the kernel of a minus lambda i? So, for so v belongs to kernel n implies v is an eigenvector of a is it not? So, that means it is as if this last fellow I can obtain by simply solving the eigen equation. I know that there are n repeated eigenvalues. I must have at least one eigenvector, we have seen the existence of eigenvalue eigenvector few lectures back. So, there must be at least one, if not more. If I have exactly n eigenvectors then it is diagonalizable, no problems at all. We are in fact interested in the case when that number is less than n. So, the point is this as I said earlier also, if you encounter an operator what are your chances that you will encounter an operator with repeated eigenvalues minimal maybe next to nothing, but from physical dynamics of a system or whatever suppose you cannot avoid repeated eigenvalues then what are the chances of diagonalizability? Very funnily enough that is also next to nothing. So, if I give any arbitrary random matrix from some process that you do not even know of the chances of encountering a situation with repeated eigenvalues is the probability is almost 0. It is 0, but if that is given that the physical process itself has certain modes such that the eigenvalues have to be repeated then the chances that such a matrix will be diagonalizable is again 0 which is why it is important to study these kind of systems. Why is this so? Just think about it, it is a bit of a digression, but just think about it. I give you a diagonal matrix lambda, remember the eigenvalues have to be the same. If you hit it with any non-singular matrix what do you get? You still get lambda i. So, the family of all diagonalizable matrices with all their eigenvalues repeated is exactly a single done. Consider the family of all matrices on an n dimensional vector space n cross n and operators on n dimensional vector space all these eigenvalues are repeated. How many matrices in such a family will be diagonalizable? It is only 1 because you hit it with any similarity transformation any change of basis it will still remain that just the diagonal fellow. You see what I mean. On the other hand if this was not diagonalizable then by hitting it with different basis change you can generate different matrices each of which and we know that similarity transformations do not change the characteristic polynomial or the minimal polynomial. So, that is the reason why I said that if you cannot help, but have matrices with all the eigenvalues repeated then you are probably stuck with the situation either the given matrix is already diagonal then there is nothing to diagonalize, but if you know that the matrix is not diagonal and yet it has all its eigenvalues at the same location then there is no way you can diagonalize it because if you could then every other matrix would have been the same diagonal matrix. So, if you encounter any non-diagonal matrix all whose eigenvalues are at the same location lambda such a matrix can never be diagonalized is that clear? So, that is why we need to understand why this Jordan form is important because this Jordan form is the next best thing to diagonalization of a matrix all whose eigenvalues are co-located located at the same location at the same place that is the beauty of this Jordan form. Anyway that was a bit of a digression I hope that serves as a motivation that this Jordan form is not just you know or sort of a rare phenomena it is rare in the sense that you are unlikely to encounter matrices all whose eigenvalues are at the same place, but if you do encounter such a matrix and the matrix is not already diagonal then you cannot diagonalize it. So, next by looking at eigenvectors we have a handle on the first fellows here. So, we know how to do this you take set it up as a minus lambda i v is equal to 0 solve for v given that you know what the lambda is you solve for v and all these solutions you call them n raised to the m v 1 acting on v 1. Notice peculiarity of this form I am not saying this is v 1 through v k, but rather the action of some nilpotent matrix on these v's. So, I am not directly telling you the answer you realize that there is some amount of inconvenience here. So, every solution that you get of this equation in terms of v have a form like this and that is the magic of this statement here. Once you have these fellows generated how do you get these for how do you think you can generate these other objects. So, let us call ok I do not want to erase the Jordan form basically. So, I want to keep that in maybe I can erase this upper part here. So, your v gives off the following solutions. Let us say the first level solution you get is n m v 1 you do not even know what m v 1 is by the way. There is only one way to find out by eventually working through this. This is v 1 n raise to m v 2 acting on v 2 until n raise to m v k acting on v k ok. But that is only the first fellows of this humongous basis apparently. How do you get from this to this rather how would you set it up as a problem of finding this fellow