 The last time we discussed about non-derogatory matrices and how one can determine commutativity of two matrices when they are non-derogatory. So specifically, a non-derogatory matrix is one for which the geometric multiplicity of every eigenvalue equals one. And we proved one theorem which said that if A is a non-derogatory matrix, then B commutes with A if and only if there exists a polynomial P of degree at most n minus 1 such that B equals P of A. Also just made a small remark that the converse is also true that is A is non-derogatory if and only if every matrix that commutes with A is a polynomial in A. So one quick question for you, if I take the identity matrix, is it non-derogatory? No, it is not non-derogatory. Obviously because the identity matrix it commutes with every other matrix but it cannot be written, any other matrix cannot be written as a polynomial of the identity matrix. Any polynomial of the identity matrix will end up being a scaled version of the identity matrix. So you cannot in general write some matrix that commutes with A as a polynomial of A where A is the identity matrix. And so the identity matrix is non-derogatory, is not non-derogatory. How about the all ones matrix? So today we will discuss a little bit about convergent matrices. Now recall that a matrix is convergent if all the elements of A power m go to zero as m goes to infinity. So we also know that a diagonal matrix is convergent if the magnitude of all the diagonal elements of that matrix are less than one and the diagonal entries are the eigenvalues of the matrix and so you can say more generally that if the magnitude of all the eigenvalues is less than one then the matrix is convergent. This result extends directly to diagonalizable matrices also and we have seen that before. Now using the Jordan canonical form we can extend this idea to non-diagonalizable matrices as well. So if A equals s j s inverse then A power m equals s j power m s inverse. So if A power m goes to zero as m goes to infinity then this is true if and only if j power m goes to zero as m goes to infinity. So the question is when does j power m go to zero? So now j is a block diagonal matrix so it suffices to consider the behavior of a single Jordan block because each of the blocks get raised to the power m when you raise the a block diagonal matrix to the power m. So if I consider a single Jordan block say j k of lambda which is basically a matrix with lambda as on the diagonal and once on the first super diagonal and zeroes everywhere else and this can be written as lambda times the identity matrix plus this nil potent matrix which has zeroes on the diagonal and zeroes below here and once in the first super diagonal zeroes everywhere else and so we will call this lambda i plus nk. This is the k cross k matrix which when raised to the power k will give you the all zero matrix. So specifically nk is equal to j k of zero. Now nk power m equals zero for every m greater than or equal to k. So that means that if I take j k of lambda power m then this is equal to lambda i plus nk power m which can be written as the summation i equal to zero to m m choose i lambda power i times nk power m minus i and using this property here we can simplify this as summation i equal to m minus k plus one to m m choose i lambda power i nk power m minus i other terms will become equal to zero because nk to that power goes to zero and this is true for all m greater than or equal to k. Now so the diagonal elements which is the i equal to m term are all lambda power m lambda power m which means that for j power m to go to zero as m goes to infinity it is necessary that lambda power m should go to zero or mod lambda is less than one. And conversely if when lambda is less than one what we need to show is that this quantity will go to zero so that means that m choose I will write it as m choose m minus j lambda power m minus j. So I am just replacing i with m minus m minus j and so this we want to show that this goes to zero as m goes to infinity and for j equal to one through k minus one zero one up to k minus one. Now this quantity here m choose m minus j lambda power m minus j is equal to the magnitude of so this this combination term is m into m minus one all the way up to m minus j plus one divided by j factorial times lambda power j then I have a lambda power m here and this can be upper bounded by my magnitude of m power j I am replacing all these terms with m lambda power m divided by j factorial lambda power j and so basically then it suffices that to show that m power j times lambda power m so this is the only part that depends on the numerator is the only thing that depends on m so it suffices to show that lambda power m magnitude times m power j goes to zero as m goes to infinity now there are many ways to do it and what you can show is that see this is lambda power m and this is m power j j is some fixed number here it takes value zero one up to k minus one so none of that is scaling with m and so this is some polynomial term this is an exponential term and so eventually if mod lambda is less than one this will eventually overwhelm this term and you will get zero so one way to do to see that is if you take logs you have j log m plus m log lambda which will go to so this is m times log lambda whereas this is j times log m and so this will go to minus infinity if as m goes to infinity because log of mod lambda if mod lambda is less than one mod log lambda is less than zero so basically what this shows is that even for non diagonalizable matrices the matrix is convergent if and only if all the eigenvalues of the matrix are less than one so to show this result note that we made essential use of the Jordan canonical form