Why this map implies the decomposition of the tensor product space?

Question

Let $V$ be a vector space and consider the tensor product of $V$ with itself, that is $V\otimes V$. Define $\alpha' : V\times V\to V\otimes V$ by

$$\alpha'(v,w)=w\otimes v.$$

In that case, $\alpha'$ is bilinear, so that by the universal property there is a linear map $\alpha : V\otimes V\to V\otimes V$ such that

$$\alpha(v\otimes w)=w\otimes v.$$

I've read that since $\alpha^2=I$, being $I$ the identity map, the tensor product decomposes as

$$V\otimes V = \operatorname{Sym}^2(V)\oplus \operatorname{Alt}^2(V),$$

being $\operatorname{Sym}^2(V)=\{z\in V\otimes V : \alpha(z)=z\}$ and $\operatorname{Alt}^2(V)=\{z\in V\otimes V : \alpha(z)=-z\}$.

Now, I do believe this is a trivial thing to prove, but how does one see that this map $\alpha$ implies that decomposition?

How can we prove that the tensor product decomposes this way using this map $\alpha$?

Answer 1

Suppose $U$ is a vector space over a field of characteristic different from $2$.

If you have any linear map $\alpha$ on $U$ such that $\alpha^{2} = I$, then either $\alpha = \pm I$, or $x^{2} - 1$ is the minimal polynomial of $\alpha$.

In the latter case the minimal polynomial has two distinct roots $1$ and $-1$, thus $U$ decomposes as the direct sum of the eigenspace relative to the eigenvalue $1$ and the eigenspace relative to the eigenvalue $-1$