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Find an orthogonal basis for $\mathbb R^3$ consisting of the eigenvectors of the matrix $\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$

Isn't this question basically just asking 'find the eigenvectors of this matrix'? And the part about finding 'an orthogonal basis' is irrelevant?

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    @KCd OK I see you are correct. I dotted those vectors are and they are not orthogonal. So how can I get an orthogonal basis?2012-04-22

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Sorry I got mixed up earlier with the statement of the Real Spectral Theorem. It tells you that there exists an orthogonal basis for $\Bbb{R}^3$ consisting of eigenvectors of your matrix $A$ with all eigenvalues real. So indeed finding the correct vectors in the eigenspace to be orthogonal is not immediate from the outset.

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    @KCd I was confused about the statement of the real spectral theorem. I have corrected that now.2012-04-22
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The symmetric matrix (call is $A$) has two eigenvalues, one of multiplicity 2 at -1, and one of multiplicity 1 at 5. The eigenspaces $\ker(A+I)$ and $\ker (A-5I)$ are orthogonal complements, so the only issue is choosing a basis for $\ker(A+I)$ that is orthogonal.

Choose $\frac{1}{\sqrt{3}}(1,1,1)^T$ as a basis for $\ker(A-5I)$ (not a huge amount of choice there).

Since $\ker(A+I) = \ker (1,1,1)^T$, we can choose a element of the null space, say $\frac{1}{\sqrt{2}} (1,-1,0)^T$, and just find an orthogonal vector (in $\ker(A+I)$, of course), for example: $\frac{1}{\sqrt{6}} (1,1,-2)^T$.

It is easy to check that these vectors are orthogonal, in fact, orthonormal.