Let \(K\) be a commutative ring with identity and \(D\) an alternating \(n\)-linear function on \(n \times n\) matrices over \(K\). Show that
\(D(A) = 0\), if one of the rows of \(A\) is 0.
\(D(B) = D(A)\), if \(B\) is obtained from \(A\) by adding a scalar multiple of one row of \(A\) to another.
Solution
Regard \(D(A)\) as a function of the ordered rows \(r_1,\dots,r_n\).
For (a), suppose \(r_i=0\). By linearity in the \(i\)th row, \[ D(r_1,\dots,0,\dots,r_n) =D(r_1,\dots,0+0,\dots,r_n) =2D(r_1,\dots,0,\dots,r_n). \] Subtracting gives \(D(A)=0\).
For (b), suppose \(B\) is obtained by replacing \(r_j\) by \(r_j+\lambda r_i\), with \(i\ne j\). Multilinearity gives \[ D(B) =D(r_1,\dots,r_j,\dots,r_n) +\lambda D(r_1,\dots,\underset{j}{r_i},\dots,r_n). \] In the second term the \(i\)th and \(j\)th rows are equal, so alternation makes that term zero. Hence \(D(B)=D(A)\).