(A linear transformation $T : V \rightarrow V$ is decomposable if $V$ can be written as the direct sum of two proper $T$-invariant subspaces)
Some background for this question: I have been going back over linear algebra recently, with the intention of pushing standard theorems and proofs to their bare essentials, so I can get an idea of when a property is "necessary" and when it is only "sufficient".
For example the theorem which states "Every finite-dimensional inner product space $V$ induces a natural isomorphism $V \rightarrow V^*$, given by $v \mapsto <v,->$" has nothing to do with symmetry or the full power of positive definiteness. So this can be generalised from an inner product to a non-degenerate bilinear form. However this theorem does depend on the finite dimensionality of $V$, and so any proof must make use of this property.
Now I have been trying to prove the following theorem (or find a counter example to it). "Given that $V$ is a vector space, and $T : V \rightarrow V$ is a linear transformation with at least two distinct eigenvalues. Then $T$ is decomposable".
This theorem can be proven for the finite dimensional case by considering the Frobenius normal form of some matrix of $T$, but this proof explicitly uses the property of finite dimensionality. I have tried hard to produce either a more general proof that doesn't use finiteness, or a counter example in the infinite case, but I feel like I have hit a brick wall.
Any help would be greatly appreciated, so thank you in advance.
We assume that i) or ii) is true
i) $V$ has finite dimension and $T$ has at least $2$ distinct eigenvalues in $K$.
ii) $V$ has infinite dimension and (*) there is $P(x)=(x-a)^r(x-b)^sQ(x)\in K[x]$ s.t. $a,b\in K,P(T)=0,Q(a)\not= 0,Q(b)\not= 0$ and $T$ does not cancel a strict divisor of $P$.
Then $V$ is $T$-decomposable.
Proof. For i), the condition (*) is also satisfied. The polynomials $(x-a)^r,(x-b)^s,Q(x)$ are pairwise prime; then
$V=\ker((T-aI)^r)\oplus[\ker((T-bI)^s)\oplus \ker(Q(T))]$ and we are done.