Proof of the formula for coordinates in an orthonormal basis

Proof

We compute the inner product of $v$ with each basis vector.

1. Let $V$ be a finite-dimensional inner product space over $\mathbb{K} = \R$ or $\C$, with inner product $\ip{\cdot}{\cdot}$. Let $(e_1, \dots, e_n)$ be an orthonormal basis of $V$, so $\ip{e_j}{e_i} = \delta_{ji}$.
2. Let $v \in V$. Since $(e_1, \dots, e_n)$ is a basis, there are scalars $a_1, \dots, a_n \in \mathbb{K}$ with $v = \sum_{j=1}^{n} a_j e_j$.
3. Fix $i$ with $1 \le i \le n$. Linearity in the first argument gives $\ip{v}{e_i} = \sum_{j=1}^{n} a_j \ip{e_j}{e_i}$.
4. By orthonormality, every term with $j \ne i$ is $0$, and the term with $j = i$ is $a_i$. So $\ip{v}{e_i} = a_i$.
5. Substituting step 4 into step 2 gives $v = \sum_{i=1}^{n} \ip{v}{e_i} e_i$.
6. The coordinates of $v$ in a basis are unique, so they are the numbers $a_i = \ip{v}{e_i}$.