Orthogonal basis for this indefinite symmetric bilinear form Find an orthogonal basis for the bilinear form over $\mathbb{R}$ given by $(\mathbf{x}, \mathbf{y})\mapsto\mathbf{x}^{t}A\mathbf{y}$ where $A=\begin{bmatrix} 1 & 4 & 4\\ 4 & 4 & 10\\ 4 & 10 & 16 \end{bmatrix}$. 
I'm not sure if this is as easy as using Gram-Schmidt, or if there is another way. I used Gram-Schmidt and obtained very complicated fractions in my vectors, so I have a feeling that this is wrong...

PREFACE: this is called Hermite reduction. It is, for example, Theorem 23 in The Arithmetic Theory of Quadratic Forms by Burton W. Jones, pages 56-59 primarily. On page 58 you clearly see an upper triangular matrix as the change of variables. The W stands for Wadsworth.  
I get 
$$ x^2 + 4 y^2 + 16 z^2 + 20 y z + 8 z x + 8 x y = (x + 4 y + 4 z)^2 - 12 \left( y + \frac{z}{2} \right)^2 + 3 z^2,  $$ so
$$ x^2 + 4 y^2 + 16 z^2 + 20 y z + 8 z x + 8 x y = (x + 4 y + 4 z)^2 - 3 \left( 2y + z \right)^2 + 3 z^2.  $$
Note that this was very, very little work. Nowhere near finding eigenvectors. 
This is called Hermite reduction.
The relevant matrix equation is
$$  
 \left(  \begin{array}{ccc}
  1  &  0 & 0  \\
   4   &  2 & 0 \\
   4 & 1 & 1  
\end{array} 
  \right)  
 \left(  \begin{array}{ccc}
  1  &  0 & 0  \\
   0   &  -3 & 0 \\
   0 & 0 & 3  
\end{array} 
  \right)  
 \left(  \begin{array}{ccc}
  1  &  4 & 4  \\
   0   &  2 & 1 \\
   0 & 0 & 1  
\end{array} 
  \right)  
 =
 \left(  \begin{array}{ccc}
  1  &  4 & 4  \\
   4   &  4 & 10 \\
   4 & 10 & 16  
\end{array} 
  \right).  
 $$
Either that, or they really want the inverses, as in
$$  
 \left(  \begin{array}{ccc}
  2  &  0 & 0  \\
   -4   &  1 & 0 \\
   -4 & -1 & 2  
\end{array} 
  \right) 
 \left(  \begin{array}{ccc}
  1  &  4 & 4  \\
   4   &  4 & 10 \\
   4 & 10 & 16  
\end{array} 
  \right) 
 \left(  \begin{array}{ccc}
  2  &  -4 & -4  \\
   0   &  1 & -1 \\
   0 & 0 & 2  
\end{array} 
  \right)  
 =
 \left(  \begin{array}{ccc}
  4  &  0 & 0  \\
   0   &  -12 & 0 \\
   0 & 0 & 12  
\end{array} 
  \right).  
 $$
Either way, less work than eigenvectors. 
EDDDIITTTT: @copper.hat has confirmed that the eigenvalues are horrible, I get the characteristic polynomial as $$ \lambda^3 - 21 \lambda^2 - 48 \lambda + 36  $$ with three irrational real roots, so this is the Casus Irrededucibilis and so the eigenvectors are horrible. Which tells me they really want Hermite. 
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