2
$\begingroup$

The transformation $A:\mathbb{R^4}\to \mathbb{R^4}$, represented by the matrix:

$A$ =$ \begin{bmatrix} 3 & 0 & 1 &0 \\ 0 & -3 & 0 & 1 \\ -1 & 0 & 1 & 0 \\ 0&-1&0&-1 \end{bmatrix} $ induces a projectivity $\varphi_A:\mathbb{P^3}\to \mathbb{P^3}$

consider the line $r = \begin{cases} x_0-x_1+x_3=0 \\ 2x_0-x_1-2x_2=0 \end{cases}$, $r\in \mathbb{P^3}$

How to find the equation of the line $s=\varphi_A^{-1}(r)$?

Could I say (?): let $P = [x_0, x_1, x_2, x_3]$ be the generic point on $\mathbb{P^3}$, $P \in \varphi_A^{-1}(r) \Leftarrow\Rightarrow \varphi_A(P)\in r.$ So I find $A(P)$ = $ \begin{bmatrix} 3x_0+x_2 \\ -3x_1+x_3 \\ -x_0+x_2 \\ -x_1-x_3 \end{bmatrix} $ and, replacing in $r$, I obtain: $s = \begin{cases} 3x_0+2x_1+x_2-2x_3=0 \\ 8x_0+3x_1-x_3=0 \end{cases}$

Thank you.

$\endgroup$
1
$\begingroup$

Your method works. Another method is to use $A$ directly: If you have a point transform given by the invertible homogeneous matrix $A$, i.e., $\mathbf p' = A\mathbf p$, then planes transform as $\mathbf\pi'=A^{-T}\mathbf\pi$ because $$\mathbf\pi^T\mathbf p = 0 \iff \mathbf\pi^T(A^{-1}\mathbf p') = (A^{-T}\mathbf\pi)^T\mathbf p'=0.$$

Your line is described as the meet of the planes $\mathbf\pi_1=[1:-1:0:1]^T$ and $\mathbf\pi_2=[2:-1:-2:0]^T$, so the inverse image of this line is the meet of $A^T\mathbf\pi_1 = [3:2:1:-2]^T$ and $A^T\mathbf\pi_2 = [8:3:0:-1]^T$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.