# $\phi_1\wedge\dots\wedge\phi_p(v_1,\dots,v_p)=\frac{1}{p!}\det[\phi_i(v_j)]$

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?

• 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 $p\ne k=\dim V$). – user500094 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? – user500094 Apr 17 '18 at 1:01
• 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 1:25
• 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 6:33