Finding the third eigenvector for a 3×3 matrix with repeated eigenvalue

[rjw5002]

Homework Statement

Let A =
[4 0 1;
2 3 2;
1 0 4]
Let n >= 1 be an integer. Compute the matrix A^n with entries depending on n.

Homework Equations

The Attempt at a Solution

First I need to show that A is diagonalizable, and find a matrix S such that D = (S^-1)(A)(S) is diagonal. I am having serious trouble finding S. First I found the eigenvalues of A to be 3,3, and 5. Next I found the eigenvectors to be for $$\lambda = 3$$: (1 0 -1) and $$\lambda = 5$$ : (1 2 1). There's a theorem in my book that says that a matrix is only diagonalizable if and only if there are n linearly independent eigenvectors. I only have 2. I'm not quite sure how to get the third.
Now, assuming I had three eigenvectors, I would combine them to form the matrix S, and if the $$det(S)\neq 0$$ then the eigenvectors would be linearly independent, and
i could find S^-1. Then A^n would just be (S^-1)(D^n)(S). Does this all sound correct? I guess I'm just a little confused about the eigenvalue/eigenvector situation...
Thanks for any help.

 

[Dick]

There is a third linearly independent eigenvector, you are just overlooking it. Hint: try (0,1,0). Then do exactly what you propose.

 

[steelphantom]

Then A^n would just be (S^-1)(D^n)(S)

Wouldn't Aⁿ = SDⁿS^-1?

 

[Dick]

Yep. Sorry, I overlooked that.