The question is as follows:
$A=\left( \begin{array}{ccc}
1 &1& 1 \\
1 & 1 & 1 \\
1 & 1 & 1 \end{array} \right) $ Find an orthogonal matrix $Q$ so that the matrix $QAQ^{-1} $ is diagonal. Verify this by direct computation.
My friend knows how to find the eigenvector for the eigenvalue of 3, but doesn't know how to find the eigenvector for the eigenvalue of 0: when he computes it he gets a different answer than given by the solution.
He found the vectors $\left( \begin{array}{ccc}
-1 \\
1 \\
0 \end{array} \right) $ and $\left( \begin{array}{ccc}
-1 \\
0 \\
1 \end{array} \right) $ by computing $Ker(A-0I)$.
However, according to the solutions, the eigenvectors for the eigenvalue 0 are $\left( \begin{array}{ccc}
0 \\
1 \\
-1 \end{array} \right) $
and $\left( \begin{array}{ccc}
-1 \\
1/2 \\
1/2 \end{array} \right) $
Best Answer
When you are talking about finding vectors as solutions to the equation $Ax = \lambda x$, the solution is a subspace, not simply a finite set of vectors.
Hence, the statement that the 'solution' is making is the same as what your friend is saying, since the subspace generated by those sets of vectors is the same (verify this).
It doesn't matter which set of vectors you use (though of course, it will change your $Q$).