# Matrix associated to this quadratic form

I have the quadratic form

$$q(x, y) = x^2 + 6xy + 9y^2 - 6x$$

How can I find the associated matrix? The $6x$ term destroys all.

I tried to write it as

$$q(x, y) = (x+3y)^2 - 6x$$

But then?

OK UNDERSTOOD PART 1

The professor told us: "treat it as a second degree without the translational term $-6x$, and then after having found the matrix and bla bla, draw and find the whole new figure".

Ok, honestly I have done how she asked us. I found the $A$ matrix of the quadratic form without $-6x$

$$\begin{pmatrix} 1 & 3 \\ 3 & 9 \end{pmatrix}$$

Eigenvalues are $\lambda = 0, 10$ and eigenvectors are (non normalized) $\{-3, 1\}$ and $\{1, 3\}$

Now there is a problem: the matrix of the bad change has NEGATIVE determinant.

So there is also a reflection, in addition to a rotation.

Anyway how can I proceed from here?

• You do realize that there’s no such thing as the eigenvectors, I hope. Multiply one of them by $-1$ and the “reflection” that’s bothering you disappears. That aside, you really ought to go back to your professor to clarify what it is she wants you to do. There are two reasonable possibilities: that she wants you to convert $q(x,y)$ into the form $(x,y)A(x,y)^T+\mathbf b^T(x,y)^T$ or the homogeneous form $(x,y,1)Q(x,y,1)^T$. – amd Dec 21 '17 at 3:15

This isn't a quadratic form precisely due to the presence of the $-6x$ term. When $q$ is a quadratic form, you always have $q(\lambda x,\lambda y)=\lambda^2q(x,y)$. That's clearly not the case here.
Consider $A=\begin{pmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{pmatrix}$, then you get $$(x,y)A\begin{pmatrix}x\\y\end{pmatrix}=(x,y)\begin{pmatrix}a_{11}x+a_{12}y\\a_{21}x+a_{22}y\end{pmatrix}=a_{11}x^2+(a_{12}+a_{21})xy+a_{22}y^2.$$ So you see, that your function can't be a quadratic form, since each quadratic form has to have the form $q(x,y)=\alpha x^2+\beta xy+\gamma y^2$.