 So, I need one definition, this is the notion of what is called Gershgorin disks. So A is a matrix and C to the n cross n and we define D i to be the set of complex numbers such that Z minus A ii is less than or equal to magnitude of Z minus A ii is less than or equal to sigma j equal to 1 to n j not equal to i Z minus sorry A ij i equal to 1 to up to n. In other words, I am defining a circle, this is the radius of the circle and this is the center of the circle and so I am defining a disk which is centered at A ii and of radius equal to the sum of the magnitudes of the other entries of A in the same row. So, that is how I define these disks. So, for instance in this matrix that I showed you as an example in this matrix, the Gershgorin disks are all the same, they are centered at on the real line at n at location n and the radius is equal to n minus 1. I will just draw that here, so this is 0, this is n and this circle is of radius n minus 1. I do not want to draw such a big circle, so let me erase this. 0, this is n, let us say this is n minus 1, then I draw a circle here here. All the, it has n disks, every matrix of size n cross n will have n disks, but all n disks are the same, they overlap in this case. So, now let us write out the theorem. This is also one of these amazing theorems. Again, one of those completely non-intuitive results at least to me. So, A is any matrix in C to the n cross n, then the eigenvalues of A lie in the union i equal to 1 to n di. So, if you take the union of these Gershgorin disks, all the eigenvalues of A will lie in the union of those disks. Further, if a union k of the n disks forms a connected region, see these disks could overlap with each other. So, suppose k of them form a connected region that is disjoint from the remaining n minus k disks, then there are exactly k eigenvalues in the connected region. So, that is what the theorem says. Any questions about the theorem? So, I will not go through the whole proof here, but I will just give you a sketch or rather just the main idea of the proof. So, first of all, lambda of A are the roots of the characteristic polynomial, P A of lambda equal to determinant of A minus lambda i. So, what that means is that if lambda 0, which is possibly a complex number, again, I emphasize that we are not dealing with Hermitian symmetric matrices. So, eigenvalues can be complex valued here is an eigenvalue of A, then if I look at the matrix A minus lambda 0 i, this, what can I say about this matrix is singular, which implies that it is not or rather it is not invertible. What that means is that the diagonal dominance condition must be violated for some index of the diagonal entries. So, by the actual name for this theorem is dominant diagonal theorem, not diagonal dominant theorem. So, let me just correct that. So, by this dominant diagonal theorem, it must be that the one of the diagonal entries in this violates the diagonal dominance condition that implies mod of the diagonal entry for the the ith diagonal entry will be mod of A i i in magnitude will be mod of A i i minus lambda 0. This is less than or equal to the sum of the magnitudes of the off diagonal entries in the same row, which is just mod of A i j for sum i, which means that lambda 0 lies in at least one of the d i s. Because if this is true, then lambda 0 satisfies this meaning that lambda 0, if I substitute lambda 0 here, mod of lambda 0 minus A i i is less than or equal to this quantity. So, it's actually lying inside. So, lambda 0 as if I replace z with lambda 0, it satisfies this condition. So, lambda 0 belongs to d i. Of course, lambda 0 may belong to other d i s as well. And that's when we say that these disks are overlapping with each other. Okay. So, this proves the first part of the theorem. Okay. So, we just showed this part. For the second part is where I'm going to sketch the proof or rather I'm just going to give you a visual proof. So, recall that we said we can write A of epsilon as d plus epsilon b where d contains the diagonal entries of A and epsilon b and b contains all the off diagonal entries in A. Now, the eigenvalues of d are easy to locate and the eigenvalues are just a i i. Okay. And now we use the fact that the eigenvalues are continuous functions of the entries within an epsilon neighborhood around the eigenvalues of d. Okay. So, in other words, so for example, if I, you know, on the two dimensional plane, if I had say A11 over here, this is the real and imaginary part. This is the two dimensional complex plane. Then A11 may be here, A22 may be here, and A33 is here, and ANN is here. And so, basically, if epsilon is very small, then these radii of these Gershgorin disks will be epsilon times summation a ij. So, there will be small circles around these points. Those will be the Gershgorin disks. And