$r$-linear Applications Vector Space as Direct Sum of two Subspaces Good morning, good evening or good night, depending on when you read this. I'm having trouble visualizing two results. Let's denote $\mathcal{L}^r(V;W)$ as the vector space of the applications $\phi : V \times \cdots \times V \rightarrow W$ that are $r$-linear over a field $\mathbb{K}$ with $\mathrm{char}\textrm{ }\mathbb{K}\neq 2$. Also, let's denote $\mathcal{L}_a^r(V;W)$ and $\mathcal{L}_s^r(V;W)$ the subspaces of alternated and symmetric $r$-linear applications.

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*The first thing I'm having trouble is to show that $$\mathcal{L}^2(V;W) = \mathcal{L}_a^2(V;W) \oplus \mathcal{L}_s^2(V;W).$$ The side $\mathcal{L}_a^2(V;W) \oplus \mathcal{L}_s^2(V;W) \subset \mathcal{L}^2(V;W)$ of the equality I did show, but the $\mathcal{L}^2(V;W) \subset \mathcal{L}_a^2(V;W) \oplus \mathcal{L}_s^2(V;W)$ side has been really difficult to visualize the general case. I've begun taking an arbitrary $\phi \in \mathcal{L}^2(V;W)$, so we can say that $\phi$ can be symmetric, alternated or none of the two. For the two first cases, it is really easy to show that $\phi \in \mathcal{L}_a^2(V;W) \oplus \mathcal{L}_s^2(V;W)$, but for the third one I don't know how to get there.


*The second one is: the  direct sum decomposition $$\mathcal{L}^r(V;W) = \mathcal{L}_a^r(V;W) \oplus \mathcal{L}_s^r(V;W)$$ is true for $r > 2$? My intuition is to try applying induction over $r$, but the induction hipothesis application for $r+1$ is tricky. But, I also think it does not hold for $r > 2$.
Any help is very appreciated! I apologize for the trivial kind of question, but when I'm stuck, I'm stuck and all I can do is: ask for help. What I want is a hint or general idea to attack the problems specified.
Thanks in advance for the help, attention and time given!
 A: You did pretty well with the first part. The map $\phi \mapsto \phi + \phi^\tau$ is indeed not idempotent, but $P : \phi \mapsto (\phi + \phi^\tau)/2$ is (recall: $\Bbb{K}$ does not have character $2$). It is a linear map, since $\phi \mapsto \phi^\tau$ is linear. The kernel consists of $\phi$ such that $\phi = -\phi^\tau$, i.e. $\phi(v_1, v_2) = -\phi(v_2, v_1)$, i.e. $\phi \in \mathcal{L}_a^2(V;W)$. Moreover, every vector in the image is symmetric, as
$$\frac{\phi + \phi^\tau}{2}(v_1, v_2) = \frac{\phi(v_1, v_2) + \phi(v_2, v_1)}{2},$$
which is a symmetric expression of $v_1$ and $v_2$. Further, if $\phi$ is symmetric, then $\phi^\tau = \phi$, so $\phi$ will map to $(\phi + \phi) / 2 = \phi$. That is, the image of our linear map is indeed all of $\mathcal{L}^2_s(V;W)$, plus we also just proved idempotency.
All of this is a lot of information. The linearity of the map tells us that the kernel and image are both subspaces (which you knew already, but it's kind of neat). It also means that we can write any $\phi$ as:
$$\phi = P(\phi) + (I - P)(\phi) = \frac{\phi + \phi^\tau}{2} + \frac{\phi - \phi^\tau}{2},$$
where $P(\phi) \in \operatorname{Im}(P) = \mathcal{L}_s^2(V;W)$, and since $P(I - P)(\phi) = (P^2 - P)(\phi) = 0$, $(I - P)(\phi) \in \operatorname{Ker}(P) = \mathcal{L}_a^2(V;W)$. This shows that they sum to $\mathcal{L}^2(V;W)$.
We can also prove the sum is direct, by showing that, if $\phi = \sigma + \alpha$, where $\sigma \in \mathcal{L}_s^2(V;W)$ and $\alpha \in \mathcal{L}_a^2(V;W)$, then by taking $P$ of both sides,
$$P(\phi) = P(\sigma) + P(\alpha) = \sigma + 0 = \sigma,$$
in other words, $\sigma$ is determined uniquely to be the image of $\phi$ under $P$. This also uniquely determines $\alpha$ to be $\phi - P(\phi)$.
All of these steps follow in more generality; you can see how useful it is to find these linear projections.
As for the second part, it doesn't hold for $r > 2$. There are a number of ways to see it, but you could show, by the same basic argument above, that the subspaces of $r$-linear maps which are symmetric or anti-symmetric in only first two arguments sum directly to $\mathcal{L}^r(V;W)$, then argue that the fully symmetric (or alternating) $k$-linear maps form a strict subspace of these two spaces. Since the larger space sum directly, shrinking either of them will stop the subspaces summing to the full space at all.
