Eigenvalues and eigenvectors

[Identity]

I have a 3x3 matrix A of the form: v1=[2 1 0] v2=[0 5 -2] v3=[0 1 2].
I have to find all eigenvalues of A and their eigenvectors.
I have the eigenvalues being; lambda1=2, lambda2=2, lambda3=5.
I think I'm having problems with my gauss-jordan elimination:
For lambda1=2, solving system (A-2I)x=0 by gauss-jordan elimination:
I get a matrix resembling v1=[0 1 0], v2=[0 0 1], v3=[0 0 0].
(That's actually an augmented homogeneous matrix but I've left out the 0's on the right hand side.)
The point is I don't know whether I've made a mistake with my row reductions or if I'm not understanding how to actually assign values to create an eigenvector, because to me the matrix is trivial and there is no solution space.
I'm having the same problem with lamda3=5..??

[HallsofIvy]

I have a 3x3 matrix A of the form: v1=[2 1 0] v2=[0 5 -2] v3=[0 1 2].
I have to find all eigenvalues of A and their eigenvectors.
I have the eigenvalues being; lambda1=2, lambda2=2, lambda3=5.
I think I'm having problems with my gauss-jordan elimination:
For lambda1=2, solving system (A-2I)x=0 by gauss-jordan elimination:
I get a matrix resembling v1=[0 1 0], v2=[0 0 1], v3=[0 0 0].
(That's actually an augmented homogeneous matrix but I've left out the 0's on the right hand side.)
The point is I don't know whether I've made a mistake with my row reductions or if I'm not understanding how to actually assign values to create an eigenvector, because to me the matrix is trivial and there is no solution space.
I'm having the same problem with lamda3=5..??

Frankly, I've never liked "Gaussian Elimination" for simple equations like these.

By definition of "eigenvalue" and "eigenvector", if 2 is an eigenvalue, then there must exist a non-zero vector (x,y,z) such that A(x,y,z)= (2x,2y,2z). Multiplying (x,y,z) by A (I'm assuming the v1, v2, v3 are the ROWS of A), we get 2x+ y= 2x, 5y- 2z= 2y, and y+ 2z= 2z. The only way the first equation can be true is if y= 0 and then the next two equations give z= 0. x can be anything so an eigenvector is (1, 0, 0) and all eigenvectors corresponding to eigenvalue 2 is a multiple of that.
If 5 is an eigenvalue, then we must have A(x,y,z)= (5x, 5y, 5z). Now the equations are 2x+ y= 5x, 5y- 2z= 5y, and y+ 2z= 5z. The second equation has the obvious solution z= 0. But that quickly leads to x= y= 0 also. Hmm, it appears that 5 is NOT an eigenvalue.

Perhaps you mean v1, v2, and v3 as the columns of the matrix. In that case, with eigenvalue 2, we would have 2x= 2x, x+ 5y+ z= 2y, and -2y+ 2z= 2z. The last equation has the obvious solution z= 0. Putting that into the second equation gives x+ 5y= 2y or x= -3y. The first equation is satisfied by any value of x so eigenvectors corresponding to eigenvalue 2 are all of the form (1, -3, 0).

For eigenvalue 5, we have 2x= 5x, x+ 5y+ z= 5y, and -2y+ 2z= 5z. The first equation has the obvious solution x= 0. In that case the second equation gives z= 0 and the third equation z= 0. No, 5 is STILL not an eigenvalue!