A fact about the Alt map on a tensor power

Let F be a field in which k! is a unit and let V be an F-vector space. Recall that S_k acts on the tensor power \mathcal{T}^k(V) by permuting the components, and that \mathsf{Alt}_k is defined on \mathcal{T}^k(V) by \mathsf{Alt}_k(z) = \sum_{\sigma \in S_k} \epsilon(\sigma) \sigma z. Prove that \mathsf{im}\ \mathsf{Alt}_k is the unique largest subspace of \mathcal{T}^k(V) on which each permutation \sigma \in S_k acts as multiplication by \epsilon(\sigma).

Suppose z \in \mathcal{T}^k(V) such that \sigma z = \epsilon(\sigma) z for all \sigma \in S_k. Then \mathsf{Alt}_k(z) = \frac{1}{k!} \sum_{\sigma \in S_k} \epsilon(\sigma) \sigma z = \frac{1}{k!} \sum_{\sigma \in S_k} \epsilon(\sigma) \epsilon(\sigma) z = \frac{1}{k!} \sum_{\sigma \in S_k} z = z, so that z \in \mathsf{im}\ \mathsf{Alt}_k.

In particular, any subspace of \mathcal{T}^k(V) upon which every permutation \sigma \in S_k acts as scalar multiplication by \epsilon(\sigma) is contained in \mathsf{im}\ \mathsf{Alt}_k.

Note also that \sigma \in S_k acts on \mathsf{im}\ \mathsf{Alt}_k as multiplication by \epsilon(\sigma) since \mathsf{im}\ \mathsf{Alt}_k \cong_F \bigwedge^k(V).

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