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I found the following sentence in my linear linear algebra book (affine and projective geometry): $Q:V \to \mathbb{K}$ is a quadric (quadratic function) and $\alpha\in Aff(V)$. $Aff(V)$ is the set of all affin and invertible functions $\alpha$.

The set of all centres of $Q\circ\alpha$ is bijective on the set of all centres of $Q$.

There is no proof so I tried to prove it, but I have not many ideas. I formulated the statement a little bit different; I need to show that $\sigma_{\alpha(m)}=\alpha\circ\sigma_m\circ\alpha^{-1}$ where $\sigma_m$ describes a reflection on $m\in V$.

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up vote 1 down vote accepted

So, we have a conic $\mathcal{C} = \begin{pmatrix} c & B^{t} \\ B & A\end{pmatrix}$, and $v$, wich is the centre of $\mathcal{C}$. We also consider $\varphi: V \rightarrow V $ affinity; $M(\varphi)=\begin{pmatrix} 1 & S \\ 0 & D^{-1}\end{pmatrix}$

What we want to prove is that $\varphi(v)$ is the centre of $\varphi(\mathcal{C})$.

I'll assume that you know that: $v$ is the centre of $\mathcal{C} \Leftrightarrow AX= -B$, ask for a prove.

Now it' time for the real prove:

$\tilde{\mathcal{C}}=M(\varphi)^{t} \space \mathcal{C} \space M(\varphi)=\begin{pmatrix} c+S^{t}B+B^{t}S+S^tAD^{-1} & B^tD^{-1}+S^tAD^{-1} \\ (D^{-1})^tB+(D^{-1})^tAS & (D^{-1})^tAD^{-1}\end{pmatrix}$.

Let's call $\varphi(X)=\tilde{X}$.

$D^{-1}\tilde{X}+S=X \rightarrow \tilde{A}\tilde{X}=\tilde{-B}$ (that's the claim). $\tilde{A}\tilde{X}=\tilde{-B} \Leftrightarrow (D^{-1})^tAD^{-1}(D(X-S))= -(D^{-1})^tB+(D^{-1})^tAS \Leftrightarrow (D^{-1})^t(AX-AS) = -(D^{-1})^t(-AX)+(D^{-1})^tAS \Leftrightarrow AX-AS=AX-AS $.

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Thanks for your answer, but I have some problems understanding your notation. What are A and B ? Where is the relation that $Q\circ\alpha$ is bijective to $Q$ ? May you could explain your answer a little bit more, at the moment I simply see no relation. – Montaigne Nov 27 '12 at 17:30
Here is the idea , my notation is a bit simplified, I did not need so much precision. – Ivan Nov 27 '12 at 18:16

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