we've already seen that the eigenvalues lie inside these Gershgorin disks. Now, as you increase epsilon, it's possible that some of these disks will end up touching each other or they end up merging. And so, if k of these circles, so as epsilon increases, the radius of these disks increases. But the eigenvalues are always inside these disks. And so, if k of these circles form a connected region, which is disjoint from k circles or disks form a connected region that is disjoint from the remaining n minus k disks, then those k disks must contain k eigenvalues. Again, this is because the eigenvalues are continuous functions of the entries. And so, they cannot suddenly jump to another disk. So, this is the part where I'm doing a bit of hand waving. It requires a little continuity argument. So, you can see the text for the full proof. So, let's see one simple example. Suppose I have a matrix A, which is equal to 2, 0, 0, 1, 0, minus 2, 0, 1, minus 1, 0, 3, 2, and 0, 1, 1, 5. Does this matrix satisfy the dominant diagonal theorem? Yes, sir. Yes. For every row, you see that the magnitude of the diagonal entry is strictly greater than the sum of the magnitudes of all other entries in the row. Okay. So, we can use this Gershkoren disk theorem to show that exactly one eigenvalue of A, one eigenvalue of A is in the left half plane. So, what are the Gershkoren disks here? The Gershkoren disks are D1. So, D1 is the Gershkoren disk corresponding to this diagonal entry. It is a set of complex numbers Z such that Z minus 2 is less than or equal to the sum of all the other entries here in magnitude, which is 1. D2 is the set of Z such that Z minus of minus 2, which is plus 2, is less than or equal to again sum of all the other off diagonal entries in the same row in magnitude, which is 1. D3, Z minus 3, that is the third diagonal entry, is less than or equal to 1 again. And D4 is a set of all Z, Z minus 5 is less than or equal to 4. So, if I were to draw these disks on the real line, so say this is 0 and minus 1, minus 2, and this is 1, 2, 3, 4, 5. So, the first disk is Z minus 2 is less than or equal to 1. So, it is the circle of radius 1 centered around 2. So, I will draw it like this. I am not good at drawing circles. Okay. And over here and let us say D3. This is centered at 3 and radius 1. So, that will look like this. And this last one using funny colors here, D4. This is centered at 5 and as of radius 4. So, in fact, it is a big circle like this. So, these three circles form a connected region. And this is one distinct region. So, there will be one eigenvalue over here and three eigenvalues and three eigenvalues in this region. And this circle lies entirely in the left half plane and is disjoint from the other three disks. So, that is why this has exactly one eigenvalue in the left half plane. Okay. So, one remark is that Gershkorin's theorem does not say that there is one eigenvalue in each disk. Okay. This is important. This, all it says is that every eigenvalue will be in the union of the disks. Okay. So, pick an arbitrary eigenvalue. It will sit inside the union of these disks. And further it says that if k of the disks form a connected region which is disjoint from the remaining disks, then there are precisely k eigenvalues in that connected region. That means that this matrix A that we wrote, it could have all three of its eigenvalues over here for example. It need not be contained in these three disks. Now, but there is a corollary which is that and call it corollary one. If the Gershkorin disks are mutually disjoint that is di intersection dj equals the empty set for all i not equal to j then it is true that there is exactly one eigenvalue in each disk. In other words, when disks merge it's possible that the eigenvalue passes into the other disk. It need not be in the first disk anymore. But as long as the disks remain disjoint then there will be one eigenvalue in each disk. This is another corollary. If the Gershkorin disks are mutually disjoint the matrix A is real. That is it has real valued entities then its eigenvalues must be real. Anyone wants to reason out why this is the case? Take a guess. What can we say about the characteristic polynomial of a real valued matrix? So the coefficients of the characteristic polynomial are real valued numbers and so the roots of the characteristic polynomial must occur in complex conjugate pairs. And so each of the Gershkorin disks are mutually disjoint and they're all centered on the real line. So if I take any particular disk like this the roots of the characteristic polynomial since they must occur in complex conjugate pairs if there's a root up here there will be a root down here. But then if I mean there must be a root which is in this circle here and if the root has a nonzero imaginary part then there will be