Diagonalisation of matrix

and welcome to Galatians 2 channel today we are going to discuss about dynamization of matrix so what is it about in our previous videos related to eigenvalues and afters we have studied that how we can form a characteristic equation then how can we find out and the values and then factors from them right proceeding with that diagonalization of a matrix means you are forming Omnitrix which has only diagonal elements okay now you are extracting it or you are forming it from parent matrix okay so how do you form it let me tell you how this is an example but before that me let me give you the generalization form that um what are the formulas what does it mean Daniel ization let's say you have a particular matrix a the parent matrix like this now if you want to find out diamond ization or you want to form or reduce this matrix into diagonalization or diamonds form what you can do find out let's say it has three eigenvalues lambda 1 lambda 2 lambda 3 okay similarly if you form the characteristic equation find out I can they lose so you will have I do not possess it let's say eigenvectors are X 1 X 2 X 3 in the form x1 y1 z1 x2 y2 z2 and x3 y3 z3 okay these are your eigen vectors then your diagonal matrix will be formed like first of all you have to form a model matrix model matrix will be like if you want let's say P be the model metrics which will be in the form of like this let's it P is the model matrix and you have to do it like this x1 x2 extreme means you have to take these scholars and make a matrix out of them okay when you will have model matrix then you have to find out diagonal matrix help if you multiply a with P so it will be a a with these three X 1 X 2 X 3 if you take it if you brought inside so these will be your three values this is this will be your whole matrix and how you can find evaluated a will be your lambda 1 to lambda 1 X 1 is this X 1 lambda 2 X lambda 1 X 2 lambda 1 X 2 okay if it is X 1 X 2 X 3 if it is X 4 minus a 1 so you can write down here x 1 y 1 said that was okay so a X 2 will be lambda 2 X 2 is this okay and your lambda 2 will be X 3 lambda 3 X 3 lambda 3 y 3 lambda 3 said okay when you split it in the form x1 y1 z1 x2 y2 z2 x3 y3 z3 ok this is this is a matrix form by your eigenvectors and what is left lambda 1 lambda 2 lambda 3 in diagonal form this will be your diagonal matrix this will be your title metrics to make a diagonal matrix means all the eigenvalues will be in the form of turbulence will be arranged in diagonal form ok so how can we calculate it with the help of example let me tell you so this is a how can you solve it your characteristic equation first so which is you know it you have seen either we use a minus lambda equal 0 right so any is because I identity matrix has only diamond elements as one so if you will do like minus 1 minus lambda 2 minus lambda 0 minus leverage rest of the elements will be C ok now solve it how can you find out it it will be a determinant form we know how to solve it minus 1 lambda hi these the row and column in which H is existing so this is minus 2 lambda cross multiply minus 2 lambda minus minus plus lambda square minus is in formula minus - that's 1 ok then - to hide these two minus lambda minus 1 is plus 1 is too high these minus 1 minus minus PR 3 minus 1 ok now solve it you can multiply it minus 1 into these minus minus plus 2 lambda minus lambda square minus 1 right then plus 2 lambda square minus lambda cube minus lambda minus minus plus 2 lambda plus minus 2 this is 2 minus 2 minus 1 is 1 and minus 2 minus minus 2 plus 2 lambda right now arrange it in decreasing order of their powers so what it will be - thunder cube then this is square squared so 2 minus 1 is 1 lambda square then lambda is 2 minus 1 is 1 3 2 5 lambda and the constant terms minus 2 minus 1 minus 3 minus 5 right now I'm how to find out the roots with the help of hidden trial method so this is the constant term if you split it it can be its factor will be root 5 minus root 5 and 1 so put any one of 10 let's say 1 minus 1 cube is minus 1 plus 1 cancel 1 5 - 5 cancels so lambda is equal to 1 is a factor and if you put lambda is equal to 1 that means lambda minus 1 and divide it you know how to find out the factors right I'm just explaining for those who do know about it so lambda how you can do it minus lambda square minus - plus lambda square this is cancelled 5 lambda minus 5 multiplied with 5/5 lambda minus 5 now this is another would let you go so now if you split it it will be lambda minus 1 lambda minus lambda square plus 5 equals 0 right so lambda is 1 and 5 can be splitted into this right so your three eigenvalues are lambda 1 lambda 2 lambda is 1 root 5 - food fight respectively now coming to makes might make service to find out either map that's right how can we find out either backtest michael simultaneously putting these values one by one so if you put this value of month here minus 1 minus 1 is minus 2 2 minus 1 is 1 minus 1 this is minus 1 and x1 y1 z1 are your first