Let $\mathbb{F}$ be a field and $V$ an $\mathbb{F}$-vector space. Let $\operatorname{T}\in\mathrm{End}(V)$ be an $\mathbb{F}$-linear operator. It is well known that if $\dim V<\infty$ then $\operatorname{T}$ has a Jordan canonical form, i.e., it is similar to the direct sum of a number of Jordan blocks. If a certain eigenvalue (i.e., root of a polynomial) is not inside $\mathbb{F}$ then in the corresponding Jordan block it is represented by its $\mathbb{F}$-matrix representation (that is, the Jordan block is a block matrix).


In infinite dimensions the existence of Jordan canonical form is not guaranteed. I consider only those operators for which it exists.


**Question:**


Let $V$ have arbitrary dimensions, and *assume* that $\operatorname{T}$ does have a Jordan canonical form, i.e., it is similar to the direct sum of (an arbitrary set of) Jordan blocks. Let now $W\subset V$ be an $\mathbb{F}$-vector subspace which is $\operatorname{T}$-invariant, $\operatorname{T}W\subset W$. Is it true that the restriction $\operatorname{T}|\_W\in\mathrm{End}(W)$ also has a Jordan canonical form?


Thank you.



Jordan form on an invariant vector subspace
The question can be restated as follows: let $F$ be a field $T$ be an endomorphism of a vector space over $F$. Suppose that it decomposes as a direct sum of finite-dimensional $T$-stable subspaces. Does the same property hold for every $T$-stable subspace?


In turn, it can be formalized as follows, in a pure language of commutative algebra: let $V$ be a $F[t]$-module. Suppose that $V$ is a direct sum of finite length submodules. Does every submodule share the same property? 


It's true and follows as a particular case of the following



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> **Theorem** Let $R$ be a PID. Let $\mathcal{C}\_R$ be the class of $R$-modules that decompose as a (possibly infinite) direct sum of finite length submodules (and hence of finite length cyclic submodules). Then $\mathcal{C}\_R$ is stable under taking submodules.
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This is a theorem of Kulikov (1941) when $R=\mathbf{Z}$, in which case the proof can be found in L. Fuchs' book "Infinite abelian groups, Vol 1", §17-18.


When $R=k[t]$, polynomial ring in one indeterminate over a field $k$, this exactly yields your claim.


I don't know if anybody wrote the result for arbitrary PID (even for such polynomial rings, for which it should not be easier), but the proof in Fuchs' book extends with essentially no change to arbitrary modules over PIDs.


The main lemma in Fuchs' book (Theorem 17.1) is the following (in the case of $\mathbf{Z}$). For an irreducible element $r$ in $R$, an $R$-module is called a $p$-module $M$ if for every $x\in M$ there exists $n$ such that $p^nx=0$. Also, the (possibly infinite) number $h(x)=\sup\{n:x\in p^nM\}$ is called the height of $x$.



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> **Lemma** Let $R$ be a PID and $p$ an irreducible element. A $p$-module $M$ is a direct sum of cyclic modules if and only if one can write $M$ as union of an ascending chain of submodules $M\_1\le M\_2\le\dots$, $\bigcup\_nM\_n=M$, such for every $n$, we have $\sup\_{x\in M\smallsetminus\{0\}}h(x)<\infty$.
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It clearly implies, for a given PID $R$, the above theorem.


So, for completeness, let me write the proof of the lemma; it essentially consists in a copy of the proof in Fuchs' book.


Up to reindexing (adding redundancies if necessary), we can first suppose that $M\_n\cap p^n M=\{0\}$ for all $n$. Second, we can suppose that $(M\_n)$ is maximal for this property (namely, among sequences $(N\_n)\_{n\ge 1}$ satisfying $N\_n\cap p^n M=\{0\}$ for all $n$ with the ordering $(N\_n)\le(M\_n)$ if $N\_n\le M\_n$ for all $n$).


