Say I have a geometric shape in $R^2$ whose vertices are defined as ordered pairs in a $2$x$n$ matrix, through which lines segments are drawn to "connect the dots" so to speak.
For examples a square with a height of 1, with a corner on the origin and the rest in Q1 would be
$ \begin{bmatrix} 0 & 0 & 1 & 1 & 0\\ 0 & 1 & 1 & 0 & 0\\ \end{bmatrix} $
Why is it the case that any shape defined in such a way, when transformed by multiplication by a singular matrix $A$ results in a line? I know it has something to do with the fact that a $2$x$2$ defines via vectors a parallelogram whose area is equal to the determinate, and so a singular $2$x$2$ matrix defines one whose area is $0$.
But I can't find out why it would hold in a more general case such as the one I described.
A $2 \times 2$ matrix can have a 0-, 1-, or 2-dimensional image. If it has a 2-dimensional image, it is nonsingular (also called invertible). If it has a 0- or 1-dimensional image, is is called singular (also, non-invertible). An example of the former is $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$, which clearly only has the vector $(0,0)$ for its image. An example of the latter is $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, which maps $(x,y)$ to $(x,0)$, i.e. to a 1-dimensional space.
Another way to think about this is that singular matrices take some non-zero vector to the zero vector, so they take all multiples of that vector to the zero vector. This is a 1-dimensional subspace that is collapsed to a point. A parallel collapsing happens everywhere else -- only a space perpendicular to that original vector survives. (And in the 0-dimensional case, that perpendicular vector doesn't survive, because it is also sent to zero.)