What are the Eigenvectors of a 3x3 Matrix with Eigenvalues 6 and 2?

[mkay123321]

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

 

[Dick]

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

No, (-1,1,1) isn't an eigenvector corresponding to the eigenvalue 2, and 6 isn't even a eigenvalue of that matrix. How you would expect anyone to be able to guess what you are doing wrong given the amount of work you've shown beats me.

 

[mkay123321]

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

 

[Dick]

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

Ok, that's better. But (-1,1,1) is still not an eigenvector corresponding to the eigenvalue 2. There are two linearly independent eigenvectors corresponding the eigenvalue 2 and one eigenvector corresponding to the eigenvalue 6. Now can you show how you are trying to find them so someone can figure out what you are doing wrong?

 

[mkay123321]

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]

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.

x-y-z=0 is correct. Good work. But [-1,1,1] doesn't solve x-y-z=0? Does it??

 

[mkay123321]

x-y-z=0 is correct. Good work. But [-1,1,1] doesn't solve x-y-z=0? Does it??

So [1 -1 -1] ? And you said there were 2 corresponding eigenvectors for L=2, how would I find the other?

 

[Dick]

So [1 -1 -1] ? And you said there were 2 corresponding eigenvectors for L=2, how would I find the other?

[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]

[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.

[2 -1 -1]?

 

[Dick]

[2 -1 -1]?

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]

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.

Ha, [2 1 1], okay so it could be [4 2 2] as well and [6 3 3] etc?

 

[Dick]

Ha, [2 1 1], okay so it could be [4 2 2] as well and [6 3 3] etc?

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]

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?

[4 3 1]?

 

[Dick]

[4 3 1]?

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]

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.

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]

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].

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.

 

[mkay123321]

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.

Oh yes I see that it is actually easier to check them, thanks a lot for that man! Much appreciated. I just didn't understand the x-y-z=0 properly but now you've clearly explained it to me.