Show that $T^*f$ is an alternating tensor if f is an alternating tensor

Question

Show that if T: V $\rightarrow $ W is a linear transformation and if f $\in A^k(W) $ then $T^*f \in A^k(V)$ where $T^*$ is the dual transformation. Attept at the solution: If $f \in A^k(V)$ and $v_1,...v_k$ are vectors in V, then $(T^*f)^\sigma$ $(v_1,...,v_k) = (f(T(v_1),...,T(v_k)))^\sigma$ = $f^\sigma (T(v_1),...,T(v_k))$ = (sgn $\sigma$)$(f(T(v_1),...,T(v_k))$ (since f is alternating) = (sgn$\sigma$)$T^*f \in A^k(V)$ as required. I'm not that great at proof, so I'm not sure if this is rigorous enough, please let me know of any suggestions\improvements I can make to this.

Thank you :)

Answer 1

To show that $T^*f$ is alternating is suffices to show that $T^*f(\ldots,v,\ldots,v\ldots) = 0$. But

$$T^*f(\ldots,v,\ldots,v\ldots) = f(\ldots,T(v),\ldots,T(v),\ldots) = 0.$$

Alternatively, you may show that $T^*f(v_{\sigma(1)},\ldots,v_{\sigma(k)}) = \operatorname{sgn}(\sigma)T^*f(v_1,\ldots,v_k)$ for any permutation $\sigma \in S_k$. (I think this is what you're trying to do.) And indeed

\begin{align}T^*f(v_{\sigma(1)},\ldots,v_{\sigma(k)}) &= f(T(v_{\sigma(1)}),\ldots,T(v_{\sigma(k)})) \\ &= \operatorname{sgn}(\sigma)f(T(v_1),\ldots,T(v_k)) \\ &= \operatorname{sgn}(\sigma)T^*f(v_1,\ldots,v_k). \end{align}