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I recently began working on some problems related to inner product spaces in linear algebra, and I ran into one that seems somewhat extensive that I could use some help on. Consider $v_1,...,v_n$ to be an orthonormal basis for inner product space $V$. I want to prove that for any $x,y \in V$, $$\langle x,y \rangle = \sum_{k=1}^n \langle x,v_k \rangle \overline{\langle y,v_k\rangle },$$

where $\overline{\langle a,b \rangle }$ represents the conjugate of the inner product $\langle a,b \rangle.$

My friend recommended that I should recognize that $$x = \sum_{i=1}^n \alpha_i v_i$$ where $$\alpha_1,...,\alpha_n \in \mathbb{C}$$ $$y = \sum_{i=1}^n \beta_i v_i$$ where $$\beta_1,...,\beta_n \in \mathbb{C}.$$

I can see that $$\langle x,y \rangle = \left\langle \sum_{i=1}^n \alpha_i v_i, \sum_{i=1}^n \beta_i v_i \right\rangle = \sum_{i=1}^n \alpha_i \bar{\beta}_i \langle v_i,v_i \rangle = \sum_{i=1}^n \alpha_i \bar{\beta}_i.$$

Hence, This really comes down to showing that $\alpha_i = \langle x,v_i \rangle \wedge \bar{\beta}_i = \overline{\langle y,v_i \rangle}.$ However, I am having some difficulties showing this step. Any recommendation on how to best argue the previous statements?

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  • $\begingroup$ Note that you can escape the angular brackets correctly by using the \langle ($\langle$) and \rangle ($\rangle$) commands. $\endgroup$ – stochasticboy321 Sep 9 '16 at 2:21
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Go the other way round:

$$ \begin{equation} \begin{split} \displaystyle\sum_{k=1}^n \langle x,v_k\rangle \overline{\langle y,v_k\rangle}= & \displaystyle\sum_{k=1}^n \bigg\langle \displaystyle\sum_{i=1}^n \alpha_iv_i,v_k\bigg\rangle \overline{\bigg\langle \displaystyle\sum_{i=1}^n \beta_iv_i,v_k\bigg\rangle} \\ &= \displaystyle\sum_{k=1}^n \Bigg(\displaystyle\sum_{i=1}^n \alpha_i\langle v_i,v_k\rangle\Bigg) \Bigg(\overline{\displaystyle\sum_{i=1}^n \beta_i\langle v_i,v_k\rangle}\Bigg)\\ & = \displaystyle\sum_{k=1}^n \alpha_k \overline{\beta_k}\\ & = \displaystyle\sum_{k=1}^n \alpha_k \overline{\beta_k} \langle v_k,v_k\rangle\\ & = \bigg\langle \displaystyle\sum_{k=1}^n \alpha_kv_k, \displaystyle\sum_{i=1}^n \beta_iv_i \bigg\rangle\\ & = \langle x,y\rangle \end{split} \end{equation} $$

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