 Solution

For the given orthonormal basis, the orthogonal projection isIt belongs to $W$, and $⟨v-\pi v,e_j⟩=0$ for every basis vector. Thus $v-\pi v\in W^{\perp }$. For any $w\in W$, decompose $v-w=(v-\pi v)+(\pi v-w)$. The summands are orthogonal, so the Pythagorean theorem in an inner-product space for the inner product givesEquality holds exactly when $w=\pi v$. Hence $\pi v$ is the unique closest point of $W$ to $v$. This proves the minimizing property as well as the formula, including $W=\{0\}$ and $W=V$.