# Can anyone explain to me why eigenvector here is like this

by sozener1
Tags: eigenvector, explain
 P: 11 Im supposed to find the eigenvectors and eigenvalues of A I found that eigenvalues are 2 12 and -6 then I found eigen vectors substituting -6 to lambda and someone has told me I get 0 0 1 for eigenvector which I cannot understand why?? can anyone plzzzzzzzzz explain why this is???????? Attached Thumbnails
 Mentor P: 11,576 Did you multiply this vector with your matrix? What do you get? Does this fit to the definition of an eigenvector?
P: 11
 Quote by mfb Did you multiply this vector with your matrix? What do you get? Does this fit to the definition of an eigenvector?
I get 0 0 0

it doesn't fit with the definition of eigenvectors

so does that mean it should be 0 0 0 instead of 0 0 1??

I mean when you try to calculate for v3 = [ v1 v2 v3]

since the last row of the matrix is 0 0 0 should v3 come out as 0?? couldn't v3 be any number??

Mentor
P: 21,216
Can anyone explain to me why eigenvector here is like this

 Quote by sozener1 I get 0 0 0
That's not what I get.

The matrix you showed in the second attachment is not A. It is [A - (-6)I]. Of course if you multiply this matrix times your eigenvector, you'll get the zero vector.
 Quote by sozener1 it doesn't fit with the definition of eigenvectors so does that mean it should be 0 0 0 instead of 0 0 1?? I mean when you try to calculate for v3 = [ v1 v2 v3] since the last row of the matrix is 0 0 0 should v3 come out as 0?? couldn't v3 be any number??
Assuming that your eigenvalue is λ and that x is an as-yet unknown eigenvector for λ, what you're doing is solving the equation Ax = λx for x. That's equivalent to solving the equation (A - λI)x = 0. In other words, of finding the kernel of the matrix A - λI. This should be something that you have already learned to do.

The kernel here should not consist of only the zero vector - an eigenvector cannot be the zero vector.

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