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Why is the solution to the normal equations ATAxA^T A x* = ATbA^T b considered the least-squares solution?

The solution to the normal equations is considered the least-squares solution because it satisfies the necessary and sufficient condition for minimizing the residual norm ∥b⃗−Ax⃗∥\|\vec{b} - A\vec{x}\|. The derivation shows that minimizing this norm is equivalent to requiring the residual Ax⃗∗−b⃗A\vec{x}^* - \vec{b} to be orthogonal to the column space C(A)C(A). This orthogonality condition translates algebraically to AT(Ax⃗∗−b⃗)=0⃗A^T(A\vec{x}^* - \vec{b}) = \vec{0}, which is exactly the normal equation. Thus, any x⃗∗\vec{x}^* solving the normal equations yields the projection of b⃗\vec{b} onto C(A)C(A), providing the best approximate solution.

Conditions

  • The original system Ax⃗=b⃗A\vec{x}=\vec{b} may be inconsistent
  • ATAx⃗∗=ATb⃗A^T A \vec{x}^* = A^T \vec{b} has a solution

Reasoning, step by step

  1. Recall that the least-squares goal is to make Ax⃗∗A\vec{x}^* the closest point in C(A)C(A) to b⃗\vec{b}.
  2. Identify that closeness implies the residual Ax⃗∗−b⃗A\vec{x}^* - \vec{b} is orthogonal to C(A)C(A).
  3. Translate orthogonality to the condition Ax⃗∗−b⃗∈N(AT)A\vec{x}^* - \vec{b} \in N(A^T).
  4. Convert null-space membership to the equation AT(Ax⃗∗−b⃗)=0⃗A^T(A\vec{x}^* - \vec{b}) = \vec{0}.
  5. Expand to ATAx⃗∗=ATb⃗A^T A \vec{x}^* = A^T \vec{b}.
  6. Conclude that solving this equation finds the specific x⃗∗\vec{x}^* that achieves the minimum distance.

Example

The speaker states, 'This right here will always have a solution, and this right here is our least squares solution,' referring to the boxed normal equations.

Common misconceptions

  • Believing that the normal equations provide an exact solution to the original inconsistent system Ax⃗=b⃗A\vec{x}=\vec{b}.
  • Thinking that the normal equations are only an approximation method rather than the exact characterization of the minimizer.

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