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we figured out the eigenvalues for a two-by-two matrix so let's see if we can figure out the eigenvalues for a 3x3 matrix I think we'll appreciate that it's a good bit more difficult just because the math becomes a little hairier so lambda is an eigen value eigen value of a by definition if and only if I'll write it like this if and only if a times some nonzero vector V is equal to lambda times that nonzero vector V let me write that for some non-zero I could call it AI ghen vector V but I'll just call for some nonzero vector V or some non-zero V now this is true if and only if if and only if this leads to alright like this this is true if and only if and this is a bit of a review but I like to review it just because when you do this ten years from now I don't remember the form and I want you just remember the logic of how we got to it so this is true if and only if let's just subtract AV from both sides the zero vector is equal to lambda instead of writing lambda times V I'm going to write lambda times the identity matrix times V this is the same thing right identity matrix times V is just V minus a V right I just protected AV from both sides we rewrote V as the identity matrix times V well this is only true this is only true if and only if the zero vector is equal to is equal to lambda times the identity matrix minus a times V I just factored the vector V out from the right-hand side of both of these guys and I'm just left with some matrix times V well this is only true all right let me rewrite this over here this equation just in a form you might recognize that lambda times the identity matrix times a this is just some matrix this matrix times V is got to be equal to zero for some nonzero vector V that means that the null space of this matrix the null space of this matrix has got to be non-trivial that means or another way to think about it is that its columns are non are not linearly independent or another way to think about is it's not invertible or it has a determinant of 0 so lambda is in the eigen value of a if and only if each of these steps are true and this is true if and only if for some non-zero vector if and only if the determinant the determinant of lambda times the identity matrix minus a is equal to 0 and that was our takeaway I think was two videos ago or three videos ago but let's apply it now to this three by three matrix a so the lambda we're going to use the 3 by 3 identity matrix so we want to concern ourselves to see lambda times the identity matrix is just going to be times the 3 by 3 identity matrix it's just going to be so this is let me write this this is lambda times the identity matrix in r3 so it's just going to be lambda lambda lambda and everything else is going to be zeros right the identity matrix had ones across here so that's the only thing that becomes nonzero when you multiply it by lambda everything else was a 0 so that's the identity matrix times lambda now what is so lambda times the identity matrix minus a is going to be equal to it's going to be equal to it's actually pretty straightforward to find everything around along the diagonal is going to be lambda minus that let's just do it lambda minus minus 1 under the diagonals here lambda minus minus 1 is lambda plus 1 and then 0 minus 2 I'll do that in a different color 0 minus 2 is minus 2 0 minus 2 is minus 2 0 minus 2 is minus 2 let's do this 1 0 minus 2 is minus 2 0 plus or minus minus 1 is 0 plus 1 which is 1 and then let's just do this 1 0 minus minus 1 that's 1 and let me finish up the diagonal and then you have lambda minus 2 lambda minus 2 and then you have a lambda minus 2 - - so lambda is an eigenvalue of a if and only if the determinant of this matrix right here is equal to zero so let's figure out its determinant and the easiest way at least in my head to do this is to use the rule of Sarrus so let's use the rule of Sarrus to find this determinant so I just rewrite these rows right there I can just copy and paste them really so I just take those two rows and then let me paste them put them right there a little bit too close to this guy but think you get the idea and now the rule of Sarrus I just take this product Plus this product Plus this product and then I subtract out this product times this product times this product we'll do that next so this product is lambda plus 1 times lambda minus 2 times lambda minus 2 that's that one there and then plus you see minus 2 times minus 2 that's plus 4 and then we have minus 2 times minus 2 plus 4 times 1 so that is plus 4 again and then we can do - this column times this column minus this column minus this column and then or I shouldn't say column but diagonal really so we say minus is C minus 2 times minus 2 let me write this minus - minus 2 times minus 2 which is 4 times lambda minus 2 lambda minus 2 that was this diagonal and then we have - what is this going to be going to be minus 1 times lambda plus 1 so minus lambda plus 1 and then you go down this diagonal C minus 2 times minus 2 is 4 so it's going to be 4 times lambda minus 2 and we're subtracting so minus 4 times lambda minus 2 and let's see if we can simplify this so this blue stuff over here see these guys right here become 1/8 and then this becomes this becomes lambda plus 1 lambda plus 1 times if I multiply these two guys out lambda squared minus 4 lambda right minus 2 lambda n minus 2 lambda so minus 4 lambda plus 4 and then I have this plus 8 here plus 8 and then I