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  1. Suppose that $\phi_1,\dots,\phi_k\in V^\ast$ and $v_1,\dots,v_k\in V$ where $\dim V=k$. Prove that $$\phi_1\wedge\dots\wedge\phi_k(v_1,\dots,v_k)=\frac{1}{k!}\det[\phi_i(v_j)].$$
  2. More generally, show that whenever $\phi_1,\dots,\phi_p\in V^\ast$ and and $v_1,\dots,v_p\in V$, $$\phi_1\wedge\dots\wedge\phi_p(v_1,\dots,v_p)=\frac{1}{p!}\det[\phi_i(v_j)].$$

Remark about definitions: If $T$ is a $p$-tensor on $V$, then $$\operatorname{Alt}(T)(v_1,\dots,v_p)=\frac{1}{p!}\sum_{\sigma \in S_p}\operatorname{sgn}\sigma\cdot T(v_{\sigma(1)},\dots,v_{\sigma(n)})$$ and if $T_i\in \Lambda^i(V^\ast)$ for $i=1,\dots, n$, then $$T_1\wedge\dots\wedge T_n=\operatorname{Alt}(T_1\otimes\dots\otimes T_n).$$

I can prove the first part: $$\phi_1\wedge\dots\wedge \phi_k(v_1,\dots,v_k)=\operatorname{Alt}(\phi_1\otimes\dots\otimes \phi_k)(v_1,\dots,v_k)=\\\frac{1}{k!}\sum_{\sigma\in S_k}\operatorname{sgn}\sigma \cdot \phi_1(v_{\sigma(1)})\dots\phi_k(v_{\sigma(k)})=\frac{1}{k!}\det[\phi_i(v_j)].$$

But how do I show the second part?

user557
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    Possible duplicate of The matrix specify an algernating $k$-tensor on $V$, and dim$\bigwedge^k(V^*)=1$ – Javi Apr 16 '18 at 23:09
  • @Javi I believe the question in the link isn't my question (that's right that I included it as part of my text, but I also gave my solution to it, and what I'm asking is a different question, namely the case when $p\ne k=\dim V$). – user557 Apr 16 '18 at 23:23
  • I'm curious why do people keep voting to close this question if it is different from the one referred to above? – user557 Apr 17 '18 at 01:01
  • Oh, they LOVE to close questions, there is a real anti-intellectual streak on this site, inherited no doubt from the educational institutions. – Rene Schipperus Apr 17 '18 at 01:25
  • I don't understand where the assumption $\dim V = p$ plays any role in your proof. Why does your proof for the first part not immediately carry over to the second? – Joppy Apr 17 '18 at 06:33
  • @Joppy As far as I understand everything is valid except the very last equality. The big sum is then over the permutations in $S_p$ with $p<k$ and this is no longer the formula for the determinant. – user557 Apr 17 '18 at 18:47
  • If you only have $p$ vectors and functionals, then you should only be forming a $p \times p$ matrix. – Joppy Apr 17 '18 at 22:58
  • @Joppy If this is a $p\times p$ matrix, then the same proof indeed carries over. But then I'm confused why these two problems are given as separate exercises. Moreover, they are followed by confusing hints... E.g. for the second part, the hint is to consider the restrictions of $\phi_i$ to the $p$-dimensional subspace spanned by $v_1,\dots,v_p$. – user557 Apr 18 '18 at 00:07

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