 of discuss the methods for solution method different methods of the first method is iterative method. So, you have a given the what is called algebraic Riccati equation that algebraic Riccati equation will arise in case of infinite time regulator problem agree. So, this is the algebraic Riccati equation. So, this term and this term we club together again p into a this one p into b r inverse b transpose p there is a another one additional term is given added with a negative sign. So, with positive sign we have used here p b r inverse b transpose p. So, this and this cancelled. So, this will remain same as the previous equation. So, this we have written it like this way now we introduce some p we have introduced p of suffix k means value of the p matrix at kth iteration and outside this bracket is this one p k plus 1 value of p at k plus 1th iteration knowing the value of p at kth iteration we can compute the value of p at k plus 1th iteration. So, we have written into like this way. So, in order to solve this equation it is a something like Lyapunov type equation that we come across in the solution of stability analysis of a autonomous system means suppose you have x dot is equal to a x and we want to study the stability of the system based on Lyapunov function. Then we consider a energy function b in quadratic form x transpose p x then we this energy function is a positive definite function positive semi definite function. And we have to show for the system a to be stable provided b dot means rate of change of energy should decrease with time that we have to show it and that condition is that one a transpose p plus p a is equal to minus q. So, this is called a Lyapunov type equation while we will study the stability of a autonomous systems. How to solve this one that you assign what is called q is a positive definite matrix then solve a transpose p a for p you solve it if p is a positive definite matrix it indicates a matrix is a stable one based on the derivative of the Lyapunov function we arrive this condition. So, here also we initial value let us call k is equal to 0. So, p of 0 you choose in such a way so that a minus b are inverse b transpose p it is nothing but a b k if you recollect k is r inverse b transpose p. So, it is b k and you have I told you you have to select p of 0 k is equal to 0 in such a way this matrix is stable. So, it is a transpose a c transpose or whole matrix I can write a k is nothing but a closed loop system matrix at k th iteration. So, it is a k transpose p k plus p k plus 1 a k this is a k is equal to q p k p b r inverse b transpose then p k. So, I know p of 0 I have assigned initially in such a way so that a minus b r inverse b transpose p k is 0 sorry stable. So, since this is a positive definite matrix q is a positive semi definite matrix result will be positive definite matrix and if the result is positive definite matrix we know that a minus b r inverse b transpose p k is the stable one. So, if you solve this one then we will get p k plus 1 k plus 1 th iteration the p value will be positive definite and next is once I know from the knowledge of p 0 if I can compute p 1 then use the p 1 in this expression a minus b r inverse b transpose p p is p 1 now you find out p of 2 by solving the Liapunap equation and in this process you go on solving recursively until and unless the p converges to a some constant value. So, next is remarks we can just write it the initial choice of p if you recollect that what I told you the initial choice of p 0 should be such that a of 0 is equal to a minus b r inverse b transpose p of 0 is stable. So, any iteration any k th iteration I can write a of k is 0 th iteration a 0 k th iteration a k is equal to a minus b r inverse b transpose p k and k is equal to 1 2 3 dot dot any iteration r. So, it is stable initial choice is such that it is stable subsequently you can write it that subsequently it can be shown it can be shown that all matrices this for k is equal to 1 2 3 a k are stable and next observation we can see it rather p k plus 1 is less than equal to p k that means this p k minus in other words you can say p k plus minus p of k this matrix is less than equal to 0 means this is a negative definite matrix. That means in other words I can say the p k plus 1 is less positive definite than the p k then I can write third one is limit k tends to infinity the iteration if you go on increasing in the like this way recursively if you solve it the Lyapunov equation then you will see p of k will equal to p which is a constant it converts to a constant matrix. So, only things is when you will start the iteration you have to select p of 0 in such a way a of 0 is stable and then start recursively solving this one until unless that p k plus 1 and p minus p of k is less than equal to norm of this one is less than equal to