Eigenvalue and Eigenvector Computations Example

so in this example we're going to look at how to compute the eigenvalues and eigenvectors of a matrix eigenvalues and eigenvectors pop up a lot especially in signal processing applications in this little video we're just going to work on the mechanics of how you compute these for a matrix so we're going to work with the matrix a the matrix a is 4 6 10 3 10 13 minus 2 minus 6 minus 8 so this the matrix we're going to be working with and we're going to first compute the eigenvalues of the matrix a and we denote the eigenvalues as lambda sub I that's the notation that's commonly used for eigenvalues 2 is 2 note each one is a lambda value since we're working with a 3x3 matrix we will have eigenvalues lambda 1 lambda 2 and lambda 3 and we know that lambda is all these eigen values that we want to compute are the roots what we call the characteristic equation of the matrix which is the determinant of a minus lambda I equals 0 so if we can compute the determinant of this quantity set it equal to 0 and find these roots then we will have found the eigenvalues of the matrix in this equation right here I've just boxed in this is the equation that you can use for any N by n matrix this is the general equation that you need to solve so we need to compute this specifically for the example we're working the determinant of a minus lambda I so here's our matrix a here is lambda times the identity matrix I and then you see we're using the lines to indicate the determinant that's common notation to spelling out Det you put vertical lines around the matrix that means take the determinant so after we take 4 6 10 our matrix a and subtract off the identity matrix times I we have this which simplifies to 4 minus lambda 6 10 3 and then 10 minus lambda in the diagonal second diagonal element there and then minus 2 minus 6 and minus 8 minus lambda because we're subtracting off the lambda from minus 8 so we need to compute the determinant of this matrix so let's just recopy that down right here so to compute the determinant we do the cofactor expansion so I'm going to have the quantity 4 minus lambda times this two-by-two determinant which comes from the lower 2x2 entries of the original matrix minus 6 times this other 2x2 matrix and then plus 10 times this other 2x2 determinant which comes from the kind of lower left-hand 2x2 matrix of the original matrix so this is just kind of standard cofactor expansion and we've now simplified this 3 by 3 determinant into doing 3 2 by 2 determinants so we can go ahead and write these out and expand them because we know the rules for doing 2 by 2 determinants it's going to be 10 minus lambda times quantity minus 8 minus lambda minus the other cross terms multiplied 13 times negative 6 so that's one term and then we'll do the exact same expansion for the second determinant we'll have 3 times quantity minus 8 minus lambda minus 13 times negative 2 and then plus 10 times 3 times 1/8 of 6 minus the other cross term so really what we have here are three different terms that I'm going to label 1 2 3 and we need to add all these together but we're going to attack them kind of one piece at a time to simplify the algebra so let's go ahead and work on the first term here is the first term it's 4 minus lambda and if I start multiplying these things out 10 times negative 8 is negative 80 10 times the negative lambda is minus 2 minus 10 lambda then the other cross terms are negative 8 times negative lambda which gives you 8 lambda and a negative lambda times the negative lambda gives you lambda squared and so on so if I do all those and then 13 times 6 is 78 I get this which I can simplify a little bit negative 80 plus 78 gives me negative 2 and then I have a negative 10 lambda and an 8 lambda sitting there so that's minus 2 lambda and lambda squared stays by itself I can't simplify that at all and I'll go ahead and multiply and distribute all of this so I'm just going to distribute each of these things 4 times a negative 2 is negative 8 4 times a negative 2 lambda is negative 8 lambda 4 times lambda squared is 4 lambda squared and then similarly distribute the minus lambda across everything so we have six terms and we can go ahead and fine all those and we get minus lambda cubed plus 6 lambda squared minus 6 lambda minus 8 so this is an expression for term one which is just one of the three terms that we need well do similar work on the other ones these are a little bit simpler they don't have quite as many terms so minus six I expanded that 26 minus 24 is 2 so we end up with 2 minus 3 lambda if I go ahead and distribute the minus 6 I get negative 12 plus 18 lambda so here's a simplified expression for term 2 and then finally term 3 same type of dis distribution of the multiplication and just algebraic simplification gives us a simplified expression of 20 minus 20 lambda so what we'd have done here is we've computed each part of this determinate expression I now know that the determinant of a minus lambda I is term 1 plus term 2 plus term 3 and I have nice simplified expressions for each of these this is term 1 this is term 2 and this is term 3 so now I just have some more algebra to do and