I just finished an introductory class in Linear Algebra and I'm a bit confused about the geometric representation of a determinant.
Let's say there exists two square matrices, ${{A_n}_\times}_n$ and ${{V_n}_\times}_n$.
From my understanding, the determinant of a matrix represents the scalar multiple of an area (or volume, depending on the dimensions of your vector space) resulting from a linear transformation with a series of vectors.
What I would like to know is if the matrix $A$ is singular, does the transformation $AV$ always exist in a lower-dimensional space than $V$? e.g. will a collection of vectors in a three dimensional space collapse into a plane, or will a collection of vectors in a two-dimensional plane collapse into a line?
Intuitively it seems like it would, since the scalar multiple would be zero, but the example below seems to not agree with that statement:
$$A=\begin{pmatrix}1 & 0 & -1 \\ 1 & 1 & 2 \\ 3 & 1 &0 \end{pmatrix} $$ $$V=\begin{pmatrix}1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 &1 \end{pmatrix} $$ $$AV=\begin{pmatrix}1 & 1 & 0 \\ 1 & 2 & 3 \\ 3 & 4 &3 \end{pmatrix} $$
In this case, components of the first and second vectors exist in all three dimensions.
So, where do my misunderstandings lie?
In this case, the range of $AV$ is$$\{(x,y,z)\in\mathbb{R}^3\,|\,3x-3y+z=0\},$$and this space has dimension $2$. This is what always happens when (and only when) the matrix is singular: the dimension of the range is smaller than the dimension of the whole space.