Diagonalisation of a 3x3 matrix

that's what where's the process to diagonalize a 3x3 matrix well to form a diagonal matrix equivalent to this one that means you have to go through a series of operations of pre and post multiplying by elementary row operations and elementary column operations to end up with this diagonal matrix but it'll only be a diagonal matrix if you can find a set of linearly independent vectors to make up the columns of this matrix those vectors being the eigenvectors so the first step is going to be getting a compile use so you get AG in vectors checking that linearly independent and then you can go through this and find the diagonal matrix which should be the matrix consisting of eigen values so step one get the eigen values so eigen vectors was too much if you do that you can either see well for that matrix here if I H matrix E was to operating sum vector and then just produce a multiple of that vector then that effectively banana backed up and that multiple would be the eigen value and then you could rearrange that if you main aslam the V should equal zero so zero matrix taking that out as emailers of the identity element V I times whoops lambda I times V would be zero and then since you're looking for vectors if the vector can't be zero then for the solution to that you're seeing that the determinant of this matrix must be zero and then you can solve it by doing that subtracting alarm that don't the me diagonal and then getting the determinant equating it to 0 rather than that does not other technique where I can find the I add the characteristic polynomial to get the eigen vector values which was this so look at lambda minus and then it just goes through the various determinants of increasing size of squared where they are determinant of the 1x1 matrices of the two-by-two matrices or busines mid angle and the 3 by 3 matrices that would be the characteristic polynomial of a curve the result well the first one this one here is going to be the trace of e this one here two by two are going to be the three main diagonal my nose that'll be positive remember so that's going to be e what one would call it a 2 2 and a 33 in the - of those particular elements and the last one is going to be simply the determinant of the whole thing that's a big 3 by 3 determinant Mike well won't forgot geez but oh he added up in there so what's the trace of a that's easy enough you just add them up and it comes to 0 over a 1 1 for all of these particular ones here 1 1 plus a 2 2 plus a 3 3 well they're going to be well here 1 1 means the manor of this so it's going to be negative 20 but plus 18 so I've got negative 20 plus 18 the manor of this one is going to be 4 take away 18 and the manor of the bottom one's going to be negative 5 plus 9 taken out on top I've got we can see developers would go away from each other there's a negative 16 negative 16 + 4 makes it a negative 12 the last part of the determinant of the whole matrix E is going to be well I'll just use the top rule for this so I've got 1 times as Manor which was that negative 20 plus 18 minus but i--but plus the 3 then times X may not that's the I mean that's the 12 take away 18 and then plus the three times this manor which is the negativity takeaway so it's going to be plus the third so it's going to be a negative 2 and that's negative 6 so it's minus 18 but there's quite a lot here that's going to be 12 cents plus 36 so it's negative 20 plus 36 which is 16 so the characteristic polynomial would be like the cubed for the eigenvalue minus the trace of lebanon was going to protect an area minus 0 lambda plus the two by twos so that would be minus 12 epsilon the squared lambda minus the last one minus 16 equals 0 so back to this thing so I need two fightin eyes this I'll use of a synthetic division aside there 1 0 negative 12 negative 16 and analysis if I'm going to write negative something one's not going to matter through try a negative 2 1 negative 2 negative 2 4 negative 8 16 zeros there we are so I've got lambda plus 2 times lambda squared minus 2 lambda minus 8 a 0 so it's going to be lambda plus 2 times lambda lambda and that's going to be 2 and 4 will have to be negative 4 plus 30 equals 0 so in fact you've only got the two you've got repeated 1 you've got a repeated eigen value of negative 2 and a separate one for a single one of lambda equals 4 so that's the phosphor get the eigenvalues but eigenvector right now they get vector means that e times eigenvector must be the same as lambda times a vector or e maintenance at lambda i times eigenvector will equal 0 say to formless then e minus lambda times identity that just simply means subtracting lambda from the main diagonal entries so for lesson over time when until the two's first of all was a pair of them unfortunately I'll have to get two vectors if I can