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This is a follow up question to the following post: How do we show every linear transformation which is not bijective is the difference of bijective linear transforms?

The previous post dealt with showing a linear transformation which was not bijective was the difference of two bijiective homomorphisms.

Let $F$ be a field and $I$ a nonempty index set (Note: $I$ can be uncountable). Consider the direct sum $V = \oplus_{i \in I} F_i $ where each $F_i$ is isomorphic to $F$. In order to rule out a pathological example we must assume furthermore that $F$ is not isomorphic to $\mathbb{Z}_2$.

How do we show if $\phi: V \rightarrow V$ is a bijective $F$-module homomorphism (linear transformation) then $\phi = f -g$ where $f,g:V \rightarrow V$ are two bijective $F$-module homomorphisms ?

I am not quite sure if the proof from the following post applies and if so does it suffice to check that $F$ is free.

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    How is this not just a special case of the previous question? If $F$ is a field, then an $F$-module is the same as an $F$-vector space, so the previous question should apply unchanged.2011-10-01

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Here's one idea. Non-zero multiples of $\phi$ are also bijective and linear. So one way to find $f$ and $g$ is to find two non-zero elements $a, b \in F$ such that $a - b = 1$. If the characteristic of $F$ is not $2$, then there's an obvious choice for $a$ and $b$. Otherwise, as long as $F \neq \mathbf{F}_2$, you should still be able to find such a pair.

Remarks. Any $F$-module is free, and the cardinality of $I$ doesn't appear to matter. Also, it seems like it would be a good exercise to construct one of these "pathological examples" that we're avoiding!

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    @user7980 $a=2$ works often. It doesn't work, when $2=0$, i.e. when the characteristic of the field is $2$. If $F=GF(2^n)$, then you cannot choose $a$ from the prime field, because either $a$ or $a-1$ will be zero.2011-10-04