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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. –  Henning Makholm Oct 1 '11 at 16:03
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up vote 2 down vote accepted

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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I think that the claim holds for $F=GF(2), \dim V=\aleph_0$ as well. If we use different bases in the domain resp. range side, then we can assume that $\phi$ is given by the identity matrix. We can pair up the basis elements (odd vs. even index) and let $f$ use 2x2 blocks of the form $\pmatrix{1&1\cr1&0\cr}$ and $g$ use blocks of the form $\pmatrix{0&1\cr1&0\cr}$. Probably uncountable dimension works the same. A finite dimensional space over $GF(2)$ can be handled with Arturo's method from the other question, because the number of eigenvalues is less than the $2^n$ except in dun, dunn, dunn... –  Jyrki Lahtonen Oct 1 '11 at 21:34
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... the pathological case which is, of course $F=GF(2)$, $\dim V=1$. In that case there exists only a single bijective linear mapping, so... –  Jyrki Lahtonen Oct 1 '11 at 21:35
    
Sorry about the typo. The mapping $g$ was, of course, supposed to use blocks of the form $\pmatrix{0&1\cr 1&1\cr}$ in order to get the identity matrix as the difference. –  Jyrki Lahtonen Oct 2 '11 at 19:46
    
@DylanMoreland Is the obvious choice for a - b = 1 an element in F just $a=2$ or is there something more complicated? –  user7980 Oct 3 '11 at 1:49
    
@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. –  Jyrki Lahtonen Oct 4 '11 at 10:55
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