$V\otimes V=\text{Sym}^2(V)\oplus\text{Alt}^2(V)$ as $\text{GL}(V)$-representations Let $V$ be a $\mathbb{C}$-vector space. $\text{GL}(V)$ acts on $V\otimes V$ via $f.(x\otimes y):=f(x)\otimes f(y)$. Define $$\text{Sym}^2(V)=\text{span}\{x\otimes y+y\otimes x| x,y\in V\}$$ and $$\text{Alt}^2(V)=\text{span}\{x\otimes y-y\otimes x| x,y\in V\}.$$
I want to show that $V\otimes V=\text{Sym}^2(V)\oplus\text{Alt}^2(V)$ as $\text{GL}(V)$-representations.
First I have shown that $\text{Sym}^2(V)$ and $\text{Alt}^2(V)$ are invariant subspaces of $V\otimes V$: This is easy as it is enough to show this on elements $x\otimes y\pm y\otimes x$ and $$f.(x\otimes y\pm y\otimes x)=f(x)\otimes f(y)\pm f(y)\otimes f(x).$$
But I have problems to show that the intersection is trivial and need help with this please.
If we have shown the trivial intersection part, are we done? Or what is a direct sum of $\text{GL}(V)$-representations?
 A: Let $W$ be any vector space (over a field of characteristic not equal to $2$; ignore this if you don't know what it means) on which the cyclic group $C_2$ of order $2$ acts: write $g$ for the generator of $C_2$ acting on $W$. Then I claim that $W$ has a natural direct sum decomposition
$$W \cong W_1 \oplus W_{-1}$$
where
$$W_1 = \{ w \in W : gw = w \}$$
is the $1$-eigenspace of $g$ and
$$W_{-1} = \{ w \in w : gw = -w \}$$
is the $-1$-eigenspace of $g$. This direct sum decomposition comes explicitly from writing every vector $w$ as the sum of its "even" and "odd" parts
$$w = \frac{w + gw}{2} + \frac{w - gw}{2}.$$
The intersection of $W_1$ and $W_{-1}$ is trivial because if $gw = w$ and $gw = -w$, then $w = -w$, so $w = 0$. This is all a special case of a more general result about how to decompose representations of any finite group.
Now, $C_2$ acts on $V \otimes V$ for any vector space $V$ via the symmetry
$$g(v \otimes w) = w \otimes v.$$

Exercise 1: Show that the $1$-eigenspace of this action is $\text{Sym}^2(V)$ and that the $-1$-eigenspace is $\text{Alt}^2(V)$.
Exercise 2: Show that the fact that the $C_2$ action above commutes with the natural $GL(V)$ action implies that the $1$-eigenspace and $-1$-eigenspace of $C_2$ are $GL(V)$-invariant.

A: To present Qiaochu's answer in a slightly different way, define a linear map $T: V \otimes V \to V \otimes V$ on pure tensors by $T(v \otimes w) = w \otimes v$ and extend linearly. A good name for this operator would be a "flip": it interchanges the sides of the tensor product.
You can show that $\mathrm{Sym}^2 V$ is contained in the $1$-eigenspace of $T$, and that $\mathrm{Alt}^2 V$ is contained in the $(-1)$-eigenspace of $T$, which gives you that $\mathrm{Sym}^2 V \cap \mathrm{Alt}^2 V = 0$.
Now all that's left is to show that $\mathrm{Sym}^2 V + \mathrm{Alt}^2 V = V \otimes V$, which can be done by taking any element on the right hand side and writing it as a sum of a symmetric and antisymmetric part.
