Let \(V\) be a finite-dimensional inner product space, and let \(\{\alpha_1, \ldots, \alpha_n\}\) be an orthonormal basis for \(V\). Show that for any vectors \(\alpha, \beta\) in \(V\) \[ (\alpha \mid \beta) = \sum_{k=1}^{n} (\alpha \mid \alpha_k)\overline{(\beta \mid \alpha_k)}. \]
Solution
Because \(\{\alpha_1,\ldots,\alpha_n\}\) is an orthonormal basis, every vector \(\alpha\) has the expansion \[ \alpha=\sum_{k=1}^n(\alpha\mid\alpha_k)\alpha_k \] [@HK71]. Likewise, \[ \beta=\sum_{j=1}^n(\beta\mid\alpha_j)\alpha_j. \] Using linearity in the first argument and conjugate linearity in the second, \[ \begin{aligned} (\alpha\mid\beta) &=\sum_{k,j} (\alpha\mid\alpha_k) \overline{(\beta\mid\alpha_j)} (\alpha_k\mid\alpha_j)\\ &=\sum_{k=1}^n (\alpha\mid\alpha_k) \overline{(\beta\mid\alpha_k)}, \end{aligned} \] because orthonormality gives \((\alpha_k\mid\alpha_j)=\delta_{kj}\). Hence \[ \boxed{ (\alpha\mid\beta) =\sum_{k=1}^n (\alpha\mid\alpha_k) \overline{(\beta\mid\alpha_k)}.} \]