Let \(F\) be a field, \(n\) a positive integer, and \(S\) the set of \(n \times n\) matrices over \(F\). Let \(V\) be the vector space of all functions from \(S\) into \(F\). Let \(W\) be the set of alternating \(n\)-linear functions on \(S\). Prove that \(W\) is a subspace of \(V\). What is the dimension of \(W\)?
Solution
If \(D,E\in W\) and \(a,b\in F\), then \(aD+bE\) is \(n\)-linear because linear combinations preserve linearity in each row variable. If two row arguments are equal, both \(D\) and \(E\) vanish, so \[ (aD+bE)(A)=aD(A)+bE(A)=0. \] Thus \(aD+bE\) is alternating. The zero function is alternating and \(n\)-linear, so \(W\) is a subspace of \(V\).
Let \(D\in W\), and let the rows of \(A\) be \(\alpha_1,\ldots,\alpha_n\). Writing each row as \[ \alpha_i=\sum_{j=1}^n A_{ij}e_j \] and expanding by multilinearity gives a sum indexed by \(j_1,\ldots,j_n\). Every term with two equal \(j_i\) vanishes by alternation. Hence only permutations survive, and moving the rows \((e_{\sigma(1)},\ldots,e_{\sigma(n)})\) back to \((e_1,\ldots,e_n)\) gives \[ D(A)=D(I)\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma) \prod_{i=1}^n A_{i,\sigma(i)} =D(I)\det A. \] Thus every element of \(W\) is a scalar multiple of the determinant. Conversely, the determinant is a nonzero alternating \(n\)-linear function, since \(\det I=1\). Therefore \[ W=F\det, \qquad \dim W=1. \]