from the same number so I forgot I've got one takeaway negative two is three leave them alone leave that alone negative five plus the two negative three these nine one six negative six take that v6 I've got that for this part times v1 said if you think I'll call it XYZ it I know could go X 1 y X 2 X 3 Valek X of Y and say it should equal zero as in zero zero zero state way applying elementary religious that you see they're all identical now we're not down to all the ones operating first line one one negative one one one negative one one just by dividing each the losses appropriate and then the second will take away the top the thought will take away the top and there's complete redundancy the elephant they got this these two rows come to zeros so I'll just got X Y Zed for that equals 0 as in 0 0 0 so all of God then is X minus y plus Z equals 0 there's only creation of God from that whole matrix because I've got a double redundancy z and y can float freely but X is tight about every bridges for them so I'll say it was that somewhere I could go through playing with different numbers that add up to 0 I don't have to get ones that will be little independent in the end oh no put it in this way well Y can be anything that might be here Z its fleet of do every like I said it could be but that means that X or have to equal the minus B which means that this vector is going to be a minus B a B or splitting into two parts e times 1 1 0 plus B times negative 1 0 1 and that would get me again there's the A's and these that can float but there's a pair of factors then that would make up this one so I'm going to use the spare leg so I'm going to see if I'm going to have v1 there's 1 1 0 and V 2 is what I can have that has 1 0 negative 1 if I wanted that go for that one zero negative 1 that just means taken out as Maness the be just to the top so this neat there's a pair of them so far no for know the case of for what's in before a minus lambda I times the eigenvector should be 0 sine several items so I've got this vector I've got this matrix minus 4 off the main diagonals so I'll be negative 3 negative 3 3 3 I'm taking 4 of that so it's negative 9 3 6 negative 6 and 10 for off of that San Ysidro that times V which I'll just call XYZ should give 0 which is 0 0 0 now and since I see 2 0 0 any operations are applied to that of what if X 0 0 0 so I'll just concentrate and messing up over this one so the first thing is obviously I'll just knock them all day 3 into that row 3 into that row 6 into that node negative inappropriate so 1 1 negative 1 1 negative 3 1 1 negative 1 0 which is what V now then get rid of those particular ones XY see it so now it's going to be good at down to 0 so to take away 1 1 0 1 0 negative 3 take away 1 negative 4 1 take connector 2 2 1 take away 1 0 1 take away 1 0 negative one take away one negative two zero take away 1 1 you can see that what's going to have the next one we'll do two steps and one is over 1 1 negative 1 that module is going to go to zeros if I do double the bottom take away the second one without 0 0 0 just to save a lane and they're not backed up up and down by dividing by negative 2 so negative 2 equals to negative 2 into that goes negative 1 which means I look these two equations in X plus y minus ed must be 0 X plus y minus CEATEC must be 0 and 2 y minus a it must be 0 because I've got one redundant row so Z it is free Z it can be anything you like and it will still give you a zero nothing times anything is 0 so Z is fully so what C is let Z do but if you like let's say de let's go with court well if said is it then that says 2y minus C is 0 so that means Y is going to be e divided by 2 y is going to be 1/2 of e so busy to Z Y is 1/2 of it then Mustangs going to be well X plus y which is 1/2 of it my inner city should equal 0 such negative 1/2 putting it over which means X is going to be 1/2 of e they put that together something's my vector which although called theta things I've got two already would be 1/2 e 1/2 e e or taking out what taking 1/2 e that would leave me 1 1 2 so let's thought I can victor could be anything from that same we have you can choose any number for e so the simplest one we make is equal to 2 so that I've got an example for V 3 which would be 1 one two they put those three things mostly vectors back up here and then just check if they are can fight independent of each other hey here's all the information original matrix the eigenvectors eigenvalues or should have emphasized that was a table and then three eigen vectors the simplest ones i could construct from them and it's just a quick tape are they actually linearly independent they'll be letter then dependent if i can't make one of them by using some combination of the other two or if you like have some combination of all of them is equal to 0 where the Arabs in AU 0 