finding an orthonormal basis of R^3 of a matrix using eigenvalues The symmetric matrix A below has eigenvalues 3 and 6 (multiplicity 2). Find an orthonormal basis B of R3 consisting of eigenvectors of A.
\begin{bmatrix}5&1&-1\\
1&5&1\\
-1&1&5
\end{bmatrix}
so far I have tried $3I-A$ that has given me: \begin{bmatrix}1\\-1\\1\end{bmatrix}
$6I-A$:
$$\begin{bmatrix} 1\\1\\0\end{bmatrix}, \begin{bmatrix} 1\\0\\1\end{bmatrix}$$
I am not sure how to continue with this questions
 A: If a (real) matrix is symmetric (more generally, if a complex matrix is normal), then eigenvectors corresponding to distinct eigenspaces are orthogonal (for a good explanation, see this post here). 
(Note: you should check over your computation of the eigenvectors, because your second eigenvector with eigenvalue $6$ is not orthogonal to the eigenvector with eigenvalue $3$, and is in fact not even an eigenvector. I think it should be:
$$\begin{bmatrix} -1 \\0 \\1\end{bmatrix}$$
The Gram Schmidt Process is an algorithm which turns a basis into an orthogonal basis, the details of the algorithm can be found here, or in any standard linear algebra textbook. 
So, running this algorithm on each of your bases will produce orthogonal bases for each eigenspace, and since eigenvectors corresponding to distinct eigenspaces are orthogonal (your matrix is symmetric), concatenating the bases will form an orthonormal basis of eigenvectors. 
Since this case is small enough, we really don't even need to use Gram-Schmidt, we just need to find two vectors in the span of the eigenvectors with eigenvalue $6$ that are orthogonal, and they'll be orthogonal to the one with eigenvalue $3$.
I set up $$\bigg(a\begin{bmatrix} 1\\1\\0\end{bmatrix} + b\begin{bmatrix}-1\\0\\1\end{bmatrix}\bigg)  \cdot \bigg(c\begin{bmatrix} 1\\1\\0\end{bmatrix} + d\begin{bmatrix}-1\\0\\1\end{bmatrix}\bigg) =0$$
and pretty quickly found choices of $a, b, c, d$ that work.
