 We looked at the Rayleigh-Ritz theorem, which says that if you have a Hermitian symmetric matrix A, then if you order the eigenvalues, so the Hermitian symmetric matrix has all real valued eigenvalues. So you can order them in increasing order and so lambda 1 is the smallest eigenvalue and lambda n is the biggest eigenvalue. Then lambda 1 times X Hermitian X is a lower bound on X Hermitian AX and lambda n times X Hermitian X is an upper bound on X Hermitian AX for any X in c to the n. Furthermore, both the lower bound and upper bound are achievable. That is, there exists an X for which this lower bound is achieved and a different X for which the lower bound is achieved and that is given by this next two statements here, namely that lambda max which is equal to lambda n is the maximum value of X Hermitian AX over X Hermitian X for X not equal to 0, which can also be written in this way. Similarly, lambda 1 is the minimum of X Hermitian AX over X Hermitian X for all X not equal to 0. So this is a maximization problem and this is a minimization problem and the smallest and largest eigenvalues can be written as solutions to maximize the here a minimization problem and here a maximization problem. So that is one thing we saw and then we also saw that by considering the fact that these eigenvalues are in increasing order if we were to maximize X Hermitian AX over X Hermitian X subject to X being perpendicular to u 1 the eigenvector corresponding to the smallest eigenvalue then in fact suppose you were to minimize X Hermitian AX over X Hermitian X subject to X being nonzero and perpendicular to u 1 then the solution to that optimization problem is actually lambda 2 the next largest eigenvalue and extending this argument we have that the minimum over all nonzero X such subject to the constraint that X should be perpendicular to the first k minus 1 eigenvectors that is the k minus 1 eigenvectors corresponding to the smallest k minus 1 eigenvalues of the matrix A of this objective function X Hermitian AX over X Hermitian X that is equal to lambda k for k equal to 2, 3 up to n and similarly coming down from the largest eigenvalue eigenvalue if you solve a maximization problem subject to X being perpendicular to the first u n u n minus 1 up to u n minus k plus 1 that is the first that is k k minus 1 k minus 1 eigenvectors corresponding to the top k minus 1 eigenvectors eigenvalues of the matrix A of the same objective function X Hermitian AX over X Hermitian X that is going to give you lambda n minus k for k equal to 1 2 up to n minus 1 so this is this is one kind of crucial result we saw another very crucial result which we will actually use many times going forward is the Kura Fisher theorem and what this says this is a min max theorem and what this theorem says is that again this sets starting point or the setting is that A is a matrix of size n cross n and is a Hermitian symmetric matrix with eigenvalues lambda 1 less than or equal to lambda 2 less than or equal to up to lambda n and let k be an integer 1 less than or equal to k less than or equal to n then there are two ways of writing lambda k the first is a min max formulation the minimum over a set of vectors w 1 up to w n minus k c to the n the maximum of x not equal to zero x perpendicular to w 1 through w n minus k of X Hermitian AX over X Hermitian X is equal to lambda k and the next formulation is a max min formulation the maximum of w 1 through w k minus 1 of the minimum x not equal to zero x perpendicular to w 1 through w k minus 1 X Hermitian AX cross function is the same is also equal to lambda k okay and we saw the proof of this result and the proof is again like I said the last time what I considered to be a somewhat clever proof there is a non-trivial step involved where we we said that setting y 1 to y k minus 1 equal to zero where y i is equal is defined to be what u Hermitian times u i Hermitian times x by setting some of these variables equal to zero you only decrease the cost function and so that's how it you obtain a lower bound on you obtain that lambda k is a lower bound on on this part here and then you so you show that this lower bound is indeed achievable and therefore they are equal okay so this I mean I am as I mentioned the last time please go over the proof of this theorem very carefully and make sure you understand it thoroughly because this is the these the ideas in that proof is the ones that are the ones that we are going to use quite frequently in the upcoming results