Assume that $V$ is a finite dimensional real vector space of dimension $n$. Is there a $\mathbb{R} -$ valued $k$- linear map $T$ on $V$ which is not an alternative form but it vanish on all $k$- tuple $(x_1,x_2,\ldots,x_k)$ with $\sum_{i=1}^k x_i =0$?
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asked Sep 27, 2017 at 13:28
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1 Answer
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No. Notice that by linearity we should have $$ f(v_1,\dots,v_{k-1},\alpha_1v_1+\dots+\alpha_{k-1}v_{k-1}) =\frac{f(\alpha_1v_1,\dots,\alpha_{k-1}v_{k-1},-\alpha_1v_1-\dots-\alpha_{k-1}v_{k-1})}{(-1)^k\alpha_1\cdots\alpha_{k-1}} =0 $$ for all nonzero $\alpha_1,\dots,\alpha_{k-1}$. Since each linear combination is a sum of two linear combinations with nonzero coefficients, this also holds for all $\alpha_i$ [this argument works also for all fields with at least 3 elements]. Thus, the linear map $f$ vanishes on all linearly dependent $k$-tuples, in which case it is well known to be alternative.
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$\begingroup$ Thanks for your answer. How does this argument work for a triple $f(v,v,w)$, for example? $\endgroup$ Commented Sep 27, 2017 at 14:52
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$\begingroup$ You may perform the same for the last two vectors instead of the first two. $\endgroup$ Commented Sep 27, 2017 at 15:05
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$\begingroup$ Yes but you implicitly assume that all $\alpha_i$ are non zero. i can imagine a resolution for three vector: $2f(v,v,w)=f(v,v.2w+2v)+f(v,v,-2v)=0$. I guess that the general case can be solved, similarly. Yes? $\endgroup$ Commented Sep 27, 2017 at 17:18
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$\begingroup$ No, I first explicitly tell that all $\alpha_i$ are nonzero, and second tell that by summing two such expression we can get an arbitrary linear combination, with each coefficient being zero or nonzero on our choice. $\endgroup$ Commented Sep 27, 2017 at 19:17
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1$\begingroup$ $f(v,v,w)=f(1\cdot v+0\cdot w,v,w)=f(2v+w,v,w)-f(v+w,v,w)=0$. $\endgroup$ Commented Sep 28, 2017 at 4:43