Show that $\big\{A(v_{i_1}\otimes \cdots\otimes v_{i_k})\in V^{\wedge k}:i_1<\cdots<i_k \big\} $ is linearly independent

Question

Let $V$ be vector space over $\mathbb{R}$ and $V^{\otimes k}$ be the $k$th tensor power of $V$.

Denote by $S_k$ the set which contains exactly all the permutations of a set with $k$ elements.

Using the universal property of the tensor product we can show that for all $\sigma \in S_k$ there's a linear transformation $P_\sigma :V^{\otimes k}\to V^{\otimes k}$ such that $P_\sigma (v_1\otimes \cdots \otimes v_{k})=v_{\sigma (1)}\otimes \cdots \otimes v_{\sigma (k)}$ for all subset $\{v_1,\cdots,v_k\}\subset V$.

Define $A:V^{\otimes k}\to V^{\otimes k}$ by $A(x):=\frac{1}{k!}\sum_{\sigma \in S_k}(-1)^{|\sigma |}P_\sigma (x)$ in which $(-1)^{|\sigma |}$ is the signature of a permutation $\sigma \in S_k$.

Denote $\color{red}{V^{\wedge k}}:=\text{Im}(A)$. We call the set $V^{\wedge k}$ the $k$th exterior product of $V$.

Define $n:=\dim (V)$ and suppose that $\{v_1,\cdots,v_n\}$ is a basis of $V$.

My question is: how can I show that $\big\{A(v_{i_1}\otimes \cdots\otimes v_{i_k}):1\leq i_1<\cdots<i_k\leq n\big\}$ is a linearly independent subset of $V^{\wedge k}$ if $p\leq n$?

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Things I was able to prove:

1. $\big\{v_{i_1}\otimes \cdots\otimes v_{i_k}:i_1,\cdots,i_k\in\{1,\cdots, n\}\big\}$ is a basis of $V^{\otimes k}$
2. $A^2=A$
3. If $x\in V^{\wedge k}$ and $\sigma \in S_k$, then $x=(-1)^{|\sigma|}P_\sigma (x)$
4. The map $f:V^k\to V^{\wedge k}$ given by $f(v_1,\cdots,v_k):=A(v_1\otimes \cdots\otimes v_k)$ is an alternating $k$-linear map
5. $P_\sigma$ is an isomorphism for all $\sigma \in S_k$.
6. $A$ is self-adjoint with respect to the inner product $\langle \cdot ,\cdot \rangle_\otimes :V^{\otimes k}\times V^{\otimes k}\to \mathbb{R}$ given by $\langle u_1\otimes \cdots \otimes u_k,w_1\otimes \cdots \otimes w_k\rangle _\otimes=k!\Pi _{i=1}^k\langle u_i,w_i\rangle $ in which $\langle \cdot,\cdot \rangle $ is a inner product of $V$.

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Thank you for your attention!

Answer 1

Since $V$ is finite-dimensional, we can view $V \approx (V^*)^*$. With this identification, view the $k$th tensor power of $V$ as the vector space of multilinear maps from $(V^*)^k$ to $\mathbb{R}$. Since $v_1, \dots, v_n$ is a basis of $V$, it is dual to some basis $\varepsilon_1, \dots, \varepsilon_n$ of $V^*$. Suppose that $\alpha = \sum_{I}a_IA(v_{i_1} \otimes \dots \otimes v_{i_k}) = 0$, where the sum is over all $I = (i_1, \dots, i_k)$ such that $i_1 < \dots < i_k$. We need to show that $a_I = 0$ for all $I$. We have $$\alpha(\varepsilon_{i_1}, \dots, \varepsilon_{i_k}) = \frac{1}{k!}a_{I}$$ since $A(v_{j_1} \otimes \dots \otimes v_{j_k})(\varepsilon_{i_1}, \dots, \varepsilon_{i_k}) = 0$ when $J = I$, and is $1$ when $J = I$. Hence $a_I = 0$ for all $I$.