Real Canonical Form Question: Let us consider the quadratic form $q: R^3 -> R$, $$q(x,y,z) = x^2+25y^2+10xy+2xz$$Find the corresponding symmetric bilinear form $f$ and a basis $B$ such that $[f]_B$ has the real canonical form. State the signature and rank of $[f]_B$.
I thought that the best way to start this question was to find an orthogonal basis, C, using the Gram Schmitd process. I got the vectors $(1,0,0),(-5,1,1),(\frac{5}{2},-\frac{1}{2},\frac{1}{2})$ as my orthogonal basis, but I dont know how to use these to find a basis B such that $[f]_B$ has the real canonical form. Any help would be greatly appreciated. Thanks.
 A: A way to proceed to get the signature and the rank is to reduce $q$ thanks to the Gauss' algorithm :
$$q\left(x,y,z\right)=x^2+25y^2+10xy+2xz
=\left(x+5y+z\right)^2-10yz-z^2
=\left(x+5y+z\right)^2-\left(z+5y\right)^2+25y^2.$$
Let then $X=\left(x+5y+z\right)$, $Y=\left(z+5y\right)$ and $Z=5y$. Then the reduced equation of $q$ is given by
$$X^2-Y^2+Z^2=0.$$
It is a cone (signature $\left(2,1\right)$, rank=$3$).
Furthermore, we have
$$\begin{pmatrix}X\\
Y\\
Z
\end{pmatrix}=\underbrace{\begin{pmatrix}1 & 5 & 1\\
0 & 5 & 1\\
0 & 5 & 0
\end{pmatrix}}_{=:P}\begin{pmatrix}x\\
y\\
z
\end{pmatrix}
 $$
and the matrix
$$P^{-1}=\begin{pmatrix}1 & -1 & 0\\
0 & 0 & \frac{1}{5}\\
0 & 1 & -1
\end{pmatrix}$$
is the passage matrix from the canonical basis to a base where $q$ as the above reduced form.
Verification :
The representative matrix in the canonical base of $\mathbb{R}^3$ of the polar bilinear form associated to $q$ is
$$\begin{pmatrix}1 & 5 & 1\\
5 & 25 & 0\\
1 & 0 & 0
\end{pmatrix}.$$
We have :
$$q\left(x,y,z\right)=\begin{pmatrix}x & y & z\end{pmatrix}\begin{pmatrix}1 & 5 & 1\\
5 & 25 & 0\\
1 & 0 & 0
\end{pmatrix}\begin{pmatrix}x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}x\\
y\\
z
\end{pmatrix}^{T}\begin{pmatrix}1 & 5 & 1\\
5 & 25 & 0\\
1 & 0 & 0
\end{pmatrix}\begin{pmatrix}x\\
y\\
z
\end{pmatrix}$$
$$=\left(P^{-1}\begin{pmatrix}X\\
Y\\
Z
\end{pmatrix}\right)^{T}\begin{pmatrix}1 & 5 & 1\\
5 & 25 & 0\\
1 & 0 & 0
\end{pmatrix}\left(P^{-1}\begin{pmatrix}X\\
Y\\
Z
\end{pmatrix}\right)=\begin{pmatrix}X\\
Y\\
Z
\end{pmatrix}^{T}\left(P^{-1}\right)^{T}\begin{pmatrix}1 & 5 & 1\\
5 & 25 & 0\\
1 & 0 & 0
\end{pmatrix}P^{-1}\begin{pmatrix}X\\
Y\\
Z
\end{pmatrix}$$
$$=\begin{pmatrix}X & Y & Z\end{pmatrix}\left(\left(P^{-1}\right)^{T}\begin{pmatrix}1 & 5 & 1\\
5 & 25 & 0\\
1 & 0 & 0
\end{pmatrix}P^{-1}\right)\begin{pmatrix}X\\
Y\\
Z
\end{pmatrix}
=\begin{pmatrix}X & Y & Z\end{pmatrix}\begin{pmatrix}1 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & 1
\end{pmatrix}\begin{pmatrix}X\\
Y\\
Z
\end{pmatrix}=X^{2}-Y^{2}+Z^{2}
 $$
hence $q$ is reduced on the basis associated to the matrix $P^{-1}$ (where we read its rank and its signature).