set of eigenvectors now multiplying it - 2 x 1 + 2 y 1 minus 2 Z 1 X 1 plus y 1 plus z1 you know how to multiply it right and then the equation will be minus X 1 minus y 1 minus Z 1 it should take - common it will be as same as these two right so consider for us to first use it what you can do in it you can multiply it with -2 so if you multiply it with minus 2 let's do it - 2 X 1 2 1 minus 2 Z 1 if you multiply it with minus 2 and then subtract for y1 and this is cancelled so y1 is 0 if you put a violin here that will be on here and result in 2 X 1 is equals to minus at 1 so your eigen vectors are in front of X 1 it is minus 1 Y minus y1 is 0 X 1 is minus 1 I hadn't said 1 is plus 1 right so first set of your eigen vector is this let me write down write down here these are your x1 y1 z1 that means set excellent okay now moving on to the next point which is find out finding our second set so if you want second value minus 1 minus root 5 2 minus root 5 minus 1 minus 1 this was lambda right so minus root 5 this is your matrix for lambda 2 and x2 y2 z2 now I'm spending it as well first row first one X 2 plus 2 I 2 minus 2 Z 2 X 2 2 minus root 5 by 2 plus Z 2 this 1 minus X 2 minus y 2 minus 1 finds it right ok now following it you can take any of these let's say these two if you add them we cancelled what is left you can be quite to common what is that 2 minus 1 is 1 minus root 5 correct and here if you think that was common what is that 1 minus root 5 right so you can take 1 minus root 5 common and what is left inside Y 2 plus Z 2 and this will be 0 so Y 2 will be mine is it correct now after moving on you put this value in any of these Miss Leefolt for instance first 1 minus 1 minus root 5 X 2 plus 2 y 2 minus 2 Z 2 if you put Z 2 as minus y 2 minus minus plus 2y right correct laughter what you can do to undo for y2 right and you can do it like minus 4y 2 minus 1 minus root 5 right now you can cancel - with - if you take - common right it will be formed and then you can visualize usually we put the coefficient in front of the variable right as I didn't factor but here just to simplify we are using rationalization we are deviation ization so for y 2 1 plus root 5 so this was like the suit fights off obviously we have to my ply with 1 minus root Phi right now taking this 1 plus root 5 1 minus 2 5 8 as the a minus B is a square minus B Square which is minus 4 whole scans beautiful so X 2 is you can take it above right take - a numerator so yo I do was - Z 2 no x2 this so x2 will be the coefficient in front the root 5 - 1 whiter will be 1 and Z 2 will be minus 1 all order doesn't matter you can do minus 1 e 1 here as well so your roots are I mean your eigen vectors are root 5 minus 1 1 and minus 1 you can write down here this is your X 2 X 2 y 2 Z 2 right similarly you can find out third one for third one the value is minus root 5 right so if you put minus root 5 it will be minus minus plus minus minus plus minus minus plus rest all like 1 minus 1 plus root 5 then when you multiply with this let's do it they Stoke this column minus 1 plus root 5 X - 2 n minus 2 are same then this 1 X 2 2 plus root 5 by 2 Z 2 minus 1 minus X 2 minus y 2 plus root 5 is 8 right now if you solve this again it will be cancelled you can take here this will be 2 minus 1 is 1 plus root 5 right and here 1 plus 1 plus and then Y 2 will be acting as the previous one because with you now coming to this point this is minus 1 plus root 5 let's just say when you put minus 2 Z 2 and that page - my - it will be blessed oh one thing yeah for white right - 1 plus root 5 so X 2 will be oh I do just the thing which is changes this now if you you can write it like root 5 minus 1 right root 5 minus 1 so when you rationalize it will be root 5 plus 1 root 5 plus 1 right so oh my - this is 1 plus root 5 a minus B a plus B 5 minus 1 is plus 4 forest answer is 4 so it is root 5 plus 1 right it will be root 5 plus 1 where's the same correct y 2 is 1 and then 2 is minus 1 all right order doesn't matter I said to you know for diagonalization I mentioned earlier what you need to do you have to write the matrix in the form of firstly your eigenvectors in the form of matrix and then I didn't when use in the form of tiny matrix so your eigen vectors are minus 1 0 1 make sure you are writing down your writing Noctis in the form of column ok then root 5 minus 1 1 minus 1 and then root 5 plus 1 1 minus 1 this is your model matrix ok and your diagonal matrix will be dinars in the form of argument savings and rest diamond matrixes itself rest all elements are 0 we will dive in my elements are there so this is your tiny matrix I hope you understand this topic and if you liked it please do not forget to like share and subscribe my channel it all till then take care and see you soon bye

Channel: Ganesh Institute

Views: 10,656

Rating: 4.8536587 out of 5

Keywords: #education, BSC, Engineering, maths, matrix, diagonalisation of matrix

Id: OPOgjawy20E

Length: 18min 29sec (1109 seconds)

Published: Thu May 09 2019