Determinant of a matrix over a field of characteristic 2

Question

I first saw a proof for the Leibniz formula for computing determinants when I was learning about tensors and the exterior product at college (a "proof" considering that the definition of determinant is an alternating multilinear map from the columns or rows of a square matrix to its field such, that the determinant of the identity matrix equals 1).

We were told at some point that any alternating multilinear map $m$ satisfies $m(\vec{v}_1,...,\vec{v}_n)=0 \iff$ some $\vec{v}_i$ is a linear combination of all the other $\vec{v}_{j\neq i}$, only when we're working in a field of characteristic different from 2.

We actually used this property in our proof of the Leibnitz formula, but maybe there exists another proof of the Leibnitz formula which doesn't need that property to hold. So... is the Leibnitz formula valid for matrices over a field of characteristic 2?, or does characteristc 2 only imply that $\mathrm{rang}(A)<n \iff \mathrm{det}(A)$ won't be true (for a $n \times n$ matrix $A$)?

Answer 1

There are two relevant notions for these maps that agree over characteristic $\neq 2$, but are not identical in characteristic 2.

A bilinear form $m$ on a vector space $V$ is alternating if $m(v,v)= 0$ for all $v \in V$, and skew-symmetric if $m(v,w) = -m(w,v)$ for all $w,v \in V$. An alternating form is always skew-symmetric, for $$m(v,w) + m(w,v) = m(v+w,v+w) - m(v,v) - m(w,w) = 0$$ by bilinearity. But the reverse implication does not hold in characteristic 2, as there skew-symmetric coincides with symmetric. For instance, the standard dot product on $k^n$ for $\mathrm{char}(k) =2$ is skew-symmetric, but not alternating.

Alternating and skew-symmetric multilinear forms are defined similarly, and again every alternating form is skew-symmetric.

Over a field of characteristic 2, one may prove the Leibniz formula thus:

Theorem. Let $k$ be a field characteristic 2. There is a unique $n$-multilinear skew-symmetric function $m$ on $k^n$ such that $m(I) = 1$. If $A \in k^{n \times n}$ is a matrix with entries $\{a_{i,j}\}_{i,j=1}^n$, then $$m(A) = \sum_{\sigma \in S^n} sgn(\sigma) \prod_{i=1}^n a_{i,\sigma(i)} =\sum_{\sigma \in S^n} \prod_{i=1}^n a_{i,\sigma(i)}. $$

The last equality follows since $k$ is characteristic 2. Let $e_1,\ldots, e_n$ denote the standard basis for $k^n$, and let $a_j = \sum_{i=1}^n a_{ij}e_i$ denote the $j$th column of $A$. Then $$m(A) = m(a_1,\ldots,a_n) = \sum_{j_1=1}^n \sum_{j_2=1}^n \cdots \sum_{j_n=1}^n a_{1,j_1}a_{2,j_2}\cdots a_{n,j_n} m(e_{j_1},e_{j_2},\cdots,e_{j_n}).$$ Since $m$ is alternating, only terms with all distinct $j_k$'s survive, and since $m$ is skew-symmetric/symmetric and $m(I) = 1$, we obtain $$m(A) = \left(\sum_{\sigma \in S_n} \prod_{i=1}^n a_{i,\sigma(i)}\right) m(e_1,\ldots, e_n) = \sum_{\sigma \in S_n} \prod_{i=1}^n a_{i,\sigma(i)},$$ as desired.