Sum over all characters $\chi$ mod $m$

Question

I'm trying to show that \begin{equation*} \sum_{\chi \text{ mod } m} \left| \sum_{i=1}^{m} t_{i}\chi(i)\right|^{2} = \varphi(m)\sum_{i'=1}^{m} |t_{i'}|^{2}, \end{equation*} where $(i', m) = 1$.

I feel like I'm supposed to use the fact that $|z|^{2} = z \bar z$ and/or the orthogonality of the Dirichlet character, but I'm not really sure how to proceed in the proof. Any hints would help, thanks! I think this can be proved using Parseval's theorem, but we haven't learned this. I think we're supposed to do it with simple manipulations of sums.

Answer 1

$$|\sum_{i=1}^{m} t_{i}\chi(i)|^{2} = |\sum_{i'=1}^{m} t_{i'}\chi(i')|^{2}=\sum_{i'=1}^{m}\sum_{j'=1}^{m}t_{i'}\bar{t}_{j'}\chi(i')\bar{\chi}(j')$$ Where $(i',m)=(j',m)=1.$So we can assume $$i'\equiv a_{i'j'}j'\mod m.$$So $$\sum_{\chi \text{ mod } m} \left| \sum_{i=1}^{m} t_{i}\chi(i)\right|^{2}= \sum_{\chi \mod m}\sum_{i'=1}^{m}\sum_{j'=1}^{m}t_{i'}\bar{t}_{j'}\chi(i')\bar{\chi}(j')=\sum_{i'=1}^{m}\sum_{j'=1}^{m}\sum_{\chi \mod m}t_{i'}\bar{t}_{j'}\chi(a_{i'j'}).$$ As we know $$\sum_{\chi \mod m}\chi(i)=\varphi(m),i\equiv 1\mod m,\sum_{\chi \mod m}\chi(i)=\varphi(m),\text{elsewise}$$ And $a_{i'j'}=1$ when $i'=j'$.Hence $$\sum_{\chi \text{ mod } m} \left| \sum_{i=1}^{m} t_{i}\chi(i)\right|^{2}=\sum_{i'=1}^{m}\sum_{j'=1}^{m}\sum_{\chi \mod m}t_{i'}\bar{t}_{j'}\chi(a_{i'j'})=\sum_{i'=1}^{m}\sum_{j'=1}^{m}\sum_{\chi \mod m}t_{i'}\bar{t}_{j'}\varphi(m)\delta_{i'j'}=\varphi(m)\sum_{i'=1}^{m} |t_{i'}|^{2}$$ Where $\delta_{i'j'}=1$ if $i'=j'$ and is 0 for any other cases.

Answer 2

$$\{\frac{\chi}{\|\chi\|},\chi\bmod m\}$$ is an orthonormal basis of the $\varphi(m)$ dimensional vector space of functions from $\{ a\in 1\ldots m, \gcd(a,m)=1\}$ to $\Bbb{C}$