if I combine like terms I end up with minus lambda cubed plus 6 lambda squared minus 8 lambda notice the only lambda terms I have our 18 lambda and minus 20 lambda and this minus 6 lambda so if you add all that up that turns into minus 8 lambda and then the only constant terms are minus 8 minus 12 which is minus 20 but then there's a plus 20 so that gives you zero so that's why it's plus zero so we've done what we needed to do we've computed this deterministic or I'm sorry this characteristic equation and then what we do is we set it equal to zero and this is now a third order polynomial in lambda but I need to solve for the roots because this will give me the eigenvalues so really what we have here is an equation lambda cubed minus 6 lambda squared plus 8 lambda 0 and I can factor that I can factor out a lambda to give me lambda times quantity lambda squared minus 6 lambda plus 8 and now I can go ahead and see one of the eigenvalues pretty easily I can see that for the value of lambda equals 0 I have 0 equals 0 so lambda 1 equal to 0 is one of my eigenvalues and then the other two eigenvalues i come from solving this quadratic equation second-order polynomial which I can do with the quadratic equation minus B plus or minus the root of b squared minus 4ac over 2a I just plug into specifics for our problem here B is a negative 6 so a negative negative 6 is a positive 6 B squared is 36 minus 4 times 1 times 8 over 2 times 1 so now I'm going to doing some simple algebra 4 times 1 times 8 is 32 so inside the root I have 36 minus 32 which is 4 so this is really 6 plus or minus 2 over 2 so when you do - you have 6 minus 2 is 4 4 over 2 is 2 so one of the roots is lambda 2 equal 2 and then the other term will be when we add 2 6 plus 2 is 8 8 over 2 is 4 so the other eigen value is lambda 3 equal to 4 so we've now solved for the 3 eigen values lambda 1 + 0 lambda 2 is 2 and lambda 3 is 4 let's go ahead and compute the eigen vectors associated with each of these eigen values the eigen vectors we denote V sub I since we're working with a 3 by 3 matrix we will have V 1 V 2 and V 3 and remember to compute the eigen vectors each eigen vector is this the null space of a minus lambda I times the identity matrix so we're going to be evaluating this for each eigen value for little I equals 1 2 & 3 so let's go ahead and compute V 1 for V 1 the corresponding eigen value is lambda 1 equals to 0 so I actually need to solve the equation a minus lambda 1 times the identity which is equal to a minus 0 times the identity because my eigen value was 0 which is just equal to a so I need to find the null space of a and this will tell me what V 1 is so to find the null space of a I write down the matrix a and I all meant it with zeros and now I'm just going to do row manipulations to solve this system of linear equations so the first thing I'll do is I'll let Ethan 3 equal 2 times III plus equation 1 so if you do that this implies to this system of linear equations now modified equation three equations one and two have stayed the same but I have reduced it a little bit to this form and now let's go ahead and do another manipulation let's let equation 1 equal 1/4 of equation 1 so everything in that first equation is going to get divided by 4 the equations two and three will stay the same and then let's do another manipulation let's let e two equal e 2 minus 3 e 1 so here equation 1 will stay the same equation 2 is going to change to this because it's now equal to e 2 minus 3 e 1 and e 3 will remain unchanged because I haven't done any manipulations on equation 3 and finally let's let e 2 equal to xi of e 2 so we're just going to multiply it by a scalar and we'll do the same thing for e3 will take III and negate it and divide by 6 so equation 1 will remain unchanged e 2 is going to turn into 0 1 1 because I multiplied everything by 2 xi so 2 xi s cancel with 11 halves and similarly 43 i'm going to divide everything by negative 6 so the negative 6 is turned into positive ones when i do the division and then finally let's let III equals e 3 minus e 2 so now the first two equations will remain unchanged but now i'm going to get a row of zeros here for my last equation so i have simplified my system of linear equations as much as possible by doing row operations and now looking at this i can go ahead and write down the solution z is obviously equal to any value that we'd like it's a free variable because the last row is all zeros by looking at the second equation we see that Y needs to equal a negative Z and then looking at the first equation we see that X plus 3 halves y plus 5 halves e has to equal 0 if I solve for X that means that X has to equal a negative 3 halves Y minus five halves e just by moving to the other side and now if we plug in the relationship for how Y and 0 related we can simplify this we know that Y has to equal a negative Z so if I replace Y with negative Z my minus three halves Y turns into three halves Z and now I just have Z three half z minus five half Z is equal to a negative C so I've now solved this system of linear equations Z can be any number Y has to be a negative of that number and X also has to be a negative of that number so since Z is a