Induction of an irreducible group representation I'm having some trouble finding the answer to the following question. Any ideas on how to get started?

Let $H$ be a subgroup of a group $G$ and let $U_{1}$, ...,$U_{k}$ be the irreducible representations of $G$. Further, assume that $\psi_{1}$, ..., $\psi_{k}$ are the ($H$-)characters of the restrictions $Res(U_{1})$, ..., $Res(U_{k})$.  If $V$ is an irreducible $H$-representation, give a formula to compute the decomposition  $Ind($V$) = U^{\oplus e_{1}}_{1} \oplus U^{\oplus e_{2}}_{2} \oplus ... \oplus U^{\oplus e_{k}}_{k}$.

 A: The binary operation $\dim\hom_G$ behaves as an "inner product" on representations of $G$. In particular, every representation of $G$ is a direct sum of irreps $U_1,\cdots,U_k$ and
$$\dim\hom_G(U_1^{e_1}\oplus\cdots\oplus U_k^{e_k},U_1^{f_1}\oplus\cdots\oplus U_k^{f_k})=e_1f_1+\cdots+e_kf_k.$$
This follows from the distributivity of $\hom$ and Schur's lemma. As a result, one can use homs to siphon out the multiplicities of known irreps in a given rep. If $V\cong U_1^{e_1}\oplus\cdots\oplus U_k^{e_k}$ for some unknown multiplicities $e_1,\cdots,e_k$ then we may equate them with the dimensions of the hom spaces via the rule $e_i=\dim\hom_G(U_i,V)$.
From here, note that $\hom(U,V)\cong U^*\otimes V$ (for finite-dim spaces). Indeed, given an arbitrary element $\varphi\otimes v\in U^*\otimes V$, it acts as a function $U\to V$ by letting the dual vector from $U^*$ act on the argument, yielding a scalar that one may multiply $v$ by, i.e. $(\varphi\otimes v)(u)=\varphi(u)v$.
Moreover, when $U$ and $V$ are $G$-reps, $G$ acts on $\hom(U,V)$ via $g:A\mapsto gAg^{-1}$. My notation here is shorthand for $\rho_{\small V}(g)A\rho_{\small U}(g)^{-1}$ technically. And of course $G$ acts on $U^*$ by the contragredient action, and so $G$ acts on $U^*\otimes V$ "diagonally." One checks the aforementioned isomorphism $\hom(U,V)\cong U^*\otimes V$ is one of representations. Thus, we may say
$$\chi_{\hom(U,V)}(g)=\chi_{\small U^*\otimes V}(g)=\chi_{\small U^*}(g)\chi(g)=\overline{\chi_{\small U}(g)}\chi_{\small V}(g).$$
Now, $\hom_G(U,V)=\hom(U,V)^G$, and one may compute $\dim W^G$ for arbitrary reps $W$ using characters explicitly. Indeed, the operator $\frac{1}{|G|}\sum_{g\in G}\rho_{\small W}(g)\in{\rm End}(W)$ one may check is a projection onto the subspace of invariants $W^G$. The image of any projection has dimension equal to the projection's trace, so $\dim(W^G)=\frac{1}{|G|}\sum_{g\in G}\chi_{\small W}(g)$.
Finally, via Frobenius reciprocity, we have
$$\begin{array}{ll} e_i & =\dim\hom_G(U_i,{\rm Ind}_H^GV) \\ & =\dim\hom_H({\rm Res}_H^G U_i,V) \\ & \displaystyle =\frac{1}{|H|}\sum_{h\in H}\overline{\psi_i(h)}\chi_{\small V}(h). \end{array}$$
