[proofplan] We define $\pi(v) = \sum_{i=1}^k (e_i, v)\,e_i$ and verify $v - \pi(v) \in W^\perp$, confirming this is the projection from the Orthogonal Decomposition. The best approximation property follows from Pythagoras: $v - w$ decomposes into orthogonal pieces $(v - \pi(v))$ and $(\pi(v) - w)$, so $\|v - w\|^2 \geq \|v - \pi(v)\|^2$. [/proofplan] [step:Define $\pi$ and verify $v - \pi(v) \in W^\perp$] Define $\pi: V \to W$ by $\pi(v) = \sum_{i=1}^k (e_i, v)\,e_i$. Then $\pi(v) \in W$ since it is a linear combination of the basis vectors of $W$. For each $j = 1, \dots, k$: \begin{align*} (e_j, v - \pi(v)) = (e_j, v) - \sum_{i=1}^k (e_i, v)\,(e_j, e_i) = (e_j, v) - (e_j, v) = 0. \end{align*} Since $(e_1, \dots, e_k)$ spans $W$, this shows $v - \pi(v) \in W^\perp$. So the decomposition $v = \pi(v) + (v - \pi(v))$ is the [Orthogonal Decomposition](/theorems/436) of $v$ with respect to $W$. [guided] The formula $\pi(v) = \sum_{i=1}^k (e_i, v)\,e_i$ is the sum of the "components" of $v$ along each orthonormal basis vector. The coefficient $(e_i, v)$ measures how much of $v$ lies in the direction of $e_i$. To verify $v - \pi(v) \perp W$, it suffices to check against the basis vectors of $W$. For each $j$: \begin{align*} (e_j, v - \pi(v)) = (e_j, v) - \sum_{i=1}^k (e_i, v)\,(e_j, e_i). \end{align*} Orthonormality gives $(e_j, e_i) = \delta_{ji}$, so the sum collapses to $(e_j, v)$, which cancels the first term. The residual $v - \pi(v)$ is therefore orthogonal to every $e_j$, hence to all of $W$. [/guided] [/step] [step:Prove the best approximation property via the Pythagorean theorem] For any $w \in W$, write $v - w = (v - \pi(v)) + (\pi(v) - w)$. Since $v - \pi(v) \in W^\perp$ and $\pi(v) - w \in W$, these two vectors are orthogonal. By the Pythagorean theorem: \begin{align*} \|v - w\|^2 = \|v - \pi(v)\|^2 + \|\pi(v) - w\|^2 \geq \|v - \pi(v)\|^2, \end{align*} with equality if and only if $\|\pi(v) - w\|^2 = 0$, i.e., $w = \pi(v)$. [guided] The best approximation property says that $\pi(v)$ is the closest point in $W$ to $v$. The proof exploits the orthogonal decomposition: the error $v - w$ splits into a "fixed" part $v - \pi(v)$ (the perpendicular distance to $W$, which does not depend on $w$) and a "variable" part $\pi(v) - w$ (which lies inside $W$). Since these two parts are orthogonal, the Pythagorean theorem gives: \begin{align*} \|v - w\|^2 = \|v - \pi(v)\|^2 + \|\pi(v) - w\|^2. \end{align*} The first term is a constant (independent of $w$), and the second term is $\geq 0$ with equality iff $w = \pi(v)$. So $\|v - w\|$ is minimised uniquely at $w = \pi(v)$. This is the geometric content of orthogonal projection: the shortest path from a point to a subspace is the perpendicular one. [/guided] [/step]