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In underdetermined system, $\mathbf{y}=\mathbf{A}\mathbf{x}$ is with $\mathbf{A}_{[M\times N]}$, $M<N$ and $\mathbf{y}_{[M\times 1]}$, $\mathbf{x}_{[N\times 1]}$. The estimation of $\mathbf{x}$ is then non-unique as

\begin{equation} \hat{\mathbf{x}} = \mathbf{A}^{\dagger}\mathbf{y}+(\mathbf{I}-\mathbf{A}^{\dagger}\mathbf{A})\mathbf{w} \end{equation} \begin{equation} \mathbf{A}^{\dagger} = (\mathbf{A}^{H}\mathbf{A})^{-1}\mathbf{A}^{H} \end{equation} where $\mathbf{w}_{[N\times 1]}$ is an arbitrary matrix and $(\cdot)^{H}$ stands for matrix pseudo-inverse. It is said that the pseudoinverse may be used to construct the solution of minimum Euclidean norm $||\mathbf{x}||_2$ among all solutions.

I would like to know why the solution of $\hat{\mathbf{x}}$ is considered to be with the minimum Euclidean norm $||\mathbf{x}||_2$? What is the rationale behind this argument?

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closed as unclear what you're asking by user91500, Dando18, Siong Thye Goh, Shailesh, Leucippus Aug 21 '17 at 0:35

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ This question confuses Hermitian transpose and Moore-Penrose pseudo-inverse. Furthermore, the use of classical inverse presupposes certain regularity of matrix $A$ which is not stated anywhere and which in fact makes the least squares solution unique. As stated it makes little sense to look for a solution with minimum norm as the solution is unique. $\endgroup$ – batman Aug 20 '17 at 16:15