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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) $

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    $\begingroup$ Do you mean diagonal instead of orthogonal for $QAQ^{-1}$? $\endgroup$ – Julian Kuelshammer Jun 10 '13 at 5:46
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    $\begingroup$ I edited some typos. See if my edit was correct. $\endgroup$ – Julian Kuelshammer Jun 10 '13 at 5:51
  • $\begingroup$ Side note: The eigenvector for eigenvalue 3 is (1,1,1) $\endgroup$ – raindrop Jun 10 '13 at 5:57
  • $\begingroup$ Try $(1,1,-2)^T$ and $(1,-1,0)^T$ for the kernel. $\endgroup$ – copper.hat Jun 10 '13 at 6:33
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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 $Q$).

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