# pseudo-inverse least square solution [closed]

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?

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

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• 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. – batman Aug 20 '17 at 16:15