The whole point of being an eigenvalue is that there exist a non-zero vector, v, such that .
In this example, that would be or, after multiplying on the left side of the equation,
And so there exist x, y, z, not all 0, satisfying , , and .
If , those equations become x+ ty= x, ty+ tz= y, and 0= z. From the last, z= 0 so the second equation becomes ty= y or (t-1)y= 0. Since , y= 0. The first equation then is x= x which is satisfied for all x. Any eigenvector corresponding to eigenvalue 1 must be of the form which is spanned by .
Why do you think you "went wrong"? With , the equations, , , and , become x+ ty= 0, ty+ tz= 0, and 0= 0. From the first, ty= -x so the second is -x+ tz= 0 or x= tz. Then y= -x/t and z= x/t. An eigenvector corresponding to eigenvalue 0 is . If you take x= 1, this is . If you take x= t, this is . Each is a multiple of the other and both are perfectly good eigenvectors. Any multiple of an eigenvector is an eigenvector. More generally, the set of all eigenvectors corresponding to the same eigenvalue is a subspace.For , I know that this is the same as just finding Ax = 0, however, my eigenvector is , & the one the chose was . I can't see where I went wrong.
With , the equations, , , and , become , , and . Both the second and third equations, since , give z= 0. The first equation is (t-1)x= ty or . Any eigenvector corresponding to eigenvalue t is of the form . Taking x= t gives .For , they have the eigenvector being I'm completely lost there.
How are all these eigenvectors found?
Thanks for your time.