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.