In Maschke's theorem, does the averaged projector $\bar P_U$ equal $P_U$? Consider a finite group $G$ represented on a $\mathbb K$-vector space $V$ with $\operatorname{char}\mathbb K$ not dividing $|G|$, and let $U$ be an invariant subspace for the representation, and $W$ another subspace such that $V=U\oplus W$.
Then, to prove that $W$ is also an invariant, given the projector $P_U$ onto the invariant $U$ we define the averaged projector $\bar P_U$ as
$$\bar P_U = \frac{1}{|G|}\sum_{g\in G}g^{-1}P_U g.$$
This can be shown to be an intertwiner for the representation, $\bar P_U g=g \bar P_U$ for all $g\in G$, and a projector itself, $\bar P_U^2=\bar P_U$.
While I think I understand while this is the case, I'm not quite clear on the relation between $P_U$ and $\bar P_U$. They clearly agree on $U$: $P_U u=\bar P_U u=u$ for all $u\in U$, and they are both projectors. Doesn't this imply that $P_U=\bar P_U$?
I say this because if $P$ and $Q$ are projectors defined on the same vector space with the same image, then $(P-Q)x=0$ both for $x\in\operatorname{im}(P)=\operatorname{im}(Q)$ and for $x\in\ker P=\ker Q$, and thus $P=Q$.
If this is so, shouldn't there be a more direct way to prove the theorem working directly with $P_U$?
 A: The best way to see this is via examples. 
First of all, as darij grinberg pointed out in the comments, two projectors with the same image need not agree : take the plane $k^2$, there is one projector onto $k\times \{0\}$ with kernel $\{0\}\times k$, and one with kernel $k (1,1)$ for instance. 
Note that over an arbitrary field, the notion of "orthogonal" projector might not make sense (and even over $\mathbb {R,C}$, if you did not specifiy the scalar product on your vector space, it doesn't make sense)
So for instance take the action of $\mathfrak S_2$ on $k^2$ that simply permutes the coordinates. An obvious invariant subspace is $U=k(1,1)$. Indeed its elements are of the form $(\lambda, \lambda)$, so if you permute the coordinates, you don't move. Now, a possible projector might be $p$ with kernel $k\times \{0\}$, indeed $k\times \{0\}\oplus k(1,1) = k^2$. 
Explicitly, you have $p(x,y) = (y,y)$. This is clearly not $\mathfrak S_2$-invariant : $p(\sigma(\lambda, \mu)) = p(\mu,\lambda) = (\lambda, \lambda) \neq (\mu,\mu) = \sigma (\mu, \mu) = \sigma p(\lambda, \mu)$, if $\lambda\neq \mu$ where $\sigma\in \mathfrak S_2$ is the transposition. 
Now if you average $p$ over $\mathfrak S_2$, you get $q= \frac{1}{2}(p + \sigma p \sigma)$ so $q(x,y) = \frac{1}{2}((y,y) + (x,x))= \frac{1}{2}(x+y,x+y)$. 
Its kernel is now $k(1,-1)$ , which is now clearly an invariant subspace (its elements are of the form $(\lambda, -\lambda)$, so if you permute the coordinates you get $(-\lambda, -(-\lambda))$, which is still in the space)
This is the simplest example, but of course you can always find examples. 
