If $\ker A = \ker B$ and $\mathrm{Im \;}A =\mathrm{Im \;} B$ show that $A = \alpha B$ if $A$ and $B$ are $1$ dimensional $\newcommand{\Image}{\mathrm{Im \,}}$ Let $A$ and $B$ be a linear transformation of range $1$ in a vector space $V(F)$ such that $\ker A = \ker B$ and $\Image A =\Image B$. Show that $A = \alpha B$ from some $\alpha$ from the field $F$
I'm stuck in this problem and I would really like some help.
What I did so far is this:
Let $x \in V$ such that $A(x) \neq0$ this implies that $B(x) \neq 0$
Since $\dim \Image A =\dim \Image B=1$ then there exists an $\alpha \in F$ such that $A(x)=\alpha B(x)$
But I don't think that this is all.
The case where $x \in\ker A =\ker B$ is trivial because it happens for every $\alpha \in F$
 A: Choose $x \in V, x \notin \ker(A)$.  Since $\ker(A) = \ker(B), ~x \notin \ker(B)$.  Because $\operatorname{image}(A)=\operatorname{image}(B)$ is $1$-dimensional we also know $Ax = \alpha Bx$ for some $\alpha \in F, \alpha \neq 0$, so $(A-\alpha B)x=0$.
Let $S$ be a basis for $\ker A$.  Then $S \cup \{x\}$ is a basis for $V$.
Claim:  $A-\alpha B=0$.  Choose $y \in V$ arbitrary.  Then $y=\beta x + v, \beta \in F, v \in \ker(A)$.  $(A- \alpha B)y = (A - \alpha B)(\beta x +v) = \beta(A- \alpha B)x = 0$.
A: Let $V$ be an $n$-dimensional vector space over a field $k$. Let $A$ and $B$ be maps from $V$ to $k$ with the same kernel. If these are $0$ then we are really done, so we may assume they are surjective. Then the kernel has dimension $n-1$. Take a basis of it $v_1, ..., v_{n-1}$, and extend it to a basis of the domain by adding a single vector $v_n$. Both matrices are determined by what they do to $v_i$ (and of course, they sent $v_i$ to $0$ for $1 \leq i < n$ by requirement). 
Now if $a = Av_n$ and $b = Bv_n$, take $c = a/b$. That scalar is a linear map $C$ from $k$ to itself. Then $C B = A$ since it has the same value on $v_n$. That is, $CB v_i = A v_i$ for each $1 \leq i \leq n$. 
