# Find an orthogonal matrix $Q$ so that the matrix $QAQ^{-1}$ is diagonal.

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 $\left( \begin{array}{ccc} -1 \\ 1/2 \\ 1/2 \end{array} \right)$

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

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.
It doesn't matter which set of vectors you use (though of course, it will change your $Q$).