some p specific values a small values of there or when p converts to a some constant value. So, this is the this method in this way we can solve the algebraic equation that is called iterative method. But one disadvantage of this method is that if your initial guess is not appropriate then it will take longer time to converge the solution of the equation p to a constant value this is the one of the disadvantage of this method. So, next method is is a eigenvalue eigenvector method that means based on the concept of eigenvalue and eigenvector one can find out the solution of solution of algebraic equations. So, method 2 method 2 is based on the eigenvalue and eigenvector method that means solution of the algebraic equation based on eigenvalue eigenvector method of the Hamiltonian matrix. So, first step step one so first you all the when you are designing a l q r problem first check first check it is there you have to see whether the system is controllable or not. So, the controllability test we have already mention it here. So, just mention first check the controllability of the of the system agree that means if you rank of b a b dot dot a n minus 1 b must be equal to n n is the number of states order of the systems and b is the input matrix a is the system matrix the rank of this matrix must be n if it is n then we will call system is completely controllable means all the states we will be able to from initial state we will we will be able to drive the state a desired state with a finite control actions agree in a finite time. So, this is the thing if this fails then this program will show the error message that means we will not be able to design the l q r problems if the system is not controllable return with an error message the error this message will be there that will not be able to design a l q r controller for this systems step 2. So, from the Hamiltonian matrix matrix and to form the Hamiltonian matrix if you recollect we need the information a matrix b matrix q matrix and r matrix. So, a minus b r inverse b transpose then q minus a transpose and this is the Hamiltonian matrix and this Hamiltonian with how we have formed it if you recollect that x dot of t and lambda dot of t this is the cost state vector it is the base state vector this equal to h m into x of t lambda of t. So, this is the augmented state vector with state vector and the cost state vector and each dimension this dimension is 2 n cross 1 this dimension is 2 n cross 1 2 n cross 1 if you see this one. Now this we form the Hamiltonian matrix this Hamiltonian matrix dimension 2 n cross 2 n so this is the m. So, thus this matrix have a how many Eigen values are there there are 2 n Eigen values are there we have shown it analytically that this matrix having a set of Eigen values that mean n Eigen values are in the left top of this plane and or you can say n Eigen values are with negative real parts and remaining n Eigen values are the write up with a positive sign with a positive sign again values with a real positive sign. That means n Eigen values are stable Eigen values and n Eigen values are unstable Eigen values and they are symmetric about the imaginary axis. So, this are the Hamiltonian matrix and all its Eigen values let us call we have define the Eigen values are mu 1 mu 2 mu 3 dot dot mu n and another set of Eigen values will be minus mu 1 minus mu 2 dot dot mu n minus this one. We told we have seen it that Hamiltonian matrix there are 2 sets of Eigen values they are symmetric about the imaginary axis of the complex plane. So, that is let us call this set of Eigen values are the stable Eigen values of H m Hamiltonian matrix and this are the since this is if this is the stable since it is a preceded with minus that will be unstable. Because this stable means the Eigen values the real term with a negative sign if it is a stable one and negative sign and there is a preceded with negative means positive sign. So, the Eigen values real term real these Eigen values are with a positive sign that means unstable Eigen values of H m Hamiltonian matrix. So, this positive real parts this is the negative real parts positive real parts of Eigen values real parts of Eigen values negative real parts one then this will be positive real parts of Eigen values of n this is the Eigen values of that. So, this we have just defined this one then if you that will just see this one that Eigen values are symmetric about the imaginary axis of the symmetry about the imaginary axis to show this one we have already discussed this one earlier at this stage we just briefly discussed how we are making comments like this way there are Eigen values of H m Hamiltonian matrix are symmetric about the real X imaginary axis. So, do the similarity transmission similarity T transmissions. So, let us call X of T lambda of T this costate