free variable I'm going to pick a specific value for it I'm going to let Z equal one and now I can construct my eigenvector the eigen vector V one is going to have a 1 for the Z coordinate X has to equal negative 1 because X must equal negative Z and similarly y must equal a negative 1 because Y also has to equal a negative Z so this is my eigenvector v 1 we can now do similar things for V 2 I need to compute a minus lambda 2 I lambda 2 is equal to 2 so I have a minus 2i I plug this in to my equation we'll go through this one a little bit faster it's just algebra you can slow it down if you need to to check the steps but we're just evaluating this matrix so we now have the matrix a minus 2i and what I what do I need to do to find the eigenvector v 2 again I just have to find the null space of this vector so we set up our system of linear equations two people two zeros and then I will do row operations until I manipulate it to where I can solve this system of linear equations so I'm going to let a 3 equal a 1 plus e 3 I'm going to add those two together so equation 1 remains unchanged equation 2 remains unchanged III is equal to the sum of e 1 and e 3 so it turns into zeros the next manipulation I'll do is I'll let a 1 equal 1/2 e 1 so I'm going to divide everything by 2 equation 2 remains unchanged and equation 3 remains unchanged and now I can go ahead and subtract let eetu equal e 2 minus 3 e 1 so e 1 will remain unchanged but e 2 turns into this and E 3 remains unchanged and now I can go ahead and look at this and write down my solution again since I have all zeros in my third row Z can be anything from the second equation we have that minus y minus 2z has to equal 0 so solving for y i get y must equal a negative 2z and then from the first equation we have x plus 3y plus 5z has to equal 0 so solving for x we get minus 3y but Y is equal to a negative 2z so I get X is equal to a negative 3 times a negative 2z minus 5z which simplifies to z so we now have a nice solution Z can be any number Y has to equal a negative 2 times that number and X has to equal that number Z so now let us pick a number for Z I'm free to choose anything that I want I'll let Z equal 1 and now I can construct a V 2 if Z is 1 then the third coordinate of V 2 has to be equal 1 the second coordinate Y has to be equal a negative 2 times that so I get a negative 2 and the first coordinate has to be equal to Z so I get 1 so here's my third eigenvector if we wanted to we can even check we could take a times this and what you'll get out if you do the computations is you get to a negative 4 2 which is equal to 2 times my starting vector so this is just the definition of the eigenvector when I put my eigenvector into my matrix a I get out that eigen vector scaled by the value of the eigen value so we now have V 2 and finally let's compute V 3 so the same approach I have to compute a minus lambda 3 times the identity matrix which is equal to this matrix and now I have to find the null space of this matrix and that will give me V 3 so I set up my system of linear equations to solve for the null space I start doing my manipulations to solve this system of linear equations the next operation I'll do is e 2 is e 2 minus 3 e 1 e 1 remains unchanged e 2 changes to this III remains unchanged I'll let a 3 equals e 3 plus 2 e 2 so e 1 remains unchanged e 2 remains unchanged III we get all zeros and now we're back in the similar situation Z can be any number from the second equation I have minus 3y minus 5z equals 0 so I solve for y to get Y in terms of Z and from the first equation I can solve for x in terms of Z which simplifies to minus Z so again I'm free to choose any value for Z the general solution is Z is any number Y is a negative 5 Z and X is a negative Z so I could just let Z equal one just like I have been but I'm going to try to make my answer just a little nicer I'm going to choose Z equals three so when I construct a v3 Z has to equal three Y will be a negative five because Y is equal to minus 5/3 Z the third and the Z equals three cancels you get minus five and then since X has to equal a negative Z the first coordinate has to be a negative three and again if I wanted to I could check I could check the a times this vector if you do the math it comes out to minus 12 minus 2012 which is equal to four times my starting vector so this is just the definition of the eigenvector what I put in for a to operate on is what comes out scaled by the eigenvalue and that concludes our little example of how to compute eigenvalues and eigenvectors of a matrix eigen values are always roots of the characteristic equation and eigen values are always solutions of the null space of the equation a minus lambda hi

Channel: Adam Panagos

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Keywords: eigenvalue computation example, Eigenvalue example, Eigenvector example, how to compute eigenvectors, eigenvalue example 3x3, how to compute eigenvalues, finding eigenvalues, finding eigenvectors, eigenvalues of 3x3 matrices, Eigenvalues and Eigenvectors, Signal Processing, Signal Processing (Field Of Study), Eigenvalues

Id: cHOsd2PhkqE

Length: 16min 38sec (998 seconds)

Published: Mon Dec 23 2013