Given linearly independent vectors $v_1, \ldots, v_n$ and linearly independent vectors $w_1, \ldots, w_n$ and a mapping $A$ such that $$ w_i = Av_i, \quad i = 1,\ldots, n $$ then if I define a mapping $B$ by $Bv_k := \lambda A v_k$ for some $1 \le k \le n$ and $Bv_i = Av_i$ otherwise, then $\det B = \lambda \det A$. I want to prove this using just the abstract properties of the determinant as for example mentioned here on wikipedia.
I know that in terms of matrices we have $$ B = CA $$ where $C$ is the matrix with $c_{kk} = \lambda$ and $c_{ii} = 1$ if $i \ne k$ and $c_{ij} = 0$ if $i \ne k$ (like the identity expect at the $k$-diagonal entry). Then $\det(B) = \det(C)\det(A) = \lambda\det(A)$, but I want to avoid talking about matrices (which in turn needs bases). Are there any elegant abstract, preferable coordinate-free, arguments?
Define $\det(A)$ by $$\underline A(v_1 \wedge \cdots \wedge v_n) = \det(A)v_1 \wedge \cdots \wedge v_n$$ Then $$\begin{align}\det(B)v_1 \wedge \cdots \wedge v_n &= \underline B(v_1 \wedge \cdots \wedge v_n) \\ &= w_1 \wedge \cdots \wedge w_{k-1} \wedge \lambda w_k \wedge w_{k+1} \wedge \cdots \wedge w_n \\ &= \lambda w_1 \wedge \cdots \wedge w_n \\ &= \lambda \underline A(v_1 \wedge \cdots \wedge v_n) \\ &= \lambda \det(A)v_1 \wedge \cdots \wedge v_n\end{align}$$
Therefore $\det(B) = \lambda\det(A)$.$\ \ \ \ \ \square$
Coordinate-free and beautiful. You gotta love exterior algebra.