vector argument vector of this one we have used the transformation like this way. So, let us call this is the equation number we have just this equation let us call equation number one again this is the equation number one from equation one or you can say equation one can be written equation one can be rewritten as rewritten as that X dot. So, it is a I is a constant matrix P I then your X bar dot of T agree then lambda bar of T lambda bar X dot lambda dot is equal to this into X dot X bar dot lambda bar dot this equal to we are writing we have a Hamiltonian matrix H m then we are writing for this one I then 0 P I this into H bar of T lambda bar of T agree. So, this we can write it for this one that what is H m we know the H m expression then we can write it left hand side lambda bar of T is equal to lambda X bar of T augmented with a lambda bar of T this equal to I 0 P I whole inverse then our Hamiltonian matrix. If you write the expression for Hamiltonian matrix this is nothing but a B R inverse B transpose that is minus Q minus A transpose multiplied by then your I just I have simplified that one X bar top lambda bar of T. So, if you use the matrix inversion lambda of this one as we discussed earlier and if you do it after simplification we will get the final results is like this way A minus B R inverse B transpose P and this is nothing but our K if you see this equal to minus B R inverse B transpose that equal to that side is equal to A transpose P plus P A minus P B R inverse B transpose P plus Q and that will get minus of that matrix A minus B R inverse B transpose P this equal to X bar T lambda bar of T. So, if you just multiply from here to here you will get this matrix. So, now see this one this is the algebraic equation this quantity is equal to 0 for this one and this one is 0. So, it is nothing but a upper triangular structure and this is nothing but a closed loop system matrix and the closed loop system matrix is stable it is enough if you consider that Q is a positive semi definite and R is a positive definite matrix then this is a stable one if it is a stable one and this is preceded with a minus sign. So, this will be a unstable this matrix Eigen values are unstable. So, this matrix whose dimension is n cross n this dimension n rows this is n columns similarly this is n columns and this is the n rows. So, n Eigen values of this stable and this block Eigen values are unstable. So, whole system Eigen values after transformation we know the Eigen values does not change it only Eigen vector with similarly transformation Eigen vector will change. So, since it is a block diagonal matrix the Eigen values of the whole matrix is nothing but Eigen values of this block which is the stable Eigen values and Eigen values of this block which is the unstable Eigen values because of that minus sign is preceded with that matrix. So, that we have shown it earlier also here also now see that this is the second step if you just see the first step is you will check it the controllability test if the system is controllable then go the second step in second step you just form Hamiltonian matrix and other things what we have discussed just to show the Hamiltonian matrix Eigen values there are 2 n Eigen values are there n Eigen values are lying to the left of the screen in the sense the l n Eigen values with a negative real parts and remaining n Eigen values are positive real parts they are symmetric about the imaginary axis. So, once you find out this h m matrix then find out the step 3 step 3 the find out the Eigen vectors of the Hamiltonian matrix find the Eigen vectors of the Hamiltonian matrix corresponding to the stable Eigen values I repeat I once again find out the Eigen values of the Hamiltonian matrix whose dimension is 2 n cross 2 n you find out the Eigen vector corresponding to the stable Eigen values of h m that is our next step. So, step f let that v i u i whose dimension is v i dimension is n row 1 column this is the 1 column this is the n row that means this dimension twice n into 1 let atho is the Eigen vector or generalized Eigen vector of the matrix h m corresponding to negative real parts corresponding to in other words corresponding to mu i and what is mu i the stable Eigen values what is stable Eigen values the Eigen values with negative real parts Eigen values with negative real parts means this corresponds to stable Eigen values Eigen value of h m that is. So, you find out the Eigen vector of that 1 corresponding to mu i. So, if you find out this mu 1 and you can find out we know the hour for i is equal to 1 2 dot dot n you see this one i i how many stable Eigen values are there we have a n stable Eigen values are there n unstable Eigen values are there we are we are concentrated or we are we are attention to find out the Eigen vector corresponding to n stable