0. Mapping one triangle to another with an affine transformation

First, consider the standard triangle in R$^ 2$ whose vertices are $(1,0)$, $(0,1)$, $(0,0)$, which we can call e$^ {(1)},$   e$^ {(2)},$   0. Let $PQR$ be any triangle.

Problem 0.1 . Find the extended matrix of an affine transformation $T$ that takes the standard triangle to the triangle $PQR$ (with $T($e$^ {(1)}) = P$, etc.). (See Figure .)

Figure: Standard triangle to a given triangle.

book/03dir/standard.eps

Solution. Let the triangle $P' Q'$   0 be the result of translating the triangle $PQR$ so that $R$ goes to the origin, by subtracting $R$ from all three vertices. Thus $P' = P-R$ and $Q' = Q-R$. The desired affine transformation consists of (1) a homogeneous linear transformation taking e$^ {(1)}$ to $P'$ and e$^ {(2)}$ to $Q'$, followed by (2) translation by $R$. Because e$^ {(1)}$ and e$^ {(2)}$ are the standard basis vectors, the rows of the desired $2 \times 2$ matrix for (1) are just their images. Putting (1) and (2) together, we see that the desired transformation has extended matrix $\left[\begin{array}{cc}(P-R)&0\\ (Q-R)&0\\ R&1\end{array}\right]$ (to use an obvious shorthand notation). (See Figure .)

Figure: Standard triangle to a given triangle: solution

book/03dir/solution.eps

For example, to map the standard triangle to the triangle with vertices $(2,1), (4,3), (2.5,4)$, the extended matrix is ${\left [\begin{array}{ccc} -.5&-3&0\\ 1.5&-1&0\\ 2.5&4&1 \end{array} \right ] }$.

(To check that this matrix works, try multiplying the extended vertices $(1,0,1)$, $(0,1,1)$, $(0,0,1)$ by it.)

Problem 0.2 . Find how to calculate the extended matrix of an affine transformation $U$ that takes the triangle $PQR$ to the triangle $P' Q' R'$ (with $U(P) = P'$, etc.). (See Figure .)

Figure: Arbitrary triangle to arbitrary triangle

book/03dir/triangle.eps

Solution. Use the standard triangle as an intermediate stage. You know how to find $T$ taking the standard triangle to $PQR$ and $T'$ taking the standard triangle to $P' Q' R'$. $T ^ {-1}$ takes $PQR$ back to the standard triangle. Therefore $U = T ^ {-1} T'$, so that $U$ has extended matrix

$\left[\begin{array}{cc}(P-R)&0\\ (Q-R)&0\\ R&1\end{array}\right] ^ {-1} \left[\begin{array}{cc}(P' -R')&0\\ (Q' -R' )&0\\ R' &1\end{array}\right]$.

(For a specific example you could put in numbers and multiply out. If $P,Q,R$ are noncollinear, so that $PQR$ is a genuine triangle, then the left matrix will be invertible as required. Computationally, one way to calculate a product $A ^ {-1} B$ is to write down an extended matrix $[A \vert B]$ and row-reduce it to get a matrix $[I \vert C]$; here $C$ will be $A ^ {-1} B$.)

Note. In R$^ 3$, you can use the same sort of method to move a tetrahedron to a tetrahedron.