Denote by $M[p]$ the kernel of $m\mapsto pm$. For every $n\ge 1$, we choose a maximal independent subset $J\_n$ in the subgroup $M\_n[p]\cap p^{n-1}M$, and define $J=\bigsqcup\_{n\ge 1} J\_n$. For every $c\in J$, choose $a\_c\in M$ with $p^{h(c)}a\_c=c$.



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> We claim that $M=\bigoplus\_{c\in J}Ra\_c$.
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The first part is that this is indeed a direct sum. First, since all nonzero elements of the $R$-module $\langle J\_n\rangle$ generated by $J\_n$ have height $n-1$, we see that the $\langle J\_n\rangle$ generate their direct sum, so $J$ is a $R/pR$-free family in $M[p]$. 


Next suppose that we have a nontrivial combination, namely $\sum\_{i\in I} r\_ca\_c=0$, with $I$ a nonempty finite subset of $J$ and $r\_c\notin p^{h(c)+1}R$ for all $i$. If $r\_c\notin p^{h(c)}R$ for some $c\in I$, we multiply everything by $p$. So we can suppose that $r\_c=p^{h(c)}R$ for all $i$, namely $r\_c=p^{h(c)}s\_c$ with $s\_c\notin pR$. Hence $\sum\_{c\in I}s\_cc=0$, contradicting that $J$ is a $R/pR$-free family in $M[p]$. 


Now let us show that $M=\sum\_{c\in J}Ra\_c$. The first step is to show that $\langle J\rangle=M[p]$. Clearly $\langle J\_n\rangle=M\_n[p]\cap p^{n-1}M$.
Suppose by induction that every element in $M\_r[p]$ belongs to $\langle J\rangle$ (which is clear for $r=1$), and consider $b\in M\_{r+1}[p]\smallsetminus M\_r$. The maximality of $(M\_n)$ and $b\in M\_r$ implies that $(M\_r+Rb)\cap p^rM\neq\{0\}$. So we can find $0\neq c=g+kb\in p^rM$, with $g\in M\_r$ and $k\in R$. Then $kb\neq 0$; hence $k\notin pR$. Multiplying by $k'$ with $kk'-1 \mod pR$, we can suppose that $k=1$, so we now assume that $c=g+b$. We have $c\in M\_{r+1}$ and $h(c)\ge r$, and thus $h(c)=r$ since $M\_{r+1}\cap p^{r+1}M=\{0\}$. Also $h(pc)\ge r+1$, and $pc=pg\in M\_r$; if nonzero it would be a contradictio to $M\_r\cap p^rM=\{0\}$, so $pc=0$. So $pg=0$ as well. By induction, $g\in\langle J\rangle$, and hence $b=c-g$ as well. Thus $\langle J\rangle=M[p]$.


By another induction, we prove that every $m\in M$ with $p^nm=0$ belongs to $M'=\sum\_{c\in J}Ra\_c$; this holds for $n=1$ and we can now consider the case of $b$ with $p^{n+1}b=0$, with $n\ge 1$; we have to show that $b'\in M'$. By the previous paragraph, we have $p^nb\in\langle J\rangle$. We write
$$p^nb=m\_1c\_1+\dots +m\_jc\_j+n\_1d\_1+\dots +n\_kd\_k,$$ 
with $c\_1,\dots,d\_k\in J$, $h(c\_i)\ge n$, $h(d\_i)\le n-1$. We can write $m\_ic\_i=p^nm'\_ia\_{c\_i}$, and hence we have
$$p^n(b-m'\_1a\_{c\_1}-\dots -m'\_ja\_{c\_j})=n\_1d\_1+\dots +n\_kd\_k\in M\_{n}.$$
Since $M\_{n}\cap p^nM=\{0\}$, we deduce that $p^nb'=0$, with $b'=b-m'\_1a\_{c\_1}-\dots -m'\_ja\_{c\_j}$. By induction, we have $b'\in M'$, and $b\in M'$ follows. 