Eigen values of h m. So, for i is equal to 1 2 n. So, the solution of the proof we will give it later the solution of algebraic Riccati equation then one can find out that means this solution A transpose P A minus P B R inverse B transpose P plus Q is equal to 0 the solution of this one is obtained as P is equal to that you form a U 1 U 2 dot dot U n then this matrix B 1 B 2 dot dot V n whole inverse. Now, see this one which one that now see this one here this is the Eigen vector corresponding to Eigen value mu i which is stable Eigen values and when you are getting the first n rows of this column vector you partition as a V i remaining n per n rows of this vector you partition as a V i. So, since we have a what is called n stable Eigen values the first partition n first n rows I am writing V 1 V 2 V 3 dot dot V n and remaining last n rows of the Eigen vector corresponding to stable Eigen values I am writing U 1 U 2 U 3 dot dot U n. So, the solution of the Riccati equation of this one. So, this is the stable Eigen vectors of h m corresponding to U 1 corresponding to mu 1 again last n rows of this Eigen vectors is U 1 the last n rows of the Eigen vector corresponding to mu is equal to mu 2 is the U 2 and so on and this is the first n rows of the Eigen vector corresponding to stable Eigen values mu 1 this is stable Eigen values mu 2 first n rows of the Eigen vectors Eigen vectors in this way. So, if you compute this that will give you the solution of the Riccati equation. So, this dimension if you see how many rows are there n rows this is column 1 column 1 in this way we have a n column 1. So, this dimension is if you see n cross n this dimension is n cross n and this inversion is exist because Eigen vectors of that one comprises that one V 1 V 1 all these things. So, let us see the proof of this one. So, these are the Eigen vectors what you can write it for this one if you see that one this what is called this is the closed loop system Eigen values for closed loop system matrix A minus B A minus B R inverse B transpose P is the closed loop system matrix. So, this way you can solve it this one. So, let us prove that how we are getting this expression. So, proof so algebraic Riccati equation we rewrite here algebraic Riccati equation A transpose P plus P A minus P B R inverse B transpose P plus Q is equal to 0. Let us call this is equation number 1 this equation we are writing into different form look mu i is the Eigen values of the matrix Hamiltonian matrix corresponding to a stable Eigen values mu i this is I have added with a A transpose multiplied by P A transpose of this is more one term is added that is subtracted like this way mu i I agree then A minus plus B R inverse B transpose P plus Q let us call this is equation number 2 now see mu i mu mu i P minus P mu i agree that if you multiply by this one mu is a scalar quantity agree. So, this and this cancels A transpose P minus and that is the minus plus P A that minus this plus minus P B P B R inverse B transpose P plus Q. So, it is same as equation number 1. So, now next is the Eigen values let us call now if the Eigen values of A minus B R inverse B transpose P R mu 1 mu 2 dot mu n and that corresponds to stable Eigen values. If the Eigen values of this matrix are mu 1 mu 2 dot this agree stable Eigen values then we can write by definition of Eigen value and Eigen vector the definition A B R inverse B transpose P this into that vector that vector is V i let us call corresponding to this vector this is the Eigen values corresponding to this vectors are V i V i i is equal to 1 to n then V 1 V 2 is the Eigen vectors corresponding to this matrix. So, this plus is equal to that mu i into V i this is the definition of Eigen values Eigen vector this is the matrix and closed loop system it is that Eigen values are mu 1 mu 2 which are the stable Eigen values we are considered. So, by definition of Eigen values Eigen vector I can write it closed loop system matrix into corresponding Eigen vector if it is V i must be equal to mu i into V i. So, this I can write it after bring it in this side B R inverse B transpose P this minus you bring it mu i i whole V i is equal to 0 push it P V i in the inside. So, it will be V i minus P B R inverse B transpose P V i again minus mu i mu i V i. So, if I take it right inside it will be mu i V i. So, it is nothing but A V i minus P B R inverse B transpose that is called this matrix this will be a matrix multiplied by a vector. So, it will be a vector which I am denoted by U i is equal to mu i V i. So, let us call this equation is equation number 3 and this is valid for n number of equations. So, this is the one equation 3 get it. Now, if you multiply by this equation 2 by V i post multiplied by V i both sides then what will get it this equal to that is this equal to 0 null matrix from equation 1 