How to find the basis for the eigenspace if the rref form of λI - A is the zero vector?

Question

This is the matrix A that I am given, and I have to find the eigenvalues, and the basis for each associated eigenspace.

$$\begin{bmatrix}2 & 2 & -4\\-1 & -1 & 2\\1 & 1 & -2\end{bmatrix}$$

I have already found the eigenvalue 0 and its eigenspace basis: {[-1, 1, 0], [2, 0, 1]}. However, for the other eigenvalue λ = -1, calculating λI - A and reducing it to reduced row echelon form gives you

$$\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}$$

Since this means that x1 = x2 = x3 = 0, this is trivial linearly dependent, and so I am confused on how to find a basis for this. Any help would be appreciated.

Answer 1

You have made a computational mistake somewhere, $\lambda I - A$ must be singular.

>> A=[2, 2, -4; -1, - 1, 2; 1, 1, -2];
>> rref(-eye(3)-A)

ans =

 1     0    -2
 0     1     1
 0     0     0