another root down here and so then it'll have two roots inside this Gershkorin disk. But then we already said in the Gershkorin disk theorem that if the disks are mutually disjoint then each disk should contain exactly one eigenvalue that's not possible. And so each disk must contain only one real valued eigenvalue. The next corollary is that since A and A transpose have the same eigenvalues why is this true? Why do A and A transpose have the same eigenvalues? They are similar matrices so they must have the same eigenvalues. So we can use the columns of A define these Gershkorin disks. And these could yield either better or additional information. There are some examples in the text you can look at. But you can see that if you define the Gershkorin disks using the rows it'll give you a set of Gershkorin disks. If you define them using the columns it'll give you a different set of Gershkorin disks because these matrices are not symmetric, not necessarily symmetric. And so that could give you some additional information about the location of the eigenvalues. Just one more remark and this is related to this thing about the continuity of eigenvalues that I've been talking about. So suppose A is in c to the n cross n and eigenvalues lambda 1 to lambda n. Given any epsilon greater than 0 there exists a delta greater than 0 such that if B in c to the n cross n is a matrix such that mod of Aij minus Bij is less than delta for every ij. So B is a matrix that is within a delta neighborhood of the matrix A. Then there exists a labeling mu 1 through mu n of the eigenvalues of B such that mod of lambda i minus mu i is less than epsilon. In other words I don't have to consider the eigenvalues of B and see here the eigenvalues of B need not be, need not even be real value and the same holds for the eigenvalues of A. So it's just saying that you can consider the eigenvalues of B in some order such that if you take corresponding entries of corresponding eigenvalues of A and B and subtract them and look at the magnitude that will be at most equal to the difference would be at most epsilon for all i equal to 1 to n. So basically this is actually known as the continuity of eigenvalues theorem, write it up here and it basically says that if the entries of A are changed by a sufficiently small amount then the change in the eigenvalues is also small. So one way to see this is that the eigenvalues are the roots of the characteristic polynomial Pa of lambda. Now this Pa of lambda depends continuously on the entries of A, the coefficients are just obtained by taking products and sums of the entries of A and so it depends continuously on the entries of A and so this theorem here the continuity of eigenvalue theorem is actually a consequence of another theorem which says that the roots of a polynomial with complex coefficients depend continuously upon these coefficients and this in turn is related to a famous theorem in analysis called the intermediate value theorem. Okay so yeah so now yeah. So how small is sufficiently small? So that is hard to say I mean that depends on the matrix so but what it's saying is that there is a delta which is which if you choose it small enough then as long as you take any B whose entries differ from the corresponding entries of A by at most delta then the eigenvalues of B when considered in some particular order are such that every eigenvalue of B is within an epsilon neighborhood of the corresponding eigenvalue of A but how small that delta needs to be is something that depends on the matrix itself. Okay sir and is there any relation between delta and epsilon? No so the idea is that I mean there is a relation of course in the sense that if you choose if you're starting out with an epsilon that is given to you and corresponding to that epsilon you're choosing a delta of course you can see that if you choose any smaller delta it will still work but there will be a biggest possible delta such that as long as you put up by smaller than delta the eigenvalues get perturbed by at most epsilon but if you choose delta by two of that or half of that that will also work that will satisfy this condition so there is a relationship but it's more like an epsilon will determine an upper bound on how big delta can be but again these are things that these are existence results and so it just says that there exists such a delta it doesn't tell you how you will find it nor does it tell you how it depends on the matrix A. Okay so now in the next class we will actually look at several interesting results that follow as a consequence of this Gershgor and disc theorem and in particular some results on the condition of eigenvalues which in turn will tell us how these eigenvalues will get perturbed when you perturb the matrix so we'll stop here for today and continue on Wednesday.