you see we have adjusted this equal to right inside null matrix. So, here I missed it is a null matrix here. So, I multiplied by equation 2 equation 2 multiplied by V i then what we get it let us say mu i i plus A transpose P V i minus P that is V i C equation 2 V i that means I then minus A plus V R inverse B transpose P multiplied by V i plus Q V i is equal to null matrix. Now, look at this expression this expression you see this expression what you can write it A V i if you multiply by A V i you see this one A V i V R inverse V transpose U i minus mu i V i is equal to 0 if you take this one this side it will be 0. So, if you see is here then A V i R inverse B transpose P V i then mu i but it is a minus with this one. So, this quantity is 0 this quantity is 0. So, I can write it now mu i i plus A transpose P V i then I am taking the minus sign common it will pass plus A minus B R inverse B transpose P V i this is the minus sign if I take it common this is the plus. So, it will be minus mu i V i plus Q V i is equal to 0. So, this quantity from equation 0 from 3. So, you got it this one this is nothing but a U i. So, you got it mu i plus A transpose U i plus Q V i is equal to 0 and I is equal to 1 2 dot dot n. So, let us call this is equation number 4 now you just write from equation 3 and 4 from 3 and this from equation 4 we can write from 3 and 4 note V i is the what if you recollect V i is the eigenvector of the matrix closed loop system matrix corresponding to a eigenvalue mu i corresponding to eigenvalue mu i this is the eigenvector and mu i is the what is mu i mu i is nothing but a the stable eigenvalues stable eigenvalues of Hamiltonian matrix that is. So, from 3 and 4 we can write it A minus B R inverse B transpose minus Q minus A transpose this is the Hamiltonian matrix and this is equal to V i U i then is equal to mu i and V i U i and that is for i is equal to 1 2 dot dot n. So, let us call equation number 5 and these are the what this is if you see this is the eigenvector this dimension is what if you see this dimension is this matrix dimension 2 cos 2 n and this dimension is 1 cross twice n this is the twice n. So, 1 column twice n. So, these are the eigenvectors corresponding to Hamiltonian matrix again this is the that what I told you earlier also you find out the eigenvectors of the Hamiltonian matrix corresponding to the stable eigenvalues of H m means Hamiltonian matrix and then you can find out the what is called solution of algebraic equation this and this is equal and write it eigenvector eigenvector eigenvector of H m corresponding to by definition of eigenvalues eigenvector you see A x is equal to lambda into x corresponding to mu i which is a stable eigenvalue eigenvalue of H m this is or you can say stable eigenvalues of A minus B R inverse B transpose P this is the stable eigenvalues of this is the stable eigenvalues of this matrix also agree this matrix also so now this I can write it now from so we know we know or we have defined you see P v is equal to u i and i is equal to 1 2 dot dot n. So, I can write it P v 1 v 2 dot dot v n is equal to u 1 u 2 dot dot u n augmenting the i is equal to 1 2 n this matrices are these vectors are augmented like this way. So, this is nothing but a this dimension is n cross n because this this is the row is n rows this is the column 1 this is the column 1. So, we have this this is also n cross n. So, therefore, P is equal to u 1 u 2 dot dot u n divided by that was inverse of post multiplied by this one v 1 v 2 dot dot v n whole inverse mind it that v 1 v 2 v 2 is the eigenvectors of the matrix eigenvectors of the matrix A B R inverse v transpose P closed loop system eigenvalues eigenvectors of this one. This are the eigenvector eigenvector of this matrix eigenvector of this matrix corresponding to the eigenvalues mu i corresponding to eigenvalue mu i. Since these vectors are it can be proved that v 1 v 2 v 3 dot dot v n is the set of eigenvectors corresponding to the set of eigenvalues mu 1 mu 2 dot dot mu n that is the n eigenvalues of that they are linearly independent. So, the inversion exist for this one agree. So, this way you can one can get the solution of this one. So, if you summarize this one then what is the step to solve the algebraic equation first you check first step is you check whether the system is controllable or not. If the system is controllable then proceed second step. If it is not controllable send a message that this we cannot design a controller based on LQR. That means solution of lickety equation does not exist or we are not be able to stabilize the systems. Second step is here that you form a Hamiltonian matrix and once you form a Hamiltonian matrix then you find out you can immediately find out because you know a Hamiltonian matrix is known to you completely because you know a you know b