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...
 A: 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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A: The trick is to use a nonzero diagonal entry to kill all off-diagonal entries below it or on its right, by simultaneous row and column reduction:
\begin{align*}
\underbrace{\begin{bmatrix} 1 & 0 & 0\\ -4 & 1 & 0\\ -4 & 0 & 1 \end{bmatrix}}_{P^T}
\begin{bmatrix} 1 & 4 & 4\\ 4 & 4 & 10\\ 4 & 10 & 16 \end{bmatrix}
\underbrace{\begin{bmatrix} 1 & -4 & -4\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}}_P
&=\begin{bmatrix} 1 & 0 & 0\\ 0 & -12 & -6\\ 0 & -6 & 0 \end{bmatrix},\\
\underbrace{\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & -\tfrac12 & 1 \end{bmatrix}}_{Q^T}
\begin{bmatrix} 1 & 0 & 0\\ 0 & -12 & -6\\ 0 & -6 & 0 \end{bmatrix}
\underbrace{\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & -\tfrac12\\ 0 & 0 & 1 \end{bmatrix}}_Q
&=\underbrace{\begin{bmatrix} 1 & 0 & 0\\ 0 & -12 & 0\\ 0 & 0 & 3 \end{bmatrix}}_D.
\end{align*}
Hence $Q^TP^TAPQ=D$, i.e. $A=S^TDS$, where
$$
S=(PQ)^{-1}
=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & \tfrac12\\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} 1 & 4 & 4\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}
=\begin{bmatrix} 1 & 4 & 4\\ 0 & 1 & \tfrac12\\ 0 & 0 & 1 \end{bmatrix}.
$$
In other words, if you define $u=Sx$ and $v=Sy$, then $x^TAy=u^TDv=u_1v_1-12u_2v_2+3u_3v_3$.
The above process is an example of diagonalisation via congruence. (See here for another example.) This example is easy, because in the course of diagonalisation, we can always use a nonzero diagonal entries to eliminate the other entries on the same row or the same column. Sometimes, when the diagonal entry is zero, we need to 'borrow' numbers from other off-diagonal entries. See here for such an example.
