This is a problem about eigenvalues and eigenvectors, right?
Well, (0, 0, 0) is never an eigenvector since the whole point of an "eigenvalue", \lambda, of T is that there exist a non-zero vector v such that Tv= \lambda v.
You don't clearly state what the original matrix is but I think you mean
M=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 &1 \\ 1 & 1 & 1\end{array}\right]
The characteristic equation for that is not at all what you give, however. I get -\lambda^3+ 3\lambda^2= 0 which has roots 0, 0, 3. That makes sense to me since the matrix is badly "singular". It's no surprise that it has 0 as a double eigenvalue. To find eigenvectors corresponding to the eigenvalue 0, we look at
M=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 &1 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]= \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]
Obviously that gives us the single equation x+ y+ z= 0. The "solution space" is two dimensional- we can pick any values for x and y and solve for z. In particular if we take x= 1, y= 0, z= -1. An eigenvector corresponding to eigenvalue 0 is [1 0 -1]T. If we choose x= 0, y= 1, we get z= -1. An independent eigenvector corresponding to eigenvalue 0 is [0 1 -1]T.
To find an eigenvector corresponding to eigenvalue 3, we look at the equation
M=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 &1 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]= \left[\begin{array}{c} 3x \\ 3y \\ 3z \end{array}\right]
That gives three equations, x+ y+ z= 3x, x+ y+ z= 3y, and x+ y+ z= 3z which are, again, dependent equations. This time they reduce to x= y= z. An eigenvector corresponding to eigenvalue 3 is [1 1 1]T.
The matrix you are looking for, that diagonalizes M, is
M=\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 &1 \\ -1 & -1 & 1\end{array}\right]
