Example of 3x3 Eigenvalues and Eigenvectors

this video is going to give you an example of how to solve for the eigenvalues and eigenvectors of a particular matrix so our basic learning objectives we need to be sure first to given a matrix we can find the characteristic equation which is a polynomial equation that allows us to find our eigenvalues once we have that equation we need to know how to factor that polynomial so that we can find our eigen values once we have all of our eigen values for each eigenvalue we will find the eigenvectors by finding the null space of the original matrix minus lambda i or some textbooks including ours say that we should really think about lambda I minus a they have the same null space so it doesn't really matter finally I want to illustrate how to check if a vector is a valid eigen vector so here's our matrix we want to find the eigenvectors and eigenvalues of this 3x3 matrix so we're going to need to start by finding the eigenvalues once we have our eigenvalues we'll find our heigen vectors for each of those alright so first we need to find the characteristic equation the characteristic equation is found by taking the determinant of a minus lambda times the identity so what that means is we're going to take our matrix our a matrix and we're going to subtract the identity matrix with lamed on each of the diagonal terms so 2 1 3 1 2 3 3 3 20 and we're going to subtract lambda 0 0 0 lambda 0 0 0 lambda and find that new matrix so that will be the determinant you can actually go straight to this if you want we're going to go the determinant 2 minus lambda 1 3 and then 1/2 minus lambda 3 and 3 3 20 minus lambda we're calculating that determinant and notice that I used straight bars not brackets because I'm calculating a determinant so how do I calculate a determinant well i expand along the row or column and i don't really have anything clear that's suggesting anything was better than anything else so I'm going to expand along that first row so we start with the first term 2 minus lambda and then we multiply by the cofactor of that one one term so we are going to temporarily eliminate the first row and column and use this minor for the determinant so we get 2 minus lambda 3 3 and 20 minus lambda and then we have our second term we're going to add that one but our determinant our cofactors now a negative sign so we'll change the sign and take the determinant of 1 3 3 20 minus lambda and then we'll have our third term again this cofactor is going to keep the same sign so we don't have to change anything and I have 1 2 minus lambda 3 & 3 now that I have this I'm going to calculate my determinants of you to the 2x2 terms using the simple 2x2 product formula we get let's see two minus lambda times two minus lambda times 20 minus lambda and let me just finish all the work and I'll be right back alright so I found the rest of the determinants here's the first two-by-two determinant the second two-by-two determinant and third two-by-two determinant now what I need to do is I need to expand all these terms and multiply it out so for example let's just get started here I'm going to have two minus lambda and then I have two times 20 gives me 40 and then two times minus lambda is that's negative to negative 22 lambda and finally plus lambda squared minus 9 and now let me finish up the rest and now that I have that expanded I need to distribute this 2 minus lambda across this multiplication and also simplify these terms so again you don't need to watch all that so pause it work through it check your answers so here I have the first term that's been distributed and then the second term and then I've expanded the other term so let's now group things so we've got lambda cubed there's only one term so I get that's equal to negative lambda cubed let's identify all of our lambda squared terms there are two of them so I get 24 plus 24 lambda squared now let's identify our lambda terms see we've got minus 44 minus 40s that is minus 84 plus 9 so that's minus 75 74 minus 65 so we have minus 65 lambda and finally we have our constant terms let's see how many of those we have see we've got 80 minus 18 so that leaves me 62 just mean 42 and those all cancel each other so 42 so plus 42 so that's my characteristic polynomial and I need to solve the equation negative lambda cubed plus 24 lambda squared minus 65 lambda plus 42 equals 0 all right now that we have a cubic polynomial we need to find all the solutions to that polynomial equals zero this requires you remember a little bit from algebra and in particular we're actually going to use factoring so if I look at there's something called the integer root theorem and the integer root theorem says we look at the constant term of our polynomial if the polynomial has integer coefficients we look at that last polynomial at the last constant term and we find all of its factors and so that's plus and minus 1 plus and minus 2/3 C 4 T 2 is 6 times 7 so 2 3 not 4 or 5 plus or minus 6 plus or minus 7 plus or minus 21 