# 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 $Q=0$

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Do you know what orthogonally diagonalization (i hate the word.) is? You can use it to take out all non-cross products. $Q^TAQ=D$ where D is what you want. Its maybe a bit more systematic. Also for $Q=0$ what does being 0 mean? Its a polynomial so its always $\geq 0$, which implies that $Q=0$ could be the lowest value possible. Diagonalization could help with that. –  WiseStrawberry Jul 7 '13 at 15:11

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

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It's weird this method is known in French as "réduction de gauss" see fr.wikipedia.org/wiki/R%C3%A9duction_de_Gauss and I have not found a similar article on wikipedia in English. –  Sami Ben Romdhane Jul 7 '13 at 16:00
@SamiBenRomdhane I've learned very recently it was called Lagrange reduction method in English... –  julien Jul 7 '13 at 16:08
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).