Is there a non-constructive proof that the determinant exists? Let $k$ be a field where $2\neq0$, and $V$ an $n$-dimensional $k$-vector space. Then there exists a unique function $\det:V^n\to k$ that has the following properties:


*

*$\det$ is multilinear (linear in each variable)

*$\det$ is alternating (permuting two arguments changes the sign)

*$\det(B)=1$ for some basis $B$
Without providing an explicit formula (à la Leibniz), how can existence be proved?
 A: A standard approach is to use the machine of wedge products (alternating tensors):


*

*Pick a basis: $B = \{ v_1,\dots,v_n \}$ for $V$ over a field $\mathbb{K}$. 

*Construct tensor product powers of $V^{\otimes k} = V \otimes \cdots \otimes V$ ($k$-copies).

*Quotient by the subspace $U$ spanned by vectors of the form: $a_1 \otimes \cdots \otimes a_k$ where $a_i=a_j$ for some $i \not= j$. Call such a space $\wedge^k V = V^{\otimes k}/U$. Let $w_1 \wedge \cdots \wedge w_k = w_1\otimes \cdots \otimes w_k + U$ (the coset represented by $w_1\otimes \cdots \otimes w_k$).

*Prove that $B_k = \{ v_{i_1} \wedge \cdots \wedge v_{i_k} \;|\; i_1 < i_2 < \cdots < i_k \}$ is a basis for $\wedge^k V$. This means $\dim(\wedge^k V) = {n \choose k} = \dfrac{n!}{k!(n-k)!}$. In particular, $\wedge^n V$ is $1$-dimensional and has basis $v_1\wedge \cdots \wedge v_n$.

*The dual space $(\wedge^n V)^* = \{ f:\wedge^n V \to \mathbb{K} \;|\; f \mbox{ is linear }\}$ is $1$-dimensional. Select the unique linear functional $D$ such that 
$D(v_1 \wedge \cdots \wedge v_n) =1$ (i.e., $\{D\}$ is the basis dual to $\{v_1\wedge\cdots \wedge v_n \}$).

*One can show that every $k$-multilinear map $\hat{f}:V \times \cdots \times V \to W$ uniquely factors as $\hat{f}(w_1,\dots,w_k) = f(w_1 \otimes \cdots \otimes w_k)$ where $f:V^{\otimes k} \to W$ is linear (this universal property essentially defines what a tensor product is). Likewise, every alternating $k$-multilinear map $\hat{f}$ factors as $\hat{f}(w_1,\dots,w_k) = f(w_1\wedge \cdots \wedge w_k)$ where $f:\wedge^k V \to W$ is linear (this universal property captures what $\wedge^k V$ is).

*Finally, the alternating, multilinear map, $\hat{D}=\det$ associated with $D$ is the unique such map such that $\det(v_1,\dots,v_n)=1$. 
Or you could just construct the determinant. :)
