6. Making projective transformations

First, we need to choose names for special points. The first three have homogeneous coordinates that are standard basis vectors and so will be handy for working with matrices.

Let $X = pt (1,0,0) _ h$ (at infinity on the $x$-axis);

let $Y = pt (0,1,0) _ h$ (at infinity on the $y$-axis);

let $O = pt (0,0,1) _ h$ (the ordinary point $(0,0)$, i.e., the origin);

let $E = pt (1,1,1) _ h$ (the ordinary point $(1,1)$).

It is important not to confuse $X$ with $(1,0)$, which is not at infinity and has homogeneous coordinates $(1,0,1)_h$.

Problem 6.1 . Find a projective transformation $T:$   P$_2 \rightarrow$   P$_2$ for which $T(X) = P$, $T(Y) = Q$, and $T(O) = R$, where $P = (2,4)$, $Q = (4,1)$, and $R = (6,3)$.

Solution. Just write down a matrix whose rows are $P,Q,R$ in homogeneous coordinates: $A = \left[\begin{array}{ccc} 2&4&1\\ 4&1&1\\ 6&3&1 \end{array}\right]$. Then $T(X)$ $=$ $pt (1,0,0)A$ $=$ $P$, and so on.

Problem 6.2 . In Problem , was that the only solution?

Solution. No: Any choice of homogeneous coordinates for $P,Q,R$ would work. In other words, if $r,s,t$ are any nonzero scalars, any matrix of the form $\left[\begin{array}{ccc} r \cdot 2& r \cdot 4& r \cdot 1\\ s \cdot 4& s \cdot 1& s \cdot 1\\ t \cdot 6& t \cdot 3& t \cdot 1 \end{array}\right]$ is an answer to Problem .

Because there is some freedom in the answer, let's see if we can use that freedom to specify where another point goes:

Problem 6.3 . Find a projective transformation $T$ for which $T(X) = P$, $T(Y) = Q$, $T(O) = R$, and also $T(E) = S$, where $P,Q,R$ are as in Problem and $S = (6,8)$.

Solution. Write down the more general solution to Problem with $r,s,t$, as described in the solution to Problem . We would like

$(1,1,1)_h \left[\begin{array}{ccc} r \cdot 2& r \cdot 4& r \cdot 1\\ s \cdot ... ...s \cdot 1\\ t \cdot 6& t \cdot 3& t \cdot 1 \par\end{array}\right] = (6,8,1)_h$.

This is the same as the set of linear equations

$2r + 4s + 6t = 6$

$4r + 1s + 3t = 8$

$1r + 1s + 1t = 1$.

In matrix form: $\left[\begin{array}{ccc} 2&4&6\\ 4&1&3\\ 1&1&1 \par\end{array}\right] \left[... ... \par\end{array}\right] = \left[\begin{array}{c} 6\\ 8\\ 1 \end{array}\right]$.

Here we see that the coefficient matrix has columns that are the extended-vector forms of $P,Q,R$. (In other words, it is the transpose of the simplest answer to Problem .) These equations also say that $\hat S$ $=$ $r \hat P + s \hat Q + t \hat R$ (extended vectors).

Calculation (say by Gauss-Jordan) shows that the solution is $r = 1$, $s = -2$, $t = 2$. In the general solution to Problem , substitute these values. We get the answer $\left[\begin{array}{rrr} 2&4&1\\ -8&-2&-2\\ 12&6&2 \end{array}\right]$,

Definition. Four points in R$^ 2$ or P$_2$ are said to be in general position if no three are collinear (i.e., on the same line). Let's say that four points in general position form a quartet¹.

For example, in Problem , $P,Q,R,S$ are in general position and so make a quartet. $X,Y,O,E$ also make a quartet, which we can call the standard quartet.

Fact 6.4 . There is a projective transformation taking $X,Y,O,E$ to any given quartet $P,Q,R,S$ in P$_2$, with $X \rightarrow P$ and so on.

The reason is that having no three of the four points be in a line is just what is needed to guarantee that in the method of Problem , the coefficient matrix is nonsingular and none of $r,s,t$ come out zero.

Problem 6.5 . Describe a general method of finding a projective transformation that takes one given quartet $P,Q,R,S$ to another $P',Q',R',S'$, with $P \rightarrow P'$ and so on.

Solution. We can use the same idea as for affine transformations, but with the standard quartet in place of the standard triangle. First find a $3 \times 3$ matrix $A$ that gives a projective transformation taking $X,Y,O,E$ to $P,Q,R,S$. Then find another matrix $B$ that gives a projective transformation taking $X,Y,O,E$ to $P',Q',R',S'$. The answer to the problem is $A^{-1} B$. (By Problem , $A$ and $B$ are guaranteed to be nonsingular, so $A^{-1} B$ exists and is nonsingular, as it should be for a projective transformation.)

Problem 6.6 . Find a projective transformation taking the four points $(1,0)$, $(0,1)$, $(0,0)$, $(1,1)$ (the vertices of the standard square) to $P,Q,R,S$ of Problem .

Solution. Use the method of Problem . To take the standard quartet to the square, we must solve

$\left[\begin{array}{ccc} 1&0&0\\ 0&1&0\\ 1&1&1 \par\end{array}\right] \left[... ... \par\end{array}\right] = \left[\begin{array}{c} 1\\ 1\\ 1 \end{array}\right]$.

The solution is easily found to be $r = 1$, $s = 1$, $t = -1$. Therefore $A = \left[\begin{array}{rrr} 1&0&1\\ 0&1&1\\ 0&0&-1 \end{array}\right]$. We can invert $A$, and for this particular $A$ it turns out that $A^{-1} = A$. $B$ was already found in Problem . We get $A ^ {-1} B = \left[\begin{array}{rrr} 1&0&1\\ 0&1&1\\ 0&0&-1 \par\end{array}... ...ht] = \left[\begin{array}{rrr} 14&10&3\\ 4&4&0\\ -12&-6&-2 \end{array}\right]$.