Let $\beta: \Bbb R^n \times \dots \times \Bbb R^n \to \Bbb R^m$ be $r$-linear. Define
$$ \operatorname{symm}(\beta)(v_{1},\dots,v_{r}) = \frac{1}{r!} \sum_{\pi}^{}\beta(v_{\pi(1)},\dots,v_{\pi(r)})$$ where $\pi$ ranges through the set of permutations of {1,..,r}, and define the antisymmetrization$$ \operatorname{asymm}(\beta)(v_{1},\dots,v_{r}) = \frac{1}{r!} \sum_{\pi}^{} \operatorname{sgn}(\pi) \beta(v_{\pi(1)},\dots,v_{\pi(r)}) $$ I'm hoping to prove $ \alpha = \beta - \operatorname{symm}(\beta) $ is antisymmetric in the sense that for any permutation $\pi$ over $\{1,...,r\}$, $\alpha(v_{\pi(1)},\dots,v_{\pi(r)}) = \operatorname{sgn}(\pi)\alpha(v_{1},\dots,v_{r})$.
I have proved that
i) $\operatorname{symm}(\beta)$ is symmetric, and $\operatorname{symm}(\beta)=\beta \iff \beta$ is symmetric
ii)$\operatorname{asymm}(\beta)$ is antisymmetric, and $\operatorname{asymm}(\beta)=\beta \iff \beta$ is antisymmetric
thus to prove $\alpha$ is antisymmetric, we can instead show $\operatorname{asymm}(\alpha) = \alpha$: $$\operatorname{asymm}(\alpha) = \operatorname{asymm}(\beta - \operatorname{symm}(\beta)) = \operatorname{asymm}(\beta)$$ as $\operatorname{asymm}(\operatorname{symm}(\beta))=0$. However, I'm having trouble showing $\operatorname{asymm}(\beta) = \alpha, $ or $ \operatorname{asymm}(\beta) + \operatorname{symm}(\beta) = \beta $ $$\begin{split}\operatorname{asymm}(\beta)+\operatorname{symm}(\beta)(v_{1},\dots,v_{r}) &= \frac{1}{r!} \sum_{\pi}^{} \operatorname{sgn}(\pi) \beta(v_{\pi(1)},\dots,v_{\pi(r)})+\frac{1}{r!} \sum_{\pi}^{}\beta(v_{\pi(1)},\dots,v_{\pi(r)})\\ &= \frac{2}{r!}\sum_{\pi \text{ even}}^{}\beta(v_{\pi(1)},\dots,v_{\pi(r)})\end{split}$$ I don't see why this would equal to $\beta(v_{1},\dots,v_{r})$.
Any hints on how to proceed? Thanks!