Gram–Schmidt vectors are orthogonal

Statement

Let $V$ be an inner product space over $\R$ or $\C$, and let $v_1, \dots, v_k \in V$ be linearly independent. Let $u_1, \dots, u_k$ be the vectors of the Gram–Schmidt process. Then $u_1, \dots, u_k$ are nonzero and pairwise orthogonal, and $$\operatorname{span}(u_1, \dots, u_j) = \operatorname{span}(v_1, \dots, v_j) \qquad (1 \le j \le k).$$

Remarks

The proof is by induction on $j$: by construction, $\langle u_j, u_i \rangle = 0$ for $i < j$. The vector $u_j$ is nonzero, because $v_j$ is not in $\operatorname{span}(v_1, \dots, v_{j-1})$. If the $v_i$ are only spanning, then the process produces some zero vectors, and dropping them still gives an orthogonal family with the same span.

Explanations

L1 human

The vectors made by Gram–Schmidt are really perpendicular to each other. Also, the first $j$ new vectors span the same space as the first $j$ old vectors. So no information is lost at any step.