Inconsistent System of Linear Equations Let $A ∈ M_{n\times n}(F)$. Suppose that the system of linear equations $AX = B$
has more than one solution. Prove that there is a column $C ∈ F^n$ such that
the system of linear equations $AX = C$ is inconsistent.
I thought it'd be clever to use this (underlined by my pencil):

 A: Hints: What does the premise imply about the kernel of $A$? Also:
$$\dim(\text{Im } A)+\dim(\ker A)=n.$$
A: 
Lemma: Let $AX=C$ be a system of linear equations. Then the system is consistent if and only if $\text{dim Im}~(A)=\text{dim Im}~(A \mid C)$.

Let $Y$ and $Y'$ be solutions to the system $AX=B$. Then
$$ A(Y) = A(Y')=B,$$
or
$$ A(Y)-A(Y')=A(Y-Y')=0,$$
so $\dim \text{Ker}~(A)\geq 1$, which means $\dim \text{Im}~(A)<n$, or simply that 
$$\text{span}~(\{A^{(1)},\dots,A^{(n)}\})\neq F^n.$$
In other words, there exists at least one $A^{(i)}$ such that
$$A^{(i)}=k_1A^{(1)}+\cdots +k_{i-1}A^{(i-1)}+k_{i+1}A^{(i+1)}+\cdots + k_nA^{(n)}\neq ce_i,$$
for some $e_i\in \{e_1,\dots,e_n\}$ the standard basis for $F^n$ with $k_i,c\in F$. Thus, by the lemma, we clearly see that such an $e_i$ is the desired $C$ for which $AX=C$ has no solutions, namely
$$\text{dim Im}~(A)\neq \text{dim Im}(A \mid e_i)=\text{dim Im}(A \mid C).$$
This is so because if $\text{dim Im}(A)=m$, then $\text{dim Im}~(A \mid e_i)=m+1$ for such an $e_i$ described above.
