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  "body": "\n## Statement\n\nLet $V$ be an inner product space over $\\R$ or $\\C$, and let $v_1, \\dots, v_k \\in V$ be linearly independent.\nLet $u_1, \\dots, u_k$ be the vectors of the [[Gram–Schmidt process|0VT6P0GNQR]].\nThen $u_1, \\dots, u_k$ are nonzero and pairwise [[orthogonal|0Y2X7P2WB7]], and\n$$\\operatorname{span}(u_1, \\dots, u_j) = \\operatorname{span}(v_1, \\dots, v_j) \\qquad (1 \\le j \\le k).$$\n\n## Remarks\n\nThe proof is by induction on $j$: by construction, $\\langle u_j, u_i \\rangle = 0$ for $i \u003c j$.\nThe vector $u_j$ is nonzero, because $v_j$ is not in $\\operatorname{span}(v_1, \\dots, v_{j-1})$.\nIf 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.\n\n## Explanations\n\n### L1 human\n\nThe vectors made by Gram–Schmidt are really perpendicular to each other.\nAlso, the first $j$ new vectors span the same space as the first $j$ old vectors.\nSo no information is lost at any step.\n",
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