Diagonalizing symmetric real bilinear form I am given the following symmetric matrix:
$$
        A=\begin{pmatrix}
        1 & 2 & 0 & 1\\
        2 & 0 & 3 & 0\\
        0 & 3 & -1 & 1\\
        1 & 0 & 1 & 4\\
        \end{pmatrix}\in M_4(\Bbb R)
$$
Let $f\in Bil(V), f(u,v)=u^tAv.$
I want to find a base $B \subset \Bbb R^4$ such that the matrix representing $f $ in respect to $B$ is diagonal.
I took $
        v_1=\begin{pmatrix}
        1 \\
        0  \\
        0  \\
        0
        \end{pmatrix}
$ and found a vector space $V_2=\{v\in\Bbb R^4$| $f(v,v_1)=0\}$, and got $V_2 =sp\{ \begin{pmatrix}
        -2 \\
        1  \\
        0  \\
        0
        \end{pmatrix},\begin{pmatrix}
        0 \\
        0  \\
        1  \\
        0
        \end{pmatrix},\begin{pmatrix}
        -1 \\
        0  \\
        0  \\
        1
        \end{pmatrix}\} =sp\{v_2,v_3,v_4\}$
Now, the matrix in repect to $B=\{v_1,v_2,v_3,v_4\}$ looks like this:$$
        \begin{pmatrix}
        1 & 0 & 0 & 0\\
        0 & -4 & 3 & -2\\
        0 & 3 & -1 & 1\\
        0 & -2 & 1 & 3\\
        \end{pmatrix}
$$
I want to continue inductively, but I'm not sure how to procceed. 

The method of repeated completing squares also leads, by nature, to rational entries in this case. However, that is not the end of the story, as the given form is $SL_4 \mathbb Z$ equivalent to a diagonal form. In brief,
$$  (w + 2x+z)^2 + (x-2y+z)^2 - (9x-9y+13z)^2 + 19 (2x-2y+3z)^2 = w^2 - y^2 + 4z^2 + 4wx +6xy +2wz +2yz  $$
$$
A =
\left(
\begin{array}{rrrr}
1 & 2 & 0 & 1 \\
2 & 0 & 3 & 0 \\
0 & 3 & -1 & 1 \\
1 & 0 & 1 & 4
\end{array}
\right)
$$
$$
Q =
\left(
\begin{array}{rrrr}
1 & 2 & 0 & 1 \\
0 & 1 & -2 & 1 \\
0 & 9 & -9 & 13 \\
0 & 2 & -2 & 3
\end{array}
\right)
$$
and
$$
D =
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 19
\end{array}
\right)
$$
The one-line formula above is just
$$  Q^T D Q = A  $$
As far as the order the question was asked, we take
$$
P = Q^{-1} =
\left(
\begin{array}{rrrr}
1 & 2 & -6 & 25 \\
0 & -1 & 4 & -17 \\
0 & -1 & 1 & -4 \\
0 & 0 & -2 & 9
\end{array}
\right)
$$
to get
$$  P^T A P = D.  $$ Indeed,
$$
P^T  =
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
2 & -1 & -1 & 0 \\
-6 & 4 & 1 & -2 \\
25 & -17 & -4 & 9
\end{array}
\right)
$$ and
$$
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
2 & -1 & -1 & 0 \\
-6 & 4 & 1 & -2 \\
25 & -17 & -4 & 9
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 2 & 0 & 1 \\
2 & 0 & 3 & 0 \\
0 & 3 & -1 & 1 \\
1 & 0 & 1 & 4
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 2 & -6 & 25 \\
0 & -1 & 4 & -17 \\
0 & -1 & 1 & -4 \\
0 & 0 & -2 & 9
\end{array}
\right) =
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 19
\end{array}
\right)
$$
