And since , that means
So I agree with your eigenvalues, of course making note that is a repeated eigenvalue.
Now for the eigenvectors, we need to solve , where
So setting we find that the corresponding eigenvector to the eigenvalue is , where you can let be any value you like.
Row 2: where can be any value you like.
So set where can be any value you like, and you find that the eigenvector corresponding to is
It is common practice to have the same number of eigenvectors as eigenvalues, and to give a value to your parameters. This is what MatLab has done.