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Answers for “A^T A x = A^T b 的解与 Ax=b 的解相同吗?”

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The residual vector r⃗=Ax⃗∗−b⃗\vec{r} = A\vec{x}^* - \vec{b} is orthogonal to the column space C(A)C(A) because Ax⃗∗A\vec{x}^* is the orthogonal projection of b⃗\vec{b} onto C(A)C(A). Orthogonality to C(A)C(A) means r⃗\vec{r} is in the orthogonal complement C(A)⊥C(A)^\perp.

Conditions: Ax⃗∗A\vec{x}^* is the orthogonal projection of b⃗\vec{b} onto C(A)C(A); C(A)⊥=N(AT)C(A)^\perp = N(A^T) (Fundamental Theorem of Linear Algebra); Matrix multiplication distributes over subtraction