Co-ordinate free orthogonality of characters Let $G$ be a finite group and $\chi_1,\chi_2$ two characters corresponding to irreducible representations $\rho_1,\rho_2: G\to V,W$. It is a basic fact that $\langle\chi_1,\chi_2 \rangle = 0$ unless $V \cong W$. The inner product is the standard one defined on $\operatorname{Hom}(G,\Bbb C^\times)$. 
The standard proof considers an arbitrary map $f:V\to W$ and shows that:
$$f' = \sum_{g\in G}\rho_2^{-1}(g)\circ f\circ\rho_1(g)$$
is $0$ unless $V\cong W$. Then, you fix basis for $V$ and $W$ and do a computation that shows what we want.
This is not really satisfactory since what we want to prove is co-ordinate free and even the setup to our proof is co-ordinate free. Is it possible to find a truly co-ordinate free proof of this fact? 
A similar proof is usually used to also show that $\langle \chi, \chi \rangle = 1$ but this follows easily since $\chi(g^{-1}) = \overline{\chi(g)} = \chi(g)^{-1}$ since the eigenvalues are roots of unity (the group is finite, hence torsion).
I am just beginning to learn representation theory, so the fewer big theorems one uses the better. However, if it is clearer to use some results (without being circular), please feel free to do so.
 A: A more conceptual way to see this is as follows:
Lemma: Let $\rho\colon G\rightarrow {\rm GL}(V)$ be a $G$-representation.
Then $\varphi\colon V\rightarrow V$, $v\mapsto \frac1{\lvert G\rvert} \sum_{g\in
G} \rho(g)v$ is a $G$-linear and ${\rm Im}(\varphi) = V^G$, the $G$-invariants.
If $\chi_V$ is the character of $V$, then it follows from the Lemma that 
$$\dim
V^G = {\rm Tr}(\varphi) = \frac1{\lvert G\rvert} \sum_{g\in G}{\rm Tr}(\rho(g)) =
\frac1{\lvert G\rvert} \sum_{g\in G}\chi_V(g).$$
This is a very useful observation: For two irreducible representations
$\rho\colon G\rightarrow {\rm GL}(V)$ and $\sigma\colon G\rightarrow {\rm
GL}(W)$, we obtain a representation $\tau\colon G\rightarrow {\rm GL}({\rm
Hom}(V,W))$ given by
$$
(\tau(g)f)(v):= \sigma(g)f\bigl(\rho(g^{-1})v\bigr),\qquad \text{for $g\in G$,
$v\in V$, $f\in {\rm Hom}(V,W)$.}
$$
With this action, the natural isomorphism ${\rm Hom}(V,W)\rightarrow V^*\otimes
W$ becomes a $G$-linear map ($V^*$ is the dual representation of $V$ and $G$
acts on $V^*$ and $V^*\otimes W$ as usual). Notice, that ${\rm Hom}(V,W)^G$ (i.
e. the $G$-invariants) are precisely the $G$-linear maps $V\rightarrow W$.
With this, we can calculate:
\begin{align*}
\dim {\rm Hom}(V,W)^G &= \frac1{\lvert G\rvert}\sum_{g\in G}\chi_{{\rm Hom}(V,W)}
(g) = \frac1{\lvert G\rvert}\sum_{g\in G} \chi_{V^*\otimes W}(g)\\
&= \frac1{\lvert G\rvert} \sum_{g\in G} \chi_{V^*}(g)\cdot \chi_W(g) =
\frac1{\lvert G\rvert} \sum_{g\in G} \chi_V(g^{-1}) \chi_W(g)\\
&= \langle \chi_V,\chi_W\rangle.
\end{align*}
This identity also extends by linearity to reducible $V$ and $W$.
In your situation, you have shown that ${\rm Hom}(V,W)^G = 0$ and hence from the
above discussion, it follows that $\langle \chi_V,\chi_W\rangle = 0$.
