diagonalize quadratic form I have this quadratic form
$Q= x^2 + 4y^2 + 9z^2 + 4xy + 6xz+ 12yz$
And they ask me:
for which values of $x,y$ and $z$ is $Q=0$?
and I have to diagonalize also the quadratic form.
I calculated the eigenvalues: $k_{1}=0=k_{2}, k_{3}=14$, 
and the eigenvector $v_{1}=(-2,1,0), v_{2}=(1,2,3), v_{3}=(3,6,-5)$ 
I don't know if this is usefull in order to diagonalize or to see when is $Q=0$
 A: You can also do that without ever seeing a matrix, by repeated square completions: that's called Lagrange reduction method. I am not saying that's the best way to answer such a question in general, although it is quite efficient in low dimensions. And here it can be done really fast: there is only one step.
$$
x^2 + 4y^2 + 9z^2 + 4xy + 6xz+ 12yz
$$
$$
=\underbrace{x^2+4x\left(y+\frac{3}{2}z\right)}_{\mbox{square to be completed}}+4y^2 +9z^2+12yz
$$
$$
=\left(x+2\left(y+\frac{3}{2}z\right)\right)^2-\left(2\left(y+\frac{3}{2}z\right)\right)^2+4y^2 +9z^2+12yz
$$
$$
=(x+2y+3z)^2.
$$
Conclusion: the quadratic form $Q$ is positive semidefinite with signature $(1,0)$. And it is zero on the hyperplane (=two-dimensional space)
$$
\{(x,y,z)\in\mathbb{R}^3\,;\,x+2y+3z=0\}.
$$
A: If $v=(x,y,z)$ is an eigenvector for the eigenvalue $0$, then $v^\mathrm{T} Q v = v^\mathrm{T} 0\cdot v = 0$; but you also have $Q(x,y,z)= v^\mathrm{T} Q v$ by definition, so you obtain a solution.
Reciprocally, if $Q(x,y,z)=0$, then $v^\mathrm{T} Q v = 0$. Here, if I recall the theorems about the kernel of a bilinear form in finite dimension, this is equivalent to saying that $v\in\ker Q$.
So to solve $Q(v)=0$, you only need the eigenvectors for the eigenvalue $0$; however, for the diagonalization, all eigenvectors are going to be useful (in an orthonormal basis of eigenvectors, the linear form $Q$ will be diagonal with the eigenvalues on its diagonal; and the transform matrix to get from the canonical basis to this basis of eigenvectors is easily derived from the eigenvectors themselves, once you have them).
