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 $Q=0$
On
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).
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) $$ \{(x,y,z)\in\mathbb{R}^3\,;\,x+2y+3z=0\}. $$