Eigenvector of matrix of equal numbers For matrix the matrix
$$A = \begin{bmatrix}
3&1&1\\
1&3&1\\
1&1&3\\
\end{bmatrix}$$
with eigenvalues $\lambda_1=5$, $\lambda_2=2$, $\lambda_3=2$, I am trying to find the corresponding eigenvector corresponding to the eigenvalue 2. I got
$$(A - 2I_3) = \begin{bmatrix}
1&1&1\\
1&1&1\\
1&1&1\\
\end{bmatrix}$$
Reducing it (row reduced echelon form), I get:
$$\left[
\begin{array}
{ccc|c}
1&1&1&0\\
0&0&0&0\\
0&0&0&0\\
\end{array}\right]$$
Ending up with $x_1 + x_2 + x_3 = 0$. How would I find the eigenvector from there? Usually, I end up getting two equations and it's easy from there. How would you do it with one?
 A: For any square matrix with one value on the diagonal and another value everywhere else, a consistent pattern of (orthogonal) eigenvectors for the $n$ by $n$ case can be read from the columns of
$$    
 \left(  \begin{array}{rrrrrrrrrr}
  1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  2  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  3  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  4  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  5  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  6  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  0  &  7  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  0  &  0  &  8  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  0  &  0  &  0  &  9   
\end{array}
  \right).
  $$
In your case, the upper left 3 by 3 corner. If you want the result orthonormal you need to divide each column by a square root of something appropriate. I have displayed the  10 by 10 version, notice how the diagonal numbers go up to 9 = 10 - 1. 
A: You have a two-dimensional eigenspace, so there should be two linearly independent eigenvectors, that each satisfy $x_1+x_2+x_3=0$ for you to choose.
More details: Every eigenvalue has an associated eigenspace.  The dimension of this eigenspace is called the geometric multiplicity of the eigenvalue, while the multiplicity of the eigenvalue as a root of the characteristic polynomial is called the algebraic multiplicity of the eigenvalue.  These two constants satisfy $1\le GM(\lambda)\le AM(\lambda)$.  For the problem at hand, $\lambda=2$, and $GM(\lambda)=AM(\lambda)=2$.  The eigenspace corresponding to $\lambda=2$ is a two-dimensional vector space.  Every vector in it satisfies $x_1+x_2+x_3=0$.  What you're trying to find is a basis for this vector space.  @Julien gave an example in the comments above of such a basis, but it's not unique.  Any two linearly independent vectors that satisfy the equation will serve as a basis.  For example, $(3,-1,-2)$ and $(5,-10,5)$.
