How do we show every linear transformation which is not bijective is the difference of bijective linear transforms? I have been reviewing some ideas about vector spaces and came upon a surprising fact. I am not quite sure how to begin the argument because the problem requires one to construct two bijective linear transformations whose difference is equal to a given linear transformation.  
Let $V$ be a vector space over a filed $F$.  Suppose $\phi:V \rightarrow V$ is a linear transformation that is not a bijection.

How do we show $\exists f,g :V \rightarrow V$ that are both bjiective linear transformations such that $\phi = f - g$.

I tried proving the fact using contradiction but have not been able to get to far so I am wondering if there is a standard constructive proof that applies directly.  
 A: Hint 1. Show that $a$ is an eigenvalue of $f$ if and only if $a+b$ is an eigenvalue of $f+bI$.
Hint 2. (Assuming $V$ is finite dimensional) Show that $f$ is bijective if and only if $0$ is not an eigenvalue of $f$.
Hint 3. (Assuming the field is infinite) Show that if $f$ is not bijective, then there is a nonzero scalar $b$ such that $f+bI$ is bijective.
A: Arturo's hints cover the case that $V$ is finite dimensional and $F$ is infinite. I add the case of $\dim V<\infty$ and a finite field $F$ with a solution that is very similar.
Let $|F|=q$ and $\dim V=n$. We fix once and for all an identification of $V$ with the field $GF(q^n)$ as vector spaces over $F$. Consider the set
$$
S=\{\phi(x)/x \mid x\in V, x\neq0\}\subseteq GF(q^n).
$$
Because $\ker\phi$ is non-trivial, $0\in S$, so the set $S$ contains at most $q^n-2$ non-zero elements. Let $\alpha\in GF(q^n)\setminus S$. Then for all non-zero $x\in V$ we have $\phi(x)\neq\alpha x$. In other words the mapping $g:V\rightarrow V$ defined by
$$
g(x)=\alpha x-\phi(x)
$$
has a trivial kernel, and is hence bijective. The mapping $f(x)=\alpha x$ is also bijective, because $\alpha\neq0$. For all $x\in V$ we have
$$
f(x)-g(x)=\alpha x-\alpha x+\phi(x)=\phi(x)
$$
as required.
