Linear Algebra Done Right · Problem 10

Inner Product Spaces — Sheldon Axler

Mathematics — Linear Algebra · gram-schmidt-procedure

Problem

Suppose v1,…,vmv_1, \ldots, v_m is a linearly independent list in VV. Explain why the orthonormal list produced by the formulas of the Gram--Schmidt procedure (6.32) is the only orthonormal list e1,…,eme_1, \ldots, e_m in VV such that ⟨vk,ek⟩>0\langle v_k, e_k \rangle > 0 and span⁡(v1,…,vk)=span⁡(e1,…,ek)\operatorname{span}(v_1, \ldots, v_k) = \operatorname{span}(e_1, \ldots, e_k) for each k=1,…,mk = 1, \ldots, m. The result in this exercise is used in the proof of 7.58.
Intuitive explanation

The Gram–Schmidt procedure builds an orthonormal basis by iteratively projecting out previously-spanned directions, ensuring each new vector stays in the same span while gaining orthogonality. The key insight is that these two constraints—matching spans at each step and maintaining positive inner products—uniquely pin down each eke_k: it must be the unique unit vector in span⁡(v1,…,vk)\operatorname{span}(v_1, \ldots, v_k) orthogonal to e1,…,ek−1e_1, \ldots, e_{k-1} with positive projection onto vkv_k. This forces the Gram–Schmidt output to be the only list satisfying all conditions.

Intuitive explanation

The Gram–Schmidt procedure builds an orthonormal basis by iteratively projecting out previously-spanned directions, ensuring each new vector stays in the same span while gaining orthogonality. The key insight is that these two constraints—matching spans at each step and maintaining positive inner products—uniquely pin down each eke_k: it must be the unique unit vector in span⁡(v1,…,vk)\operatorname{span}(v_1, \ldots, v_k) orthogonal to e1,…,ek−1e_1, \ldots, e_{k-1} with positive projection onto vkv_k. This forces the Gram–Schmidt output to be the only list satisfying all conditions.

Related theory

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