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recently I have a linear system $A\mathbf{x}=\mathbf{b}$ to solve. Suppose $A^\dagger$ is the Moore-Penrose inverse of $A$. If $A$ is overdetermined, i.e. more rows than columns, then $\mathbf{x}=A^\dagger\mathbf{b}$ is the unique least squares solution. If $A$ is underdetermined, i.e. more columns than rows, though there are infinitely many solutions, then $\mathbf{x}=A^\dagger\mathbf{b}$ is the unique solution with minimum norm.

Is my understanding right? Thanks for help!

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  • $\begingroup$ Yes, that's right. $\endgroup$ – Omnomnomnom Sep 6 '16 at 3:19
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Details worked out in Is the unique least norm solution to Ax=b the orthogonal projection of b onto R(A)?.

The picture shows the uniqueness condition of a trivial nullspace $\mathcal{N}\left( \mathbf{A} \right).$ The least squares minimizers are an affine space.

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