Query about the Moore Penrose pseudoinverse method I have recently discovered the Moore-Penrose psuedoinverse method, and I am currently testing the waters with it. I noticed if I have a system, say
$$a_1x_1=0$$
$$a_2x_1+a_3x_2=0$$
$$\vdots$$
$$a_5x_1+a_6x_2=0$$
we can write it in matrix form as 
$$AX=\hat{0}$$
where 
$$A=\begin{bmatrix}a_1 & 0\\
a_2 & a_3 \\
\vdots & \vdots\\
a_5& a_6
\end{bmatrix},\qquad X= [x_1\;\; x_2],\quad \hat{0}=\begin{bmatrix}0\\ \vdots\\ 0\end{bmatrix}$$
after doing all the necessary calculations it does indeed find an (left) inverse. My question is upon observation if $a_1\neq 0 $ then clearly $x_1=0$ is a solution and this would fix $x_2=0$, but the solutions obtained via the Moore-Penrose pseudoinverse method doesn't give these trivial solutions. Would the left inverses obtained via this method indeed be the correct solutions to this system?  
 A: There is confusion in either my understanding or the question. Start with a full column rank matrix, 
$$
  \mathbf{A} \in \mathbb{R}^{m\times 2}_{2}
$$
with $m>2$. The span of the row space is the entire plane:
$$
\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} = \mathbb{R}^{2}.
$$
The colors distinguish $\color{blue}{range}$ spaces from $\color{red}{null}$ space.
Next, you construct the Moore-Penrose pseudoinverse, $\mathbf{A}^{\dagger}$. Recall this pseudoinverse is a projector onto the range space $\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)}$.
Problems:


*

*It seems you are looking for a $\color{red}{null}$ space solution with a $\color{blue}{range}$ space tool.

*There is no $\color{red}{null}$ space $\color{red}{\mathcal{N} \left( \mathbf{A} \right)}$ for this problem.


Perhaps this toy problem will help.
$$
\begin{align}
  \mathbf{A} x & = b \\
%
\left[
\begin{array}{cc}
 1 & 0 \\
 0 & 0 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 x_{1} \\
 x_{2} \\
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
 b_{1} \\
 b_{2} \\
\end{array}
\right]
%
\end{align}
$$
The Moore-Penrose pseudoinverse is
$$
  \mathbf{A}^{\dagger} =
%
\left[
\begin{array}{cc}
 1 & 0 \\
 0 & 0 \\
\end{array}
\right].
$$
Case 1: No existence
$$\left[
\begin{array}{c}
 b_{1} \\
 b_{2} \\
\end{array}
\right]
=\left[
\begin{array}{c}
 0 \\
 b_{2} \\
\end{array}
\right]$$
The data vector is in the null space:  $b\in\color{red}{\mathcal{N}\left( \mathbf{A}^{*}\right)}$. There is no solution.
Case 2: Existence and uniqueness
$$\left[
\begin{array}{c}
 b_{1} \\
 b_{2} \\
\end{array}
\right]
=\left[
\begin{array}{c}
 b_{1} \\
 0 \\
\end{array}
\right]$$
The exclusion is $b_{1}\ne0$. There is no $\color{red}{null}$ space: $\color{red}{\mathcal{N}\left( \mathbf{A}^{*}\right)} = \mathbf{0}$. The data vector is in the $\color{blue}{range}$ space: $b\in\color{blue}{\mathcal{R}\left( \mathbf{A}\right)}$. The solution exists and is unique. The direct solution is also the least squares solution
$$
 x = x_{LS} = 
\left[
\begin{array}{c}
 b_{1} \\
 b_{2} \\
\end{array}
\right]
$$
Case 3: Existence, no uniqueness
$$\left[
\begin{array}{c}
 b_{1} \\
 b_{2} \\
\end{array}
\right]
=\left[
\begin{array}{c}
 b_{1} \\
 b_{2} \\
\end{array}
\right]$$
he exclusion is $b_{1}\ne0$, $b_{2}\ne0$. Now there is a $\color{red}{null}$ space: $\color{red}{\mathcal{N}\left( \mathbf{A}^{*}\right)}$. The data vector inhabits both $\color{blue}{range}$ and $\color{red}{null}$ space: 
$$
b = 
\color{blue}{b_{\mathcal{R}}} + 
\color{red}{b_{\mathcal{N}}}
$$
A solution exists and it is not unique. There is no direct solution. The least square solution includes an arbitrary vector $y\in\mathbb{R}^{2}$:
$$
 x_{LS} = 
\color{blue}{\mathbf{A}^{\dagger}b} + 
\color{red}{\left( \mathbf{I}_{2} - \mathbf{A}^{\dagger}\mathbf{A}\right)^{-1} y}
=
\color{blue}{
\left[
\begin{array}{c}
 b_{1} \\
 0 \\
\end{array}
\right]}
+
\alpha
\color{red}{
\left[
\begin{array}{c}
 0 \\
 1 \\
\end{array}
\right]}
$$
