Exercise 18

An alternate proof of Theorem 1.5 uses the definition of determinant functions.

• Note that the vectors forming $S$ make a linearly dependent set if and only if $\mathchoice{\left|S\right|}{|S|}{|S|}{|S|}=0$, and check that the result holds in this case.

• For the $\mathchoice{\left|S\right|}{|S|}{|S|}{|S|}\neq 0$ case, to show that $\mathchoice{\left|TS\right|}{|TS|}{|TS|}{|TS|}/\mathchoice{\left|S\right|}{|S|}{|S|}{|S|}=\mathchoice{\left|T\right|}{|T|}{|T|}{|T|}$ for all transformations, consider the function $d\colon\mathcal{M}_{n\!\times\!n}\to\mathbb{R}$ given by $T\mapsto\mathchoice{\left|TS\right|}{|TS|}{|TS|}{|TS|}/\mathchoice{\left|S\right|}{|S|}{|S|}{|S|}$. Show that $d$ has the first property of a determinant.

• Show that $d$ has the remaining three properties of a determinant function.

• Conclude that $\mathchoice{\left|TS\right|}{|TS|}{|TS|}{|TS|}=\mathchoice{\left|T\right|}{|T|}{|T|}{|T|}\cdot\mathchoice{\left|S\right|}{|S|}{|S|}{|S|}$.