So, suppose I had the position vector $\begin{bmatrix}x\\y\end{bmatrix}$ and I wanted to translate it so that it's tale is at $\begin{bmatrix}a\\b\end{bmatrix}$. I know I want my translation matrix yo yield $\begin{bmatrix}a+x\\b+y\end{bmatrix}$. However, I'm having trouble coming up with a $2×2$ matrix for this. I can come up with the folloiwng $3×3$ matrix: $T=\begin{bmatrix}1 & 0 & a\\0 & 1 & b \\ 0 & 0 & 0\end{bmatrix}$
Then, multiplying the point $\begin{bmatrix}x\\y\\1\end{bmatrix}$ by $T$ yields $\begin{bmatrix}a+x\\b+y\\0\end{bmatrix}$. However, the best $2d$ matrix I could come up with is $\begin{bmatrix}1 & \frac{a}{y} \\ \frac{b}{x} & 1 \end{bmatrix}$. Of course, this doesn't work if my position vector is on the $x$ or $y$ axis. I feel like this should be an easy problem to solve, but I seem to hit a road block. Any suggestions?
Best Answer
It's not possible to write the transformation you're looking at as a two-dimensional linear transformation. Consider what happens to the zero vector: We know that for any $2\times 2$ matrix $M$, the we should have $M\cdot \mathbf{0} = \mathbf{0} \cdot M = \mathbf{0}$. The trick you've found to augment the vector with an extra column and then use $3 \times 3$ matrices instead is exactly one of the standard ways of representing affine transformations.