How can I prove that the determinant function satisfying the following properties is unique:
$\det(I)=1$ where $I$ is identity matrix, the function $\det(A)$ is linear in the rows of the matrix and if two adjacent rows of a matrix $A$ are equal, then $\det A=0$.
This is how Artin has stated the properties.I find Artin's first chapter rough going and would appreciate some help on this one.
Best Answer
I have thought about your problem for a while now and I think there is a nice slick way to do this. Consider the space $W$ of all multilinear alternating forms $f$ in $k$ - variables
$$f : V \times \ldots \times V \to \Bbb{C}.$$
We claim that there is a canonical isomorphism between $W$ and $(\bigwedge^k V)^\ast$. Indeed, this should be clear because given any $f \in W$, the universal property of the $k$ - th exterior power tells us that there is a unique linear map $g \in (\bigwedge^k V)^\ast$ such that $f = g \circ \iota$ where $\iota : V \times \ldots \times V \longrightarrow \bigwedge^k V$ is the canonical mapping that sends the tuple $(v_1,\ldots,v_k)$ to $v_1 \wedge \ldots \wedge v_k$. Conversely given any $h \in (\bigwedge^k V)^\ast$ we can precompose it with $\iota$ to give us a mapping from $V \times \ldots \times V \to \Bbb{C}$.
In summary, we can use these facts to give us a canonical isomorphism between $W$ and $(\bigwedge^k V)^\ast$. If we put $k = n$, where $n = \dim V$ then
$$ 1= \dim_{\Bbb{C}} \bigwedge\nolimits^{\!k}V = \dim_{\Bbb{C}} \left(\bigwedge\nolimits^{\!k} V\right)^\ast $$
from which it follows that $W$ is one dimensional. In other words, any $f \in W$ is a scalar multiple of $\det$, where
$$\det : V \times V\times \ldots \times V \longrightarrow \Bbb{C}$$
is the mapping that sends the tuple $(v_1,\ldots, v_n)$ to the determinant of the matrix whose columns are the vectors $v_1, v_2, \ldots, v_n$. Now here comes the killer blow: Suppose we demand that an alternating multilinear $f$ be such that $f(e_1,\ldots,e_n) = 1$ where the $e_i$ are the standard basis vectors of $\Bbb{C}^n$. Then because
$$f(v_1,\ldots,v_n) = c\cdot \det(v_1,\ldots,v_n)$$
for some constant $c$, shoving in $(v_1,\ldots,v_n) = (e_1,\ldots,e_n)$ we must have that
$$\begin{eqnarray*} 1 &=& f(e_1,\ldots, e_n) \\ &=& c\cdot \det(e_1,\ldots,e_n) \\ &=& c \end{eqnarray*}$$
because the determinant of the identity matrix is $1$. Consequently we have shown:
$$\hspace{6in} \square$$