would imply that I've just multiplied by e B and C all being 0 so I'd be the same as saying less then 1 1 0 1 0 negative 1 1 1 2 that's just these three the donor has the columns of that matrix times ABC should equal 0 there's a lot system to solve just define what a B and C are hopefully st. a B and C come to 0 and I'm sorry because you've got an E and then the BBC and 8 the DC and then 0 e as e times that it's just have been combined differently well again since is equal to 0 I can just concentrate reducing this part so 1 1 1 so Opera top take away 2 take away 1 0 negative 1 0 and then for this one of subtle you that appears just better and then dropping this one down you see this is a little bit messy that's what happens in haste so 0.1 1 0 negative 1 0 and then 3 take away 2 that's going to be 0 to t 1 to 0 0 and from that I get that sucker to C equals 0 0 0 negative b 0 d r0 a plus 0 plus 0 is 0 which means a is 0 so that our little independent because to form the homogeneous equation the only way I could get zeros an answer would be if a B and C or o0 satisfy I know finally I can get my day I can matrix by pre-multiplying a post multiplying by the metrics composed of those address of PEP where P is equal to just a thing that I don't before one one zero one zero a negative one one one two one problem is now I'm going to have two main tenders of P so I take a little bit of jiggery-pokery or leaders determinants so what's the determinant of T could use this condom this idea so one times as zero and it's just going to be one minus one times that's going to be two plus one is three so I guess me 1 minus 3 is negative 2 then no to get the envious I'm going to have to form the matrix of saint- being very careful where the earth may take note so that one is going to replace by nothing take away a negative one source a positive one not one is going to place by two but it's in a naked position negative two and that one is getting replaced by a negative one this one is going to be replaced by two plus one is three but that's in a negative position that zero is going to replace by 2 plus T of 2 and that one which is in a negative position is going to be replaced by a negative one which will put it into a positive one this 0 is going to be replaced by one the negative one is going to be replaced by a zero was handy and the 2 is going to be replaced by a negative one then tear the envious this is a very case of a transpose the matrix formed by the same minor and divide by the determinant so it's going to be negative 1/2 times that so it's going to be negative 1/2 times flip those round 1 negative 2 negative 1 negative 3 2 1 1 0 negative 1 what amazed well football the same is over so I'm going to have 1 a negative 1 3 negative 1/2 negative 2 0 and 1 negative 1 1 if that's the inverse I can fit all that back into here then all right so it's just a case of copying it down carefully so the inverse P is going to be learned first so coconut endings should all the numbers and signs are correct get rid of other further space here then e copy Eden carefully again and then copy down P here so that book is copied and clearly and then I can all go so starting off then half of leave that form alone I'm going to do the multiplication of these two rays over go I've got one take away 3 is negative 2 over negative 3 is negative 2 1 negative 3 and 6 is 4 we've got 3 take away 5 is negative 2 3 take away 3 is 0 3 take away 5 and 6 is 4 6 degree 6 is 0 6 and nothing and dictaphones to 6 degree 6 and 8 is 8 now these are first one alone you can use that half up to simplify that second matrix up in extra point negative 2 negative 1 0 2 0 1 4 right now final multiplication I know what's expected respecting these eigen value so I've got negative 1 negative 3 0 is negative 2 I've got 1 0 negative 1 is 0 and I've got negative 2 6 and I negative 4 is 0 I've got negative 2 2 0 0 negative 2 0 0 is negative 2 4 tick away 4 0 0 a negative 1 plus a 1 and 0 0 and negative 1 0 1 0 and a 2 take away 2 and a 4 s 4 then as think of expecting all along a diagonal matrix is just the matrix formed by having that I can now use in the main diagonal where you have P and the inverse of P here in case you want to do anything further after that by using the day agonal factorization which would just be reversing those and then you can carry operations and II like finding a poet over and so on just by cutting that on the diagonal one which is an easy thing to take square softened cubes off and so on late that's that part

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Keywords: diagonalize a 3x3 matrix, characteristic polynomial, trace of a matrix, determinant of a matrix, inverse of a 3x3 matrix, linearly independent vectors, eigenvalue, eigenvector

Id: Sf91gDhVZWU

Length: 19min 46sec (1186 seconds)

Published: Mon Oct 08 2012