[proof] The key idea is to start from a vector-space complement — which always exists by linear algebra — and "symmetrise" it into a $G$-invariant complement using group averaging. From linear algebra, $W$ has a vector-space complement $W'$ with $V = W \oplus W'$ as vector spaces (not yet as $G$-spaces). Let $q: V \to W$ be the projection onto $W$ along $W'$: for $v = w + w'$ with $w \in W$, $w' \in W'$, set $q(v) = w$. The problem is that $q$ need not be $G$-equivariant. Define the averaged projection \begin{align*} \bar{q}(v) = \frac{1}{|G|} \sum_{g \in G} g \cdot q(g^{-1} v). \end{align*} (Here $\frac{1}{|G|}$ exists because $\operatorname{char} \mathbb{F} = 0$.) One checks three properties: (1) $\operatorname{im} \bar{q} \subseteq W$: for any $v \in V$, the vector $q(g^{-1}v)$ lies in $W$, and $W$ is $G$-invariant so $g \cdot q(g^{-1}v) \in W$; the average of elements of $W$ lies in $W$. (2) $\bar{q}(w) = w$ for all $w \in W$: since $W$ is $G$-invariant, $g^{-1}w \in W$ and $q$ fixes $W$, so each summand contributes $g \cdot g^{-1}w = w$; dividing by $|G|$ and summing gives $w$. (3) $G$-equivariance, $\bar{q}(hv) = h\bar{q}(v)$ for all $h \in G$: substitute $g' = hg$ in the sum defining $\bar{q}(hv)$ and use that $g \mapsto hg$ permutes $G$; the sum becomes the definition of $h\bar{q}(v)$. Properties (1) and (2) say $\bar{q}$ is a $G$-equivariant projection onto $W$. Setting $U = \ker \bar{q}$: equivariance implies $U$ is $G$-invariant (if $\bar{q}(v) = 0$ then $\bar{q}(hv) = h\bar{q}(v) = 0$). Since $\bar{q}$ is a projection, $V = \operatorname{im}\bar{q} \oplus \ker\bar{q} = W \oplus U$ as $G$-spaces. [/proof]