Theorem: If the matrix A has a right inverse then for each b the equatin AX=b has at least one solution. If A has a left inverse then the equation has at most one solution.
Proof:
Existence of right inverse:
Let A be an $m \times n$ matrix and X be an $n \times 1$ column vector.
Observe:
$AX=b$
$\Rightarrow AXX^{-1}=bX^{-1}$
$\Rightarrow AI = bX^{-1} = A$
$\Rightarrow X = bA^{-1}$
Indeed, $\forall b \in \mathbb{R}^{n \times 1}$, there exists an $X\left ( b \right )$.
The one solution that must exists is the vector $b = \vec{0}.$
Existence of left inverse:
Let A have a left inverse.
Then, $A^{-1}A=I$.
Observe:
$AX=b$ for a fixed column vector b.
$\Rightarrow A^{-1}AX = A^{-1}b$
$\Rightarrow IX=A^{-1}b$
$\Rightarrow X = A^{-1}b$
Indeed, since b and A are both fixed, so is X. Evidently, one and only one solution exists to the equation AX=b.
I would appreciate if anyone could verify the validity of my proof.
Best Answer
The first part does not make sense, because we cannot know if $X$ is invertible.
For the second part, you have to consider the case that $AX=b$ cannot be achieved.
If $A$ has a right inverse $A^{-1}$, we can set $X=A^{-1}b$ and we easily verify that, with an $X$ chosen like this, we get $AX = AA^{-1}b = Ib = b.$ Hoewever, we do not know if this is the only way of obtaining an $X$ that satisfies $AX=b.$ Therefore, we say that $AX=b$ has at least one solution.
If $A$ has a left inverse $A^{-1}$, we can multiply $AX=b$ with $A^{-1}$ from the left and we get $X=A^{-1}b.$ Therefore, if $AX=b$ can be achieved at all (which is not guaranteed!) the solution must be $X=A^{-1}b.$ However, this $X$ is not guaranteed to satisfy $AX=b.$ Therefore, we say that $AX=b$ has at most one solution.