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Why is the least-squares fit Ax* identified with the orthogonal projection of b onto C(A)C(A)?

The least-squares fit Ax⃗∗A\vec{x}^* is identified with the orthogonal projection of b⃗\vec{b} onto C(A)C(A) because the goal of least squares is to find the vector in the subspace C(A)C(A) that is closest to b⃗\vec{b}. A fundamental geometric property of subspaces states that the unique closest point in a subspace to an external vector is its orthogonal projection. Therefore, Ax⃗∗=proj⁡C(A)b⃗A\vec{x}^* = \operatorname{proj}_{C(A)} \vec{b}.

Conditions

  • C(A)C(A) is a subspace of Rn\mathbb{R}^n
  • b⃗∈Rn\vec{b} \in \mathbb{R}^n
  • Distance is measured by the Euclidean norm

Reasoning, step by step

  1. Define the least-squares objective: minimize ∥b⃗−Ax⃗∗∥\|\vec{b} - A\vec{x}^*\| subject to Ax⃗∗∈C(A)A\vec{x}^* \in C(A).
  2. Invoke the geometric theorem: For any subspace WW and vector b⃗\vec{b}, the vector w⃗∈W\vec{w} \in W minimizing ∥b⃗−w⃗∥\|\vec{b} - \vec{w}\| is proj⁡Wb⃗\operatorname{proj}_W \vec{b}.
  3. Apply this theorem with W=C(A)W = C(A).
  4. Conclude that the optimal image vector Ax⃗∗A\vec{x}^* must equal proj⁡C(A)b⃗\operatorname{proj}_{C(A)} \vec{b}.

Example

The video boxes the equation Ax⃗∗=proj⁡C(A)b⃗A\vec{x}^* = \operatorname{proj}_{C(A)} \vec{b} after explaining that the closest vector in the subspace is the projection.

Common misconceptions

  • Thinking that the projection is onto the null space instead of the column space.
  • Believing that any vector in C(A)C(A) is equally close to b.

Watch the explanation

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.