Matrix of singular transformation

Question

I'm dealing with an affine map $g$ that maps $A= (0,1)$ to $A'=(0,0)$, $B=(2,0)$ to $B'=(0,1)$ and $C=(2,1)$ to $C'=(0,2)$, and I'm being asked to find the matrix representation of this map. I figured finding the translation vector would be a good place to start, so I found the coefficients $\alpha, \beta, \gamma$ such that $\alpha (0,1)+\beta(2,0)+\gamma(2,1)=(0,0)$ and $\alpha+\beta+\gamma=1$. This gave me $g(0,0)=(0,-1)$.

Now my current problem is finding the matrix representation of $g$. The transformed points do not form an affine basis for $\mathbb{R}^2$, so the matrix consisting of the vectors $\overrightarrow{A'B'}$ and $\overrightarrow{A'C'}$ in its columns does not give the correct answer. Surely something like this can be done, or am I stuck solving for each coefficient of the matrix?

The answer that seems to work is the matrix with the vector $(0,1)$ in both its columns.

Edit: perhaps I figured out why this isn't giving me the correct answer. Would I need to check to where $g$ maps the standard basis vectors $(1,0)$ and $(0,1)$ instead of my affine basis $(A,B,C)$ vectors? This seems to give the correct answer. Will this always work?

Answer 1

The map can be expressed as $$ \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} b_1 \\ b_2 \end{bmatrix} $$ For the non-singular case, you should be able to solve for the 6 unknowns ($a_{ij}, b_i$) using the 6 equations that you have: $$ \begin{align} 0 &= a_{11}(0) + a_{12}(1) + b_1\\ 0 &= a_{21}(0) + a_{22}(1) + b_2\\ 0 &= a_{11}(2) + a_{12}(0) + b_1\\ 1 &= a_{21}(2) + a_{22}(0) + b_2\\ 0 &= a_{11}(2) + a_{12}(1) + b_1\\ 2 &= a_{21}(2) + a_{22}(1) + b_2 \end{align} $$ In matrix form: $$ \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 2 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 2 & 0 & 0 & 1 \\ 2 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 2 & 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} a_{11} \\ a_{12} \\ a_{21} \\ a_{22} \\ b_1 \\ b_2 \end{bmatrix} $$ For the singular case you will need to compute a pseudoinverse to find the unknowns. For the data you have provided an inverse appears to exist and you don't need a pseudoinverse.

Answer 2

Let $g(x) = Mx+b$. If $g$ is affine, it can always be written in this form.

We just need to determine $M,b$.

Since $A+B-C = 0$ (and $1+1-1 = 1$) we have $g(0) = b = g(A+B-C) = g(A)+g(B)-g(C) = (0,-1)^T$.

Note that $Mx = g(x)-b$ and $A,B$ are linearly independent, so $M$ is completely specified by its behaviour on $A,B$.

$M(0,1)^T = (1,1)^T, M(2,0)^T = (0,2)^T$, hence $M=\begin{bmatrix} 0 & 0 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 1 & 0 \end{bmatrix}^{-1} = \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$.

It is straightforward to check that this gives the correct results.

Answer 3

The columns of a transformation matrix are the images of the basis vectors. That makes things easy enough: assemble the image vectors into a matrix. The catch is that in problems like this one the resulting transformation matrix is not expressed relative to the standard basis: vectors in its domain are expressed relative to the basis $(\overrightarrow{AB},\overrightarrow{AC})$, that is, instead of mapping the coordinate tuple $(2,-1)$ to $(0,1)$, it maps the tuple $(1,0)$ to $(0,1)$. So, you instead have to either determine what $g$ does to the standard basis vectors, or perform a change of basis on the matrix that you’ve constructed.

If you work in homogeneous coordinates, which is what I believe you have to do anyway to obtain a matrix representation of a transformation that involves a translation, you can write down an expression for the transformation matrix directly from the given conditions: $$\begin{bmatrix}A'&B'&C'\\1&1&1\end{bmatrix} \begin{bmatrix}A&B&C\\1&1&1\end{bmatrix}^{-1}.$$ The matrix on the right maps the three given source points to the standard basis vectors, and the matrix on the left maps the standard basis vectors to the appropriate outputs.

Answer 4

Here is a more geometric approach. You can find the linear part of an affine transformation by looking at how it transforms vectors between pairs of points - the translation part of the affine transformation then cancels out.

The vector $B \rightarrow C$ is $(0,1)$. This transforms into the vector $B' \rightarrow C'$ which is also $(0,1)$. This tells us that $a_{12}=0$ and $a_{22}=1$.

The vector $A \rightarrow C$ is $(2,0)$. This transforms into the vector $A' \rightarrow C'$ which is $(0,2)$. This tells us that $a_{11}=0$ and $a_{21}=1$.

So we have

$x' = b_1$

$y' = x + y + b_2$

for some $b_1$ and $b_2$. Finally, since $A=(0,1)$ transforms to $A'=(0,0)$ we can see that $b_1=0$ and $b_2=-1$.