Why do these eigenvalues give the zero vector for 3 eigenvectors? Given the quadratic form $q(x_{1}, x_{2}, x_{3}, x_{4}) = x_{1}^2 + x_{2}^2 - x_{3}^2 - x_{4}^2 + 4x_{1}x_{2} - 2x_{1}x_{4} + 2x_{3}x_{4}$, the eigenvalues I am getting are $\lambda = -1, -2.27, .14, 3.13$. However, all of these eigenvalues except $\lambda = -1$ are resulting in the zero-vector for the their respective eigenvectors, which is not possible by definition. What is wrong here?
Edit: The characteristic equation I am getting is $\lambda^4 - 8\lambda^2-6\lambda + 1 = 0$.

As far as the eigenvalues, the other answers are fine. 
A milder thing to do is find out how to make your symmetric matrix "congruent" to a diagonal matrix, then invoke Sylvester's Law of Inertia. Below, $PQ=QP=I.$
Two eigenvalues are positive and two negative. 
$$ P^T H P = D  $$
$$\left( 
\begin{array}{rrrr} 
1 & 0 & 0 & 0 \\ 
 - 2 & 1 & 0 & 0 \\ 
0 & 0 & 1 & 0 \\ 
 -  \frac{ 1 }{ 3 }  &  \frac{ 2 }{ 3 }  & 1 & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrrr} 
1 & 2 & 0 &  - 1 \\ 
2 & 1 & 0 & 0 \\ 
0 & 0 &  - 1 & 1 \\ 
 - 1 & 0 & 1 &  - 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrrr} 
1 &  - 2 & 0 &  -  \frac{ 1 }{ 3 }  \\ 
0 & 1 & 0 &  \frac{ 2 }{ 3 }  \\ 
0 & 0 & 1 & 1 \\ 
0 & 0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrrr} 
1 & 0 & 0 & 0 \\ 
0 &  - 3 & 0 & 0 \\ 
0 & 0 &  - 1 & 0 \\ 
0 & 0 & 0 &  \frac{ 1 }{ 3 }  \\ 
\end{array}
\right) 
$$
$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rrrr} 
1 & 0 & 0 & 0 \\ 
2 & 1 & 0 & 0 \\ 
0 & 0 & 1 & 0 \\ 
 - 1 &  -  \frac{ 2 }{ 3 }  &  - 1 & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrrr} 
1 & 0 & 0 & 0 \\ 
0 &  - 3 & 0 & 0 \\ 
0 & 0 &  - 1 & 0 \\ 
0 & 0 & 0 &  \frac{ 1 }{ 3 }  \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrrr} 
1 & 2 & 0 &  - 1 \\ 
0 & 1 & 0 &  -  \frac{ 2 }{ 3 }  \\ 
0 & 0 & 1 &  - 1 \\ 
0 & 0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrrr} 
1 & 2 & 0 &  - 1 \\ 
2 & 1 & 0 & 0 \\ 
0 & 0 &  - 1 & 1 \\ 
 - 1 & 0 & 1 &  - 1 \\ 
\end{array}
\right) 
$$
