Let \(V\) and \(W\) be finite-dimensional vector spaces over a field \(k\). Let \(f\colon V^n\to W\) be a function such that
- For each fixed \(i\in\{1,\ldots,n\}\) and fixed \(v_1,\ldots,v_{i-1},v_{i+1},\ldots,v_n\in V\), the map
\[ \begin{aligned} &V\to W\\ &x\mapsto f(v_1,\ldots,v_{i-1},x,v_{i+1},\ldots,v_n) \end{aligned} \]
is a \(k\)-linear transformation; and
- \(f(v_1,\ldots,v_n)=0\) whenever \(v_i=v_{i+1}\) for some \(i\in\{1,\ldots,n-1\}\).
Prove that either \(\dim V\geq n\) or \(f\) is identically zero.
Solution
Fix \(i\), and \(v_1,\ldots,v_{i-1},v_{i+2},\ldots,v_n\in V\), and define \(g(x,y)=f(v_1,\ldots,v_{i-1},x,y,v_{i+2},\ldots,v_n)\). Then
\[ \begin{aligned} 0&=g(x+y,x+y)\\ &=g(x+y,x)+g(x+y,y)\\ &=g(x,x)+g(y,x)+g(x,y)+g(y,y)\\ &=g(y,x)+g(x,y) \end{aligned} \]
so interchanging adjacent arguments changes the sign of the value of \(f\).
Suppose \(v_1,\ldots,v_n\in V\) are such that \(v_i=v_j\) for some \(i<j\). Then we can interchange arguments repeatedly to move \(v_j\) to the \(i+1\) position, possibly changing the sign of the value of \(f(v_1,\ldots,v_n)\) as we go along. Since at the end the result is zero, we must have had \(f(v_1,\ldots,v_n)=0\) originally. Thus \(f(v_1,\ldots,v_n)=0\) whenever \(v_i=v_j\) for some \(i\neq j\).
We now solve the problem. If the conclusion fails, we have \(\dim V<n\) and there exist \(v_1,\ldots,v_n\in V\) with \(f(v_1,\ldots,v_n)\neq0\). Since \(\dim V<n\), the vectors \(v_1,\ldots,v_n\) must be linearly dependent. Thus for some \(i\), we can write \(v_i=\sum_{j\neq i}c_jv_j\) for some constants \(c_j\in k\) for \(j\neq i\). By linearity of \(f\) in the \(i\)-th argument,
\[ \begin{aligned} f(v_1,\ldots,v_n)&=\sum_{j\neq i}c_jf(v_1,\ldots,v_{i-1},v_j,v_{i+1},\ldots,v_n)\\ &=\sum_{j\neq i}c_j\cdot0 \end{aligned} \]
by the previous paragraph, since in each term some \(v_j\) appears twice as an argument. Thus \(f(v_1,\ldots,v_n)=0\), a contradiction.