you know q and you know r. So, Hamiltonian matrix is known to you completely then you find out the eigenvalues of this matrix and consider only these stable eigenvalues of the Hamiltonian matrix and find out the corresponding eigenvector of the Hamiltonian matrix corresponding to the stable eigenvalues agree and once you got the eigenvectors of the Hamiltonian matrix corresponding stable eigenvalues mu1 mu2 dot dot mu n. then first n rows of this vector you picked up that is denoted by v i and remaining n rows or last n rows you denoted by u i then you form a matrix like this way take the inverse then this will be the solution of the recalculation that is the proof of that one but this this is our method what we have considered this is the eigenvalue and eigenvector method this method is numerically is not much sound in the sense when you are finding out the eigenvalues eigenvalues or eigenvectors and when the eigenvalues are the repeated eigenvalues then we have to look for what is called generalized eigenvectors and competition of generalized eigenvectors may create what is called numerically some problems can be not it is not reliable much to when we will compute the eigenvectors by this method ordinary method so the next method is numerically stable methods method 3 is a numerically more stable method while we will computing the eigenvectors of the Hamiltonian matrix corresponding to stable eigenvalues of h m in semi-numerically matrix so this method is called sure sure method to solve a re algebraic equation and the sure method is based on q r decomposition method decomposition method and that is numerically stable method is numerically stable or more reliable when you will compute the eigenvalues of eigenvectors of this matrix when the eigenvalues having a repeated eigenvalues are there so one can see the reference of this one good paper you can see a sure method for solving algebraic equation that is appeared in IEEE transaction AC volume 24 volume 24 number 6 June issue and 1971 79 even though it is a very old paper but it is a basic things is given here only and page 119 292 101 author is l n j lam so we will just briefly discuss the solution of this Riccati equation algebraic equation based on what is called the sure complement my sure method so step algorithmic steps so step 1 but in this method what will do it will just find out the eigenvectors of the matrix h m corresponding to stable eigenvalues me 1 me 2 me 1 that is the method only once you know the eigenvectors of this one and we know how to find out the solution of the Riccati equation p that means first n rows we picked up and will form v 1 v 2 v 3 v n and this is nothing but a the eigenvalue eigenvector corresponding to closed loop system eigenvalues and the remaining n eigenvalues that of this Hamiltonian matrix eigenvectors that eigenvectors such that 2 n cross 1 eigen dimension the first n rows is denoted by v 1 and the remaining n rows that means last n rows is denoted by u 1 so from that information we can find out the solution of p and in this method sure method is giving the competition of eigenvector corresponding to eigenvectors of h m corresponding to stable eigenvalues me 1 me 2 me 1 so first step is form Hamiltonian matrix h m is equal to a b r inverse b transpose minus q minus a transpose this this is the equation number 1 whose dimension is n cross twice n into twice n this step 2 with the help of with the help of q r algorithm the matrix h m is transform into upper sure form first strike structure is the s is equal to w transpose h m of w and w is the orthogonal matrix orthogonal matrix and orthogonal matrix means if you if we inverse of this matrix is nothing but a transpose of that same matrix this is orthogonal matrix and orthogonal w w transpose is equal to identity matrix that the properties of orthogonal matrix so this is the orthogonal matrix and the structure of a if you look at this expression this dimension is h n dimension twice n cross twice n and this orthogonal matrix also dimension is twice n into twice n this because this dimension is twice n into twice n agree so the structure of s the structure of s you will get it s 1 1 s 1 2 dot dot s 1 k s 2 2 s 2 3 dot dot s 2 k and in this way if you just go on s k k so this whole matrix dimension twice cross twice n just like here this dimension of this one is also twice n into twice n this orthogonal matrix now what will do it you do the sequence of orthogonal transformation on h m such that the structure of s matrix when you are premultivate by w transpose and w the structure of p matrix will be this form upper triangular structure this will be a upper triangular structure agree this form and look at this expression each block either s 1 1 s 2 2 s 2 3 3 4 4 s k k 4 k k this block will be either 1 by 1 block or 2 by 2 block this