 This theorem, the Kurofischer theorem is very powerful and so we will actually discuss many results that come out as a consequence of this theorem. The first is a result which relates the eigenvalues of A plus B with those of A or B. This is a theorem due to somebody called Weil. So it's called Weil's theorem. So let A be C to the n cross n be Hermitian symmetric matrices. Then as usual, we'll arrange the eigenvalues in increasing order. So I'm introducing notation here. So lambda i of A is the ith largest eigenvalue of A. So lambda 1 of A is the smallest eigenvalue, lambda 2 of A is the second, is the next bigger eigenvalue and so on. And lambda i of A plus B is the ith eigenvalue of A plus B when the eigenvalues are arranged in increasing order. Then for each K equal to 1, 2 up to n, we have lambda K of A plus lambda 1 of B is less than or equal to lambda K of their sum and which is in turn less than or equal to lambda K of A plus lambda n of B. Now when you look at this result, I think at least some of you will see that the result is obvious. The kth eigenvalue of A plus B is going to be at least lambda K of A plus lambda 1 of B and at most lambda K of A plus lambda n of B. So for example, if B was the identity matrix, then what you see from this result is that the eigenvalues of A plus the identity matrix are actually you know this already that all the eigenvalues of A will get shifted up by 1 and therefore this result is also saying that lambda K of A plus the identity matrix is at least equal to lambda K of A plus 1 and at most equal to lambda K of A plus 1. In other words, lambda K of A plus B is equal to lambda K of A plus 1. So let us see, let us just write out how this thing is proved. Sir, A and V are can be interchanged here. Yes, of course. So there is nothing special about B here. B is not in any way different from A and B are both emission symmetric matrices. So in fact, before I proceed, I will write that also. So you can also say K of B plus lambda 1 of A is less than or equal to lambda K of A plus B lambda K of B plus lambda 1 of A lambda n of A. Okay, so basically if you want to obtain bounds for lambda K of A plus B, you can choose the min of these two quantities lambda K of A plus lambda 1 of B lambda K of B plus lambda 1 of A. And sorry, you can choose the max of these two that will still be a lower bound on lambda K of A plus B. And you can choose the min of these two. And that will also be an upper bound on lambda K of A plus B. So proof. So we know that for any zero not equal to x in c to the n by the Rayleigh-Ritz theorem, lambda 1 of B is less than or equal to x emission Bx over x emission x is less than or equal to lambda n of B. So as a consequence for any K 1, 2 up to n, if I look at lambda K of A plus B by my Kurov-Fischer theorem, this is equal to the min over W1 through Wn minus K. So I am using the min max formulation. So you should pay attention to this because you will see that for some results, we will use this min max formulation and for some other results will start from the max min formulation. Okay, and it's actually a very interesting exercise to see if you can prove the same result by starting from the say from the max min formulation instead of the min max formulation. Okay, in some cases it'll turn out that the proof kind of works out the same way. But in some other cases, it'll turn out that the proof going one way is much easier than the proof going the other way. So max over x not equal to 0, x perpendicular to all these vectors of x emission times A plus B times x over x emission x. So this is just Kurov-Fischer theorem just written out for this case. And this itself I can write as x emission Ax plus x emission Bx over x emission x. So I can just instead of writing a whole step, I'll just say that this is equal to x emission Ax over x emission x plus x emission Bx over x emission x. Now, this x emission Bx over x emission x is at least equal to this. So if I replace the x emission Bx by lambda 1 of B, I'm only decreasing the value of whatever this thing is. So this is greater than or equal to, in fact, I'll write that here itself so that it's clear. This is always for any x not equal to 0, this is greater than or equal to x emission Ax over x emission x plus lambda 1 of B. And so that means that this lambda K of A plus B is greater than or equal to, because I've replaced this by its lower bound, min over W1 through Wn minus K, max x not equal to 0, x perpendicular to W1 through Wn minus K, x emission Ax over x emission x plus lambda 1 of B. And this part now doesn't depend on x anymore, but this part by Kurov-Fischer theorem, this is directly lambda K of A. And so this is exactly equal to lambda K of A plus lambda 1 of B. And in a similar way, instead of replacing this by lambda 1, I could replace it by lambda n. And then I get an upper bound. And then I substitute lambda n here. And this is still equal to lambda K of A. So I get the upper bound at lambda K of A plus B is equal to the min over something of the max over something, it's all these things that I'm not writing again and again x emission Ax over x emission x plus x emission Bx over x emission x. And this is less than or equal to this min over all these things of the max over all these things x emission Ax over x emission x plus lambda n of B. And this is just lambda K. That proves the result. Okay, so some small thing that you can think about is when would equality be obtained in the bounds of the Weyl theorem? X permission Bx by X permission X equal to lambda n B and lambda 1 B. So X is a variable of optimization here. Okay, so all lambdas are equal. Now you should think about it. So for example, suppose B was equal to some alpha times. Okay, I'll write it as Ui, Ui Hermitian. Okay, where Ui is an eigenvector of A. Okay, so suppose it was like this. Then, yeah, so by choosing B of this form and alpha being some positive number, you can actually attain these bounds. I don't want to give you the whole answer here, but do think about it. But this is the form of the B that will attain these bounds.