Finding mapping transform in homogenous coordinate system using vanishing points

Question

This question has been bothering me for some time, help would be appreciated! Suppose we have an image of a building facade with vanishing points at Vx = (x,0) and Vy = (0,y) which are horizontal and vertical respectively. We know two points P = (0,0) and Q = (1,1). We want to find a transform that maps this facade onto a rectangle that keeps P and Q fixed.

Since this is a homogenous coordinate system I set P = (0,0,1) and Q = (1,1,1), Vx = (x,0,0) and Vy = (0,y,0). Then I tried to find a transform that maps this onto the following: (0,0,1), (1,1,1,), (1,0,1), (0,1,1). However when I tried this I was wrong according to the answers. The answer ending up being:

$\begin{bmatrix}y-x+xy & 0 & 0\\ 0 & y-x+xy & 0 \\ y & -x & xy\end{bmatrix}$

I feel like I'm approaching this question wrongly so any bit of advice that can help me understand how to tackle this question would be much appreciated!

Answer 1

I'd use this computation to find the transformation matrix. Applied to your situation:

1. Start with the matrix $$ A = \begin{pmatrix}x&0&0\\0&y&0\\1&1&1\end{pmatrix} $$ $A$ is formed by the points $V_x$, $V_y$ and $P$, correcting for the fact that $V_x$ and $V_y$ presumably are finite and therefore have a $1$ in their last coordinate.

2. The adjoint of this is $$ \operatorname{adj}A = \begin{pmatrix}y&0&0\\0&x&0\\-y&-x&xy\end{pmatrix} $$

3. Multiplying that with $Q$ you get $$ \operatorname{adj}A\cdot Q=\begin{pmatrix}y\\x\\xy-x-y\end{pmatrix} $$

4. Scaling the columns of $A$ by these coefficients you get $$ B = \begin{pmatrix}xy&0&0\\0&xy&0\\y&x&xy-x-y\end{pmatrix} $$ This matrix will map the projective basis $(1,0,0)$, $(0,1,0)$, $(0,0,1)$, $(1,1,1)$ to $V_x$, $V_y$, $P$, $Q$.

5. So you want the inverse of this operation, or the adjoint of this matrix. \begin{align*}\operatorname{adj}B&=\begin{pmatrix} x^{2} y^{2} - x^{2} y - x y^{2} & 0 & 0 \\ 0 & x^{2} y^{2} - x^{2} y - x y^{2} & 0 \\ - x y^{2} & - x^{2} y & x^{2} y^{2}\end{pmatrix} \\&=xy\cdot\begin{pmatrix} x y - x - y & 0 & 0 \\ 0 & x y - x - y & 0 \\ - y & - x & x y \end{pmatrix}\end{align*} You can safely drop that factor $xy$ and take that final matrix as your transformation matrix. It looks very similar to what you have, except for some sign changes.