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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?

user1892304
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  • In an elementary course it is proved many times by means of induction and development by rows (columns) of minors, proving directly for the case $\;1\times1\;$ and more often that not also the case $\;2\times2\;$ and then some basic induction. I suppose it could also be proved by exterior algebra arguments but this requires way more sophistication and knowledge. – DonAntonio Nov 09 '18 at 20:45
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    Note that the assumption that $2\neq 0$ is pretty artificial. You can remove it as long as you fix the definition of "alternating" (the definition should be that $\det$ is zero whenever two of the inputs are equal). – Eric Wofsey Nov 09 '18 at 20:46
  • @DonAntonio I was precisely hoping for a sophisticated and knowledgeable argument! – user1892304 Nov 09 '18 at 20:48
  • This might be helpful: https://math.stackexchange.com/a/21617/587142 – twnly Nov 09 '18 at 20:49

1 Answers1

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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. :)

Bill Cook
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  • The rub is step 4, though. I don't know a way to prove that $B_k$ is linearly independent which doesn't amount to proving that some formula for determinants really is multilinear and alternating. – Eric Wofsey Nov 09 '18 at 21:07
  • Not surprising that at some point you have to face the fact that alternating, multilinear things can *only* be computed using determinants. – Bill Cook Nov 09 '18 at 21:10
  • Given the strong uniqueness, one really shouldn't be able to completely avoid it. – Bill Cook Nov 09 '18 at 21:12
  • Caveat emptor: Proving that $B_k$ is a basis for $\Lambda^k V$ requires the notion of the sign of a permutation, or another tool (e.g., [Newman's lemma](https://en.wikipedia.org/wiki/Newman%27s_lemma)) that allows one to keep track of how a wedge product changes when its factors are being swapped. But the sign of a permutation is already 90% of the work required to define a determinant. – darij grinberg Nov 10 '18 at 05:45