Suppose that $A$ is a $5\times 3$ matrix and there exists....

Question

Suppose that $A$ is a $5\times 3$ matrix and there exists a $3\times 5$ matrix $C$ such that $C$$A$ = $I$. Suppose further that for some $b$ in $R^5$, $Ax = b$ has at least one solution. Show that this solution is unique.

Down below is my attempt at solving this, and I need someone to check my work by giving me hints.

If there exists a matrix $C$ such that $C$$A$ = $I$, then we know that the matrix is invertible.

If the matrix is invertible, then $Ax = b$ has a unique solution for ALL $b$, this implies that the $b$ in $\Bbb R^5$ MUST be unique

Answer 1

Try to rephrase the problem in terms of linear maps:

$A:\mathbb R^5 \to \mathbb R^3$ and $C:\mathbb R^3 \to \mathbb R^5$.

If $CA=I$, with $I$ the identity map on $\mathbb R^5$, then $A$ must be injective.

This is a property of functions in general.

Answer 2

Your argument is wrong. For a matrix to be invertible, it must be square. If it is square and a matrix $C$ with $CA = I$ exists, then it turns out that $A$ is invertible. But if it's not square...then it doesn't make sense to say it's invertible, and you need a rather different argument.

Answer 3

Hint:

Assume that there is more than one solution, say $x$ and $y$ are solutions, and compute $$CA(x-y)$$ in two ways.