have this you have minus 4 times lambda let me just multiply everything out so I have minus 4 lambda plus 8 minus lambda minus 1 minus 4 lambda minus 4 lambda plus 8 plus 8 and then let me let me simplify this up a little bit so this guy over here will Cu constant terms I have an 8 I have a minus 1 I have an 8 and I have an 8 so that's 24 minus 1 so that is a 23 and then the lambda terms I have a minus 4 lambda I have a minus lambda and have a minus 4 lambda so it's minus 8 minus 1 so minus 9 lambda minus 9 lambda plus 23 and now I have to simplify this out so first I can take lambda and multiply it times this whole guy right there so it's going to be lambda cubed lambda cubed minus 4 lambda squared plus 4 lambda and then I can take this one and multiply it times that guy so plus lambda squared plus lambda squared minus 4 lambda plus 4 and now and of course we have these terms over here so we're going to have to simplify it again so what are all of our constant terms we have a 23 and we have a plus 4 so we have a 27 plus 27 and then what are all of our lambda terms we have a minus 9 lambda and then we have a C we have a minus 9 lambda we have a plus 4 lambda and then we have a minus 4 lambda so these two cancel out so I just have a minus 9 lambda and then what are my lambda squared terms I have a plus lambda squared and I have a minus 4 lambda school so if you add those two is going to be minus 3 lambda squared minus 3 lambda squared and then finally I have one only one lambda cubed term that right there so this is the characteristic polynomial for our matrix so this is the characteristic polynomial and this is represents the determinant for some for any lambda the determinant of this matrix for any lambda and we said this has to be equal to 0 if and only if lambda is truly an eigenvalue so we're going to set this equal to 0 and unlucky or lucky for us there is no real trivial you know there's no quadratic well there is actually 4 try but it's it's very complicated and so it's usually a waste of time so we're gonna have to do kind of the art of factoring a quadratic polynomial I got this problem out of a book and I think it's fair to say that if you ever do run into this and in an actual linear algebra class or really in an algebra class generally doesn't even have to be in the context of eigen values you probably will be dealing with integer solutions and if you are dealing with integer solutions if you are dealing with integer solutions then your roots are going to be factors of this term right here especially it was especially if you have a 1 coefficient down here so your potential roots in this case what are the factors of 27 so 1 3 9 and 27 so all of these are potential roots so we could just try them out 1 cubed is 1 minus 3 so let me write try it one if we try 1 it's 1 minus 3 minus 9 plus 27 that does not equal 0 I'm going to feel like it's minus 2 minus 9 is love - 11 plus it's 16 that does not equal 0 so 1 is not a root if we try 3 if we try 3 we get 3 cubed which is 27 minus 3 times 3 squared is minus 3 times 3 which is minus 27 minus 9 times 3 which is minus 27 plus 27 that does equal 0 so lucky for us on our second try we're able to find one 0 for this so 3 is a zero that means that X minus 3 is one of the factors of this so that means that this is going to be X minus three times something else or I should say lambda minus three so let's see what the other root is so if I take lambda minus three and I divide it into this guy up here into lambda cubed minus 3 lambda squared minus 9 lambda plus 27 what do I get see lambda goes into lambda cubed lambda squared times lambda squared times that lambda squared times lambda is lambda cubed lambda squared times minus 3 is minus 3 lambda squared you subtract this these guys you get 0 you get 0 and then we could put here we can well we could do it either way we could put it down the minus 9 we could bring down everything really we could so now you have minus 9 lambda plus plus 27 you can almost imagine we just subtracted this from this whole thing up here and we're just left with these terms right here and so lambda minus 3 goes into this well lambda minus 3 goes into 9 lambda it goes into 9 lambda minus 9 times so we're going to write I'll just write it minus 9 here minus 9 times lambda minus 3 is minus 9 lambda plus 27 plus 27 so it went in very nicely so you get to 0 so this our our characteristic polynomial has simplified to our characteristic polynomial has simplified to lambda minus 3 lambda minus 3 times lambda squared minus 9 and of course we're going to have to set this equal to 0 if lambda is truly an eigen value of our matrix and this is very easy to factor so this becomes lambda minus 3 times lambda squared minus 9 is just lambda plus 3 times lambda minus 3 and all of that equals 0 these roots we already know one of them we know that three is a root and actually this tells us that three is a root as well so the possible eigenvalues the possible eigenvalues of our matrix a are three by three matrix a that we had way up there this matrix a right there the possible eigenvalues are lambda is equal to three or lambda is equal to minus three those are the two values that would make our characteristic polynomial or the determinant for this matrix equal to zero which is a condition that we need to have in order for lambda to be an eigenvalue of a for some non-zero vector v in the next video we'll actually solve for the eigenvectors now that we know what the eigenvalues are