## Over a field of characteristic not two, the tensor square of a vector space decomposes as the direct sum of the symmetric and exterior tensor squares

Let $F$ be a field of characteristic not 2, and let $V$ be an $n$-dimensional vector space over $F$. Recall that in this case, we can realize $\mathcal{S}^2(V)$ and $\bigwedge^2(V)$ as subspaces of $V \otimes_F V$. Prove that $V \otimes_F V = \mathcal{S}^2(V) \oplus \bigwedge^2(V)$.

Recall that $\mathsf{dim}_F\ V \otimes_F V = n^2$.

Suppose $z \in \mathcal{S}^2(V) \cap \bigwedge^2(V)$. Then we have $\mathsf{Alt}_2(z) = z = \mathsf{Sym}_2(z)$. Expanding these, we see that $z + (1\ 2) z = z - (1\ 2) z$, so that $2(1\ 2) z = 0$, so that $z = 0$. That is, $\mathcal{S}^2(V)$ and $\mathsf{Alt}_2(V)$ intersect trivially.

Counting dimensions, recall that $\mathsf{dim}_F\ \mathcal{S}^2(V) = \frac{n(n+1)}{2}$ and $\mathsf{dim}_F\ \bigwedge^2(V) = \frac{n(n-1)}{2}$.

Thus $V \otimes_F V = \mathcal{S}^2(V) \oplus \bigwedge^2(V)$.