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Suppose I have a $2\times 2$ matrix $M$ in implicit equation form, where $(x,y)$ is transformed to $( xOut,yOut )$: $$ ax + by = xOut, \quad cx + dy = yOut $$ Now additionally I have a line $L$, also in implicit equation form: $$ ex + fy = g $$ I want to take line $L$ and transform it into the line that results from all its points being transformed by matrix $M$. One way to do it is to parameterize (assuming $f \neq 0$) and solve equations: $$ ( x,y ) = ( x,(ex-g)/f ) $$ $$ ax + b(ex-g)/f = xOut, \quad cx + d(ex-g)/f = yOut $$ $$ (a + be/f)x - g/f = xOut,\quad (c + de/f)x - g/f = yOut $$ $$ (c+de/f)xOut - (a+be/f)yOut = (-(c+de/f)g/f - (a+be/f)g/f) $$ That gives me the new line equation, which is good, but there is that pesky $f$ in the denominator. A line that is just $x = c$ should be just as valid after a transform. Of course I could go through the whole process again with the requirement that $e \neq 0$ instead, but that seems wasteful. Is there a more direct way of combining these three equations without parameterizing? They are all linear but I can't see how to put them all together in one matrix, where the output of one system is the input of another system.

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Can you format your post using the TeX commands for all the math? Otherwise, it's hard to read. –  JohnD Jan 8 '13 at 23:24
Yeah I know. I wish it would just format simple equations... –  Dan Jan 8 '13 at 23:26
Cheers to whoever just centered and italicized the equations. –  Dan Jan 8 '13 at 23:49

1 Answer 1

You will not necessarily get a new line by acting $M$ on $L$. Consider for example the zero matrix $M=O_{2\times 2}$.

Let the line $L=\{x\in{\Bbb R}^2:a\cdot x=c\}$ , where $0\neq a\in{\Bbb R}^2$ and $c\in{\Bbb R}$ are constants. Then the image of $L$ under $M$ (the output) is $$ M(L)=\{Mx:a\cdot x=c,x\in{\Bbb R}^2\}. $$ If $M$ is invertible, $M(L)=\{y\in{\Bbb R}^2:a\cdot(M^{-1}y)=c\}.$

If $M$ is not invertible, you get $M(L)=\{M(x_0+kv):k\in{\Bbb R}\}$ where $x_0$ is a point on $L$ and $v$ is such that $a\cdot v=0$. Let $(x_1',x_2')^T=Mx_0+kMv$ and cancel out $k$, you would get the equation for your new line (or a point only).

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