Quadratic form in canonical form Reduce the quadratic form $q(x,y) = 6xy$ using the orthogonal reduction (i.e, find a orthogonal basis such that the matrix of the bilinear form is diagonal and $a_{ii} = 0$ or $a_{ii} = ^+_-1$)
What is exactly the process of "orthogonal reduction"? I mean, to find this basis, Which vector should I use to start my orthogonal basis? If I take $(0,1)$, for example, then $q(0,1) = 0$. Should I ever start with the canonic basis of $R^2$ if it doesnt say anything (should I admit using the canonic base)?
thanks in advance!
 A: This is known as orthogonal diagonalisation, and the process is outlined on Wikipedia.
First write $q(x, y)$ as
$$q(x, y) = [x\quad y]A\left[\begin{array}\ x\\ y\end{array}\right]$$ 
where
$$A = \left[\begin{array}\ 0 & 3\\ 3 & 0\end{array}\right].$$
Note that $A$ is symmetric. As such, it is orthogonally diagonalisable. That is, there is an invertible matrix $P$ satisfying $P^T = P^{-1}$ such that $P^TAP$ is a diagonal matrix. To find $P$, you need to find an orthonormal basis of eigenvectors; these form the columns of $P$. The eigenvectors of $A$ are $\lambda_1 = 3$ and $\lambda_2 = -3$ with eigenvectors $v_1 = [1\quad 1]^T$ and $v_2 = [-1\quad 1]^T$ respectively. As eigenvectors from distinct eigenvalues are orthogonal, normalising $v_1$ and $v_2$, we obtain an orthonormal basis: $\left\{\frac{1}{\sqrt{2}}v_1, \frac{1}{\sqrt{2}}v_2\right\}$. Therefore
$$P = \frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -1\\ 1 & 1\end{array}\right].$$
Note that
$$P^TAP = \frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & 1\\ -1 & 1\end{array}\right]\left[\begin{array}\ 0 & 3\\ 3 & 0\end{array}\right]\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -1\\ 1 & 1\end{array}\right] = \left[\begin{array}{cc} 3 & 0\\ 0 & -3\end{array}\right]$$
which is diagonal.
Now make the variable substitution
$$\left[\begin{array}\ \hat{x}\\ \hat{y}\end{array}\right] = P^{-1}\left[\begin{array}\ x\\ y\end{array}\right].$$ 
Then we have
\begin{align*}
q(x, y) &= 6xy\\
&= [x\quad y]A\left[\begin{array}\ x\\ y\end{array}\right]\\ 
&= \left[\begin{array}\ x\\ y\end{array}\right]^TA\left[\begin{array}\ x\\ y\end{array}\right]\\ 
&= \left(P\left[\begin{array}\ \hat{x}\\ \hat{y}\end{array}\right]\right)^TA\left(P\left[\begin{array}\ \hat{x}\\ \hat{y}\end{array}\right]\right)\\ 
&= [\hat{x}\quad \hat{y}]P^TAP\left[\begin{array}\ \hat{x}\\ \hat{y}\end{array}\right]\\ 
&= [\hat{x}\quad \hat{y}]\left[\begin{array}{cc} 3 & 0\\ 0 & -3\end{array}\right]\left[\begin{array}\ \hat{x}\\ \hat{y}\end{array}\right]\\
&= 3\hat{x}^2 - 3\hat{y}^2.
\end{align*}
Geometrically, $P$ corresponds to a rotation of the coordinate system by $\frac{\pi}{4}$. To see this, note that $P$ is the rotation matrix for $\theta = \frac{\pi}{4}$.
