Why is $\mathbf{x}$ yielding a product of $\mathbf{y}$ and the right inverse for $\mathbf{A}$?

When you solve the equation $$\mathbf{Ax}=\mathbf{y}$$ where $$\mathbf{x} \in \mathbb{R}^{5}$$, $$\mathbf{y} \in \mathbb{R}^{4}$$ and $$\mathbf{A}$$ is a $$4 \times 5$$ matrix with a right inverse, I understand that the $$\mathbf{x}$$ is $$\mathbf{A}^{-1}\mathbf{y}$$ but how come it can happen that $$\mathbf{x}$$ is not $$\mathbf{A}^{-1}\mathbf{y}$$ but some right inverse times $$\mathbf{y}$$? So if $$\mathbf{X}$$ were a right inverse to $$\mathbf{A}$$, $$\mathbf{x}$$ is $$\mathbf{X}\mathbf{y}$$?

I would like some algebraic proof such as for the fact that $$\mathbf{x}$$ is $$\mathbf{H}^{-1}\mathbf{y}$$ for some square matrix $$\mathbf{H}$$ $$\mathbf{H}^{-1}\mathbf{Hx} = (\mathbf{H}^{-1}\mathbf{H})\mathbf{x} = \mathbf{I}_4 \mathbf{x} = \mathbf{x} = \mathbf{H}^{-1}\mathbf{y}$$

I do not think this is valid statement $$\mathbf{AXx} = (\mathbf{AX})\mathbf{x} = \mathbf{I}_4 \mathbf{x} = \mathbf{x} = \mathbf{X}\mathbf{y}$$ so that does probably not that why $$\mathbf{x}$$ in my case is a the result of post-multiplying a right inverse of $$\mathbf{A}$$ by $$\mathbf{y}$$

To give details about my case where $$\mathbf{x}$$ is $$\mathbf{X}\mathbf{y}$$?, I have $$\mathbf{A} = \left[\begin{array}{rrrrr} 2&-4&-1&-3&2\\ -1&2&1&0&1\\ 1&-2&-1&-3&-1\\ -1&4&-1&0&5 \end{array}\right]$$ and $$\mathbf{y}$$ is a column vector of four unknowns. Then I have the solution for $$\mathbf{x}$$ as $$\mathbf{x} = \left[\begin{array}{r} x_1\\ x_2\\ x_3\\ x_4\\ x_5 \end{array}\right] = \left[\begin{array}{r} 3y_1+y_4-3y_3+y_2-15t\\ \frac{1}{2}y_4+y_1-y_3+\frac{1}{2}y_2-6t\\ -y_3+y_1+y_2-4t\\ -\frac{1}{3}y_2-\frac{1}{3}y_3\\ t \end{array}\right] \quad \text{where } t \in \mathbb{R}$$ For exapmle when $$t = 0$$, we have that $$\mathbf{x} = \left[\begin{array}{r} x_1\\ x_2\\ x_3\\ x_4\\ x_5 \end{array}\right] = \left[\begin{array}{r} 3y_1+y_4-3y_3+y_2-15 \cdot 0\\ \frac{1}{2}y_4+y_1-y_3+\frac{1}{2}y_2-6 \cdot 0\\ -y_3+y_1+y_2-4 \cdot 0\\ -\frac{1}{3}y_2-\frac{1}{3}y_3\\ 0 \end{array}\right] = \left[\begin{array}{rrrr} 3&1&-3&1\\ 1&\frac{1}{2}&-1&\frac{1}{2}\\ 1&1&-1&0\\ 0&-\frac{1}{3}&-\frac{1}{3}&0\\ 0&0&0&0 \end{array}\right] \left[\begin{array}{r} y_1\\ y_2\\ y_3\\ y_4 \end{array}\right]$$ We now have $$\mathbf{x} = \mathbf{Xy}$$ where $$\mathbf{X}$$ is a right inverse to $$\mathbf{A}$$ - so why this relationship ($$\mathbf{x} = \mathbf{Xy}$$ with $$\mathbf{X}$$ being a right inverse)?

• So your question is - If $Ax=y$ then does there exist $X$ such that $x=Xy$ and $AX=1$? – Hagen von Eitzen May 17 '19 at 18:18
• In my case there does exist such an $\mathbf{X}$ and I want to understand why. – Fac Pam May 17 '19 at 18:40

Claim. Let $$V,W$$ be vector spaces, $$A\colon V\to W$$ and $$B\colon W\to V$$ be linear maps with $$AB=1_W$$. Also suppose that $$Ax=y$$ for certain $$x\in V$$, $$y\in W$$ with $$y\ne 0$$. Then there exists a linear map $$B'\colon W\to V$$ such that $$AB'=1_W$$ and $$x=B'y$$.
Proof: It suffices to find $$C\colon W\to V$$ such that $$AC=0$$ and $$Cy=x-By$$ as we then can take $$B'=B+C$$. Let $$v:=x- By$$. Note that $$v\in\ker A$$. By extending $$y$$ to a to a basis of $$W$$, we find a linear map $$W\to \ker A$$ that maps $$y\mapsto v$$. Composed with the inclusion $$\ker A\to V$$, we obtain a linear map $$C\colon W\to V$$ such that $$AC=0$$ and $$Cy=x-By$$. $$\square$$
Remark. If we allow $$y=0$$, the claim is no longer true because there may exist non-zero $$x$$ with $$Ax=0$$.
• I am sorry but I do not really understand. How does this explain why $\mathbf{x} = \mathbf{Xy}$ with $\mathbf{X}$ being a right inverse to $\mathbf{A}$? – Fac Pam May 17 '19 at 19:30