given you know this fellow. What is the relation between this and this? You cannot have an n inverse come on n is a singular matrix right. But that is the beauty you are on the track track you have found this. This it turns out is equal to n acting on n to the power m v 1 minus 1 times v 1. But now this is my unknown. So, as your friend suggested that take n inverse. But how do you actually get to that n inverse right? It turns out there is no guarantee of a solution in the first place. But because of this Jordan canonical form see I just want you to appreciate why this might not seem like a very obvious thing. n does not have an inverse. And yet this theorem tells you that the proposition the statement the premise of this theorem is that once you have found this v by solving for an eigenvector just go ahead and equate that eigenvector as. So, suppose this v is your first eigenvector that you obtained all right. Now, set it up as the right hand side of an equation set of the n as the a matrix or the like you know you want to set it up as an a x is equal to b form. So, this now becomes your x and because this is an non invertible matrix you cannot just invert it. But the funny thing is that this eigenvector that you found continues to remain in the column span of n that is why you will always have a solution that is the beauty of Jordan canonical form. And you can successively keep repeating this process. So, once you first you obtain the eigenvector at the first stage you obtain the eigenvector then what do you do? You take the eigenvector as the constant in a x is equal to b that is a b set of the a as the main proton matrix n. Let the n act on some unknown vector that leads me to the eigenvector that unknown vector will always have a solution. Once you have got this unknown vector now now put this unknown vector and stack it up on the right hand side of the equation again set up another equation all right. So, I mean maybe I will run out of symbols, but nonetheless. So, first you set up a minus lambda i v is equal to 0 you got your v solve for v ok. This is your first step second step set up v is equal to n w solve for w. The Jordan form merely guarantees that such a solution that you will not run into trouble at in this algorithm at this step that is the beauty of this. The Jordan form assures you as we will see in the proof next day that the Jordan form assures you that this equation will have a solution a aspective of the fact that n is not invertible. Then you carry on like this solve for w next you set up p is equal to n sorry w is equal to n p. How long do you carry this on? Now it is one thing to say that there exists some numbers m v, but in practice when you are trying to get a matrix to its Jordan form no one will tell you that number. But it is this algorithm you see as I said it will run into a stopping criteria when there will be no solution. So, you keep repeating this repeat till no solution exists. So, although I have written it down in the opposite manner like this it is actually this fellow that happens to be your eigenvector. See that is why it is so special these are the fellows in the kernel of n fellows in the kernel of n are exactly the eigenvectors. Now subject to this ordering the first fellow here is an eigenvector once you have got the eigenvector set up the eigenvector as a constant vector and try to solve for a matrix n x or n v n w n p whatever is equal to the eigenvector. So, these subsequent vectors are called the generalized eigenvectors. The first fellow is the eigenvector the subsequent fellows are the generalized eigenvectors ok. We have run out of time, but we will discuss this in the next module in the next lecture, but if you have some time just try to figure out that subject to this choice of ordered basis that is the reason I have written it down in this form. If you use this as not just a basis, but an ordered basis ok try to see the representation of n in terms of this basis that will basically be the main crux of the Jordan canonical form representation. That will tell you why in the Jordan canonical form you will have no more in non-zero entries apart from the ones in the diagonal and the super diagonal ok. So, it is not just any arbitrary block diagonal matrix in that larger scheme of things it is a very special kind of a decomposition where even the part that contains repeated eigenvalues can be blocked down to as thin a representation around the diagonal as possible in that only the diagonal entries exist and only the first super diagonal. So, it is not just any arbitrary upper triangular matrix, but a very special kind of an upper triangular matrix with a minimum coupling because in solving every system of equation you are only going to be requiring one variable and a variable just below it right. So, that is the benefit of Jordan canonical form we will see that representation you can try this in terms of this ordered basis and then we will see a proof of this Jordan canonical form in the next lecture. Thank you.