will be s i i where s i i i is equal to 1 2 dot dot k is either 1 cross 1 mean scalar or 2 cross 2 matrix again this will be the this matrix this matrix so this is the a sequence of orthogonal transformation we can make it on a h m matrix such that this matrix will be converted into this form either 1 by 1 block or 2 by 4 if the 2 by 2 block is coming in there then this Eigen values of that one s you see Eigen values one can conclude the Eigen values of a is same as the Eigen values of h m because you are doing the some similarity transformation w transpose is nothing but a w inverse of this one you can think of it like this way so Eigen values are remain same so this Eigen values are either it will be 1 1 this book or it will be 2 2 block that means this Eigen values will be either complex or real this if it is a 2 by 2 blocks are there so now see how it is done it so this is just to get some idea that what we do it in this equation that suppose given given a our problem is convert into a decompose a is equal to a into q r again so let us call the matrix a is decompose into q into r q is a orthogonal matrix an r is the upper triangular matrix upper triangular matrix again now this is a is equal to this this is the a matrix can be decompose into this q into r form this can be done by using a sequence of orthogonal transformation or in other words one can do a sequence of householder transformation which will decompose a matrix into this q into r let us call we know how we assume that this technique we know how to decompose a matrix q into r so if it is so then how will you get in sure form into this form a is equal to I am writing is the given matrix is written assigned with a 0 which is equal to our q 0 r 0 now I have divided a 1 if you reverse the order of this one because both are a dimension if it is n cross n this each each is n cross n this is also n cross n so if you reverse this order the dimension this matrix will be different from a 0 or a so this matrix I am now considering is a 1 is a r 0 q 0 so what is r 0 if you from here you take the inverse of both side so it will be a q 0 inverse means transpose because q is orthogonal matrix q 0 is orthogonal matrix so this will be this one into a 0 into q 0 so what is r 1 r 1 is nothing but a r 0 q 0 and whatever the decomposition we do rate q 0 r 0 reverse order you write it r 0 q 0 is a 1 similarly I can write it now a 2 a 2 that that a 1 is just like if you decompose this matrix decompose into a q 1 r 1 I decompose in write the reverse order which is denoted by a 2 is equal to r 1 q 1 so what is our r 1 r 1 is nothing but a q 1 transpose a 1 r 1 is nothing but a q 1 transpose a 1 into you see r 1 is nothing but a q 1 from here r 1 below I am writing r 1 below I am writing q transpose a 1 multiplied by q 1 again so what is a 1 just now you got it a 1 is this one so it is a q 1 transpose q 0 transpose a 0 q 0 q 0 q 0 q 1 again so if you write it like this way then I can write it is nothing but a 2 is equal to q 1 transpose q 0 transpose a 0 q 0 q 1 this so which is I can write it if you see q 0 q 1 whole transpose a 0 q 0 q 1 so next is what I will write it I will write it a 2 a 3 is equal to because this I can this matrix a 2 matrix I can decompose a 2 is what I am written into r 1 into q 1 so I can decompose this a 2 is equal to q 2 r 2 agree q 2 r 2 I can I decompose this matrix and how I will write it this simply in this way that q 2 is this and what you can write it for this one after simplification in this way then I am writing a 3 a 3 is q 2 a 3 what will can write it a 3 similarly I can write finally q 0 q 1 q 2 whole transpose a 0 q 0 q 1 q 2 and ultimately if you see if you do the kth iteration a k what you can write it a k that q 0 q 1 q 2 dot dot q minus 1 whole transpose a 0 q 0 q 1 q k minus 1 in this agree so ultimately it is nothing but a q transpose a 0 and q the 2 orthogonal matrix multiplication results is orthogonal matrix agree so it is something like this way if you do this sequence like this way ultimately I will get a is in the sure form into sure form means into this form just like into this form into this structure this is all r 0 that means either a 1 s 1 1 may be 2 by 2 block or 1 by 1 block this may be 1 by 1 block or 2 by 2 block but it cannot be more than 2 by 2 block it is may 1 by 1 block or 2 by 2 block so ultimately if you proceed like this way you will get it a k into sure form so I just leave it this is an exercise please do it a 2 this one agree this you just proceed that one and next class I will just complete the algorithm how to solve that algebraic equation using the sure complement is sure form and sure form is nothing but a numerically you are finding out the eigenvectors of a of Hamiltonian matrix corresponding to a stable eigenvalues that is all only agree so we will stop it here