Oh let's see 3 to get the 3 in there 14 I guess plus or minus 21 and plus or minus 42 those are the only possible integer routes and one way that I could do this if I have a graphing calculator and you're not going to have a calculator on a test but for practice problems you do you can actually look at a graph so I plug this polynomial into my calculator here's the polynomial I had to change all my lambda sin taxes and I graph it and I see that my graph crosses the axis and what appears to be lambda equals 1 and lambda equals 2 so we've got lambda equals 1 and lambda equals 2 and I need to use factorization to be able to see if that's really true now if lambda equals 1 is in fact a root lambda minus 1 is going to be a factor and if lambda equals 2 is a root then lambda minus 2 will be a factor so I need to see if I can factor that polynomial one way to do that is long division and the fast way is actually to use synthetic division so we're going to see about the lambda equals 1 by doing synthetic division so we're going to test the root of 1 with our coefficients negative 1 24 negative 65 and 42 all right so always start with zero we're going to add and multiply 23 23 negative 42 42 zero and since this is zero we know that it is in fact a factor it's lambda minus 1 times negative lambda squared plus 23 lambda minus 42 and that's equal to zero now I knew that lambda equals two is also a root so we can write lane - one and I guess we should check our polynomial negative 123 negative 42 we're testing land equals two so negative 1 negative 2 21:42 zero sure enough so we get lambda minus 2 and we read off the last polynomial negative lambda plus 21 so the factorization by peeling off these two these two roots lambda equals 1 and lambda equals 2 factoring them out we find in fact our third root so we have lambda equals 1 lambda equals 2 and lambda equals 21 these are my eigenvalues next we'll need to find out what the eigenvectors associated with each of those eigenvalues are so let's start with the eigenvalue lambda equals 1 what we need to do is we need to find the null space of a minus lambda I so with lambda equal 1 we need to find the null space of the matrix a minus lambda I so here we're going to subtract the identity and see what that new matrix is so I get 1 1 3 1 1 3 3 3 19 now the null space is asking so this is a matrix and the null space is saying what vector can I multiply by and get back the zero vector and this is a system of equations so when I want to solve my problem I'm going to take this matrix and I'm going to find the reduced row echelon form of the a minus lambda I matrix so recall that was 1 1 3 1 1 3 3 3 19 we'll do row operations so I can start by taking Row 2 and replacing it by Row 2 minus Row 1 and I can do Row 3 and turn it into Row 3 minus 3 Row 1 so here's the first step now what we're going to do is going to swap Row 2 and Row 3 so Row 2 and Row 3 will switch places and then in the next step I'm going to replace Row 2 the new Row 2 by 1/10 Row 2 so that I can get a 1 in that spot now finally with a 1 in this spot I can eliminate the 3 by taking Row 1 and replacing it by Row 1 minus 3 Row 2 and this is going to give me my final reduced row echelon form matrix my reduced row echelon form matrix gives me a new system of equations that's in a reduced form so if I think of this as x y and z that i want to multiply and have this equal to 0 0 0 that would be the system of equations I get a system X plus y plus 0 Z equals 0 Z equals 0 and I have an equation that starts with X I have the equation starting with Z but I have no equation starting with y so Y is my free variable so Y is free and we can give it a parameter say T because it's free and solve for X X is going to be equal negative Y so that's negative T and this allows me to solve my system my vector X Y Z has to be of the form negative T t0 and so I can choose any T I want and scale it I like to factor it out negative 1 1 0 and any multiple of this vector is my eigenvector that's what the T stands for it tells me I can multiply by anything I want so some possible vectors might be negative 1 1 0 or 2 negative 2 0 as long as my first and second coordinates are opposite one another and I found my eigen vectors so let's indicate this so I have an eigenvector negative 1 1 0 is an eigenvector with eigenvalue lambda equals 1 and now let's go the next one all right we do the same thing we start with our matrix we subtract lambda I so here lambda is 2 so we're going to take our original matrix 2 1 3 1 2 3 3 3 and 20 and we're going to subtract twice the identity and we get 0 1 3 1 0 3 3 3 18 and I need to find the reduced row echelon form of this matrix because I'm solving the system a minus lambda I times the vector equals the zero vector so I want to find the reduced row echelon form equivalent of that system of equations so what do we need to do we're going to start with our original matrix and swap Row 1 and Row 2 so I get 1 0 3 0 1 3 3 3 18 and now we're going to take Row 3 and replace it