The discussion revolves around finding the eigenvectors of a 3x3 matrix given its eigenvalues, specifically 6 and 2. The matrix in question has been corrected multiple times during the conversation, indicating some confusion about its entries.
Participants are actively engaging with the problem, sharing their attempts and questioning the validity of their results. Some guidance has been offered regarding the nature of eigenvectors and the relationships between them, but no consensus has been reached on the final eigenvectors.
There are indications of confusion regarding the matrix entries and the definitions of eigenvectors. Participants express uncertainty about the methods used and the results obtained, particularly in relation to the eigenvalue of 6.
mkay123321 said:Homework Statement
So the 3x3 matrix involved is [3 -1 -1:-4 6 4:-1 1 1], The eigenvalues are L=6 and L=2.
Homework Equations
(A-LI)e=0
The Attempt at a Solution
I stuck the eigenvalues into the matrix and got (-1 1 1)(not sure if its right) for L=2 but when I use L=6 in I can't seem to get the answer, I can't bring it down to echelon form that easily either. Thanks for any help
mkay123321 said:Sorry the matrix is [3 -1 -1:-4 6 4:1 -1 1] and one of the eigenvalue is 6 and the other is 2
mkay123321 said:Okay, So for L=2 I am getting [1 -1 -1:-4 4 4:1 -1 -1] and when I try using G.E I get [1 -1 -1:0 0 0:0 0 0].
From there x - y -z=0
x=y+z, that's why I thought it was [-1 1 1], I've never done this before so I am not sure if what I am doing is right. I am trying to learn off youtube but I am not understanding how they are getting the eigenvectors.
Dick said:x-y-z=0 is correct. Good work. But [-1,1,1] doesn't solve x-y-z=0? Does it??
mkay123321 said:So [1 -1 -1] ? And you said there were 2 corresponding eigenvectors for L=2, how would I find the other?
dick said:[1,-1,-1] means x=1, y=(-1) and z=(-1). That doesn't solve x-y-z=0 either. Just put the numbers in and try it. There are an infinite number of solutions of x-y-z=0. Try and find one of them first.
mkay123321 said:[2 -1 -1]?
Dick said:x=2, y=(-1), z=(-1) so x-y-z=2-(-1)-(-1)=2+1+1=4! Not 0. Can you try just once more, please? Find me ANY correct solution of x-y-z=0.
mkay123321 said:Ha, [2 1 1], okay so it could be [4 2 2] as well and [6 3 3] etc?
dick said:there you go. Much better. But [4,2,2]=2*[2,1,1] and [6,3,3]=3*[2,1,1]. So they aren't really 'different'. They are linearly dependent, they all lie along the same line. The eigenvectors of 2 form a plane. To get a full set of eigenvectors you need to find one off that line. Any ideas?
mkay123321 said:[4 3 1]?
Dick said:That works. So you could give the eigenvectors as the space spanned [2,1,1] and [4,3,1]. It's pretty likely your book will express it as something like the span of [1,1,0] and [1,0,1]. That also works and looks simpler. But it's equivalent to the span of your two vectors.
mkay123321 said:Yes I was getting confused with the 1's and 0's everywhere in the notes. So for L=6 will it be easier to use Gaussian Elimination or simultaneous? I am trying to use G.E but then the matrix goes into fractions and makes it harder. What I got so far is [1 -4 1].
Dick said:You can do it any way that feels easier to you. You can test your answer of [1,-4,1]. If it's an eigenvector with eigenvalue 6 then if you multiply it by the original matrix, then you should get 6*[1,-4,1], right? It's easier to check them than to solve for them. Try it, you'll find you are correct.