I am studying the wedge product for multivariable analysis, but I feel that the operations are not so intuitive in general. I am looking at the following question which deals with linear transformations.
Let $V,W,Z$ be vector spaces over a field $k$
If $f:V \rightarrow W$ and $g:W \rightarrow Z$ are linear transformations. Prove that $\wedge ^r (g\circ f)=\wedge ^r(g)\circ \wedge ^r(f)$.
I've seen examples of linear transformation in $R^n$, and I think that this is either a little too simple and need only involve a basis for each vector space, or this is too messy to write. I tried to get a little more comfortable with the notation around this problem by making up an example with two different dimensions. I used a linear transformation such as
$T_1: R^2\rightarrow R^3$ defined by $T_1(x,y)=(x-y,x,y)$ and
$T_2: R^3\rightarrow R^3$ defined by $T_2(x,y,z)=(2x, 2z, y-x)$
and I considered r=2.
As a result I got the following:
$\wedge^2 f = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$
$\wedge^2 g = \begin{bmatrix} 0 & 4 & 0 \\ 2 & 0 & 0 \\ 0 & 2 & -2 \end{bmatrix}$
When I take the product of both matrices I get
$(\wedge^2 g) \circ (\wedge^2 f)= \begin{bmatrix} 4\\ 2\\ 0 \end{bmatrix}$
If instead I first take the composition and try to compute the wedge product also I get
$\wedge^2(g\circ f)= \begin{bmatrix} 4 \\ 2 \\ 0 \end{bmatrix}$
I just feel that I am not really manipulating the abstract generalized notion for the definition used over wedge products. I initially thought that maybe it might help to assume certain dimensions $l,m$ and $n$ for the vector spaces given and write out a basis for each of them, but yet when writing things out, I don't really know what else to do (the definition itself looks a little intimidating).
I appreciate any help given in order to understand a little more what is happening with the wedge product when considering an arbitrary $r$ and dealing with both linear transformations.