Address #186

tex/linalg/dets.tex

@@ -106,9 +106,24 @@ \section{Wedge product}
 it's an area zero parallelogram!
 
 The \textbf{miracle of wedge products} is that the only additional condition
-we need to add to the tensor product axioms is that $v \wedge w = -(w \wedge v)$.
+we need to add to the tensor product axioms is that $v \wedge v = 0$.
 Then suddenly, the wedge will do all our work of interpreting volumes for us.
 
+\begin{remark*}
+	[Side digression on definitions in mathematics]
+	This ``property-based'' philosophy is a common trope in modern mathematics.
+	You have some intuition about an object you wish to define,
+	and then you write down a wishlist of properties that ``should'' follow.
+	But then it turns out the properties are sufficient to work with,
+	and so for the definition, you just define an abstract object
+	satisfying all the properties on your wishlist.
+	Thereafter the intuition plays no ``official'' role;
+	it serves only as cheerleading motivation for the wishlist.
+
+	For wedge products,
+	the wishlist has only the single property $v \wedge v = 0$.
+\end{remark*}
+
 In analog to earlier:
 \begin{proposition}
 	[Basis of $\Lambda^2(V)$]
@@ -181,16 +196,19 @@ \section{Wedge product}
 	in increasing order.
 \end{proof}
 
-
 \section{The determinant}
 \prototype{$(ae_1+be_2)\wedge(ce_1+de_2) = (ad-bc)(e_1\wedge e_2)$.}
+
 Now we're ready to define the determinant.
 Suppose $T \colon V \to V$ is a square matrix.
 We claim that the map $\Lambda^m(V) \to \Lambda^m(V)$ given on wedges by
 \[ v_1 \wedge v_2 \wedge \dots \wedge v_m
-	\mapsto T(v_1) \wedge T(v_2) \wedge \dots \wedge T(v_m) \]
-and extending linearly to all of $\Lambda^m(V)$ is a linear map.
-(You can check this yourself if you like.)
+	\mapsto T(v_1) \wedge T(v_2) \wedge \dots \wedge T(v_m). \]
+and extending linearly to all of $\Lambda^m(V)$ is a
+well-defined  linear map
+(Here ``well-defined'' means that equivalent elements of the domain
+get mapped to equivalent elements of the codomain.
+This, and linearity, both follow from $T$ being a linear map.)
 We call that map $\Lambda^m(T)$.
 \begin{example}
 	[Example of $\Lambda^m(T)$]