by Row 3 minus 3 Row 1 to give me 1 0 3 0 1 3 0 3 9 and I'm going to replace Row 3 by subtracting 3 Row 2 and then actually eliminates that second row so 1 0 3 0 1 3 0 0 0 and I'm done this is my reduced row echelon form I'm going to interpret it if I think of this as three variables times that matrix equals 0 0 0 then I get three equations I get X plus 3z equals zero y plus 3z equals zero and the third equation 0 X 0 y 0 Z equals 0 we don't need to write that down so x and y are not free but Z is free so we can say all right my Z is free we'll give it a parameter that allows me to solve for y and for X in my solution vector X Y Z has to be of the form negative 3 T negative 3t + T that I can factor out the T negative 3 negative 3 1 and that tells me what my eigen vector is so X any multiple of negative 3 negative 3 1 is an eigenvector with eigenvalue lambda equals 2 now we find the eigenvectors for eigenvalue lambda equals 21 we do basically the same thing a - 21 identity so we're going to I'm just going to subtract that 21 off the diagonal terms so I get negative 19 1 3 1 negative 19 3 and then 0 3 3 negative 1 see well how will I approach this I'm going to move Row 1 and Row 2 then I'm going to take Row 2 and add 19 row ones I'm going to take Row 3 and subtract 3 row ones and now I want to just notice that these rows are multiples of each other I need to get something that cancels and in fact it's easiest here to say we're going to replace Row 2 by rho 2 plus 6 Row 3 and then after that I'm going to replace Row 3 by Row 3 divided by 60 to get rid of that leading term now I can get rid of this negative 19 Row 1 is Row 3 plus 19 row so Row 1 plus 19 Row 3 and then I'm going to switch Row 2 and Row 3 and I'll be in reduced row echelon form now I can interpret this reduced row echelon form this is my reduced row echelon form it tells me that X minus one-sixth Z equals 0 and y minus one-sixth Z equals 0 and Z is free so let's give it a parameter T and I can now solve for y Y is equal to 160 and so is X so my solution vectors X Y Z all have the form some sixth of a T sixth of a T and T if i factor that out they're all multiples of 1/6 1/6 one is the end in particular these are all multiples if I think of multiplying everything six these are all multiples of 1 1 6 so this gives me a simple vector that I can think of as my eigenvector 1 1 6 is an eigen vector with corresponding eigen value of lambda equals 221 that gives me my third collection of eigenvectors any scalar multiple of that so we get to summarize what we've done already we found our eigenvalues by solving the determinant of a minus lambda I equals 0 then for each of the three eigenvalues we found the null space of the matrix a minus lambda I and we got our three eigen vectors now to show how to check our work what we're going to do is we're going to multiply each of these vectors by our matrix a and we should find that the result is this eigen value multiple of our original vector so here's our set up for our first example we take our matrix a we take our eigenvector and we do our matrix multiplication so our first row 2 1 3 times our column negative 1 1 I get negative 2 plus 1 let's move negative 1 and I'll go along the second row 1 2 3 times our column we get negative 1 plus 2 that gives me 1 then I go along my third row 3 3 20 times my column I get negative 3 plus 3 that gives me 0 and so what I find is a times this vector equals our original vector and that's exactly what it means to say we have an eigen value of lambda equals 1 now I've set up the second example if I take my matrix times my vector negative 3 negative 3 1 I should get a multiple advancing vector so we start with our first row and I'll get negative six minus three sort negative 9 negative 9 plus 3 gives me negative 6 we go along the second row 1 2 3 so negative 3 minus 6 that gives me negative 9 plus 3 so again negative 6 and now I do the third row 3 329 negative 9 so that's negative 18 negative 18 plus 20 gives me positive 2 so what we see is the effect of that matrix multiplication any times our second vector was it actually double our second vector that's exactly what it means to say we have an eigenvector with eigenvalue lambda equals 2 finally here's the set up for our third we'll take each of our rows times our our vector and when we take our first row we get 2 plus 1 is 3 18 3 and 18 is 21 we'll get that again on the second now 3 3 20 I get 3 plus 3 6 plus 120 gives me 126 and this is 21 times our original vector a X 3 is 21 X 3 and again we found an eigen vector corresponding to an eigen value of lambda equals 21 if you give it an eigen vector and you just need to check it just multiply it by the matrix and see if it's expanded by a multiple so we're done we found and verified our eigen values and eigen vectors

Channel: DrBrainWalton

Views: 397,495

Rating: 4.641212 out of 5

Keywords: Eigenvalues And Eigenvectors, Linear Algebra (Field Of Study)

Id: r5dIXpssvrA

Length: 28min 4sec (1684 seconds)

Published: Mon Sep 15 2014