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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)?

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  • $\begingroup$ So your question is - If $Ax=y$ then does there exist $X$ such that $x=Xy$ and $AX=1$? $\endgroup$ May 17, 2019 at 18:18
  • $\begingroup$ In my case there does exist such an $\mathbf{X}$ and I want to understand why. $\endgroup$
    – That Guy
    May 17, 2019 at 18:40

1 Answer 1

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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 $Ax=0$.

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  • $\begingroup$ 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 $\mathbf{A}$? $\endgroup$
    – That Guy
    May 17, 2019 at 19:30
  • $\begingroup$ I am sure you are correct but I can't really make sense of it $\endgroup$
    – That Guy
    May 17, 2019 at 19:31

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