Proof for $u · v = (u · w_1)(v · w_1) + ... + (u · w_n)(v · w_n)$ - Parseval's identity

Question

Suppose {w₁, w₂, ... w_n} was an orthonormal basis for Rⁿ and u and v were vectors in Rⁿ.

I'm trying to prove that u · v = (u · w₁)(v ·w₁) + ... + (u · wn_n)(v · w_n)

I know that since {w₁, w₂, ... w_n} is an orthonormal basis that spans a subspace that contains u and v, u can be rewritten as

(u · w₁)w₁ + (u · w₂)w₂ + ... + (u · w_n)w_n

and v can be rewritten as

(v · w₁)w₁ + (v · w₂)w₂ + ... + (v · w_n)w_n

However when I plug in these new values for u and v, I end up getting

(u · w₁)w₁(v · w₁)w₁ + (u · w₂)w₂(v · w₂)w₂ + ... + (u · w_n)w_n(v · w_n)w_n

I could try factoring the w's out but that won't lead to the desired result. Perhaps I'm approaching this problem the wrong way? Any feedback is appreciated, thanks in advance.

Answer 1

$$u=\sum_{j=1}^n (u\cdot w_j)w_j$$ $$v=\sum_{k=1}^n (v\cdot w_k)w_k$$ $$u\cdot v=\sum_{jk} (u\cdot w_j)(u\cdot w_k)(w_j\cdot w_k)\\ =\sum_{j=k} (u\cdot w_j)(u\cdot w_k)1+\sum_{j\ne k} (u\cdot w_j)(u\cdot w_k)0\\ =\sum_{k=1}^n (u\cdot w_k)(u\cdot w_k)$$