Some remarks/questions from Primary Decomposition Theorem to get verified In course of self-studying the Canonical Form in Linear Algebra I'm trying to put some remarks from the concept I acquired from Primary Decomposition Theorem which reads as follows: 


*

*Let $T:V_F\to V_F~(\dim V<\infty)$ be a linear transformation and let $m(t)$ be the minimal polynomial (existence of such polynomial is guaranteed due to Cayley-Hamilton Theorem) of $T$ over $F$ which is expressed in it's totally factored form (uniqueness follows from the observation that $F[x]$ is a UFD) as $m(t)=\prod_{i=1}^k [f_i(t)]^{n_i}$ for some $k\in\mathbb Z^+$ where $f_i$'s are monic irreducible polynomial over $F.$ Then $V=\oplus_{i=1}^k \ker (f_i(T))^{n_i}$ ($(f_i(T))^{n_i}\in L(V,V)$ since $L(V,V)$ is a linear algebra). 


Here're my remarks/questions which I want to get verified/answered:


*

*If $T$ is non-decomposable then for every collection of $T$-invariant subspaces $\{W_i\}_{i=1}^m$ of $V$ satisfying $V=\oplus_{i=1}^m W_i$ one has $W_i=(0)$ for some $i$. (A linear operator $T:V\to V$ is called decomposable if $\exists$ a collection of non-zero $T$-invariant subspaces $\{W_i\}_{i=1}^m$ (with $m>1$) of $V$ such that $V=\oplus_{i=1}^m W_i$)

*However if $T$ is decomposable then $\exists$ a collection of non-zero $T$-invariant subspaces $\{W_i\}_{i=1}^m$ of $V$ such that $V=\oplus_{i=1}^m W_i.$ 


*

*Q. Does for decomposable $T,$ the collection $\{\ker (f_i(T))^{n_i}\}_{i=1}^k$ serve the purpose, i.e. whether $\ker (f_i(T))^{n_i}\neq (0)$ for all $i$?
Of course, even if $T$ is decomposable, for every collection $\{W_i\}_1^n$ of $T$-invariant subspaces of $V$ satisfying $V=\oplus_1^n W_i$ the condition the $W_i\ne(0)~\forall ~i$ doesn't hold since $V=V\oplus(0)$ where both $V,(0)$ are $T$-invariants.


*I can see that (for the sake of simplicity I use $W_i=\ker (f_i(T))^{n_i}$) $(f_i(T|_{W_i}))^{n_i}=\underline 0:W_i\to W_i~\forall~i=1(1)k$ whence $\ker (f_i(T|_{W_i}))^{n_i}=W_i.$ Consequently, $\ker ((f_i.T)^{n_i}|_{W_i}))=\ker (f_i(T|_{W_i}))^{n_i}=\ker (f_i(T))^{n_i}.$ So $(f_i(T))^{n_i}$ can't send any element outside $W_i$ to $0.$
 A: $\newcommand{\ker}{\operatorname{ker}}$
0) Claiming the existence of the minimal polynomial from Cayley-Hamilton is overkill.  The powers $T^k$ of $T$ lie in the finite-dimensional vector space $L(V,V)$ so they cannot all be linearly independent.  A nontrivial linear dependence relation between them gives a nonzero polynomial satisfied by $T$.
1) Writing $[f_i(t)]^{n_i}$ is unnecessarily notationally heavy.  Better would be $f_i^{n_i}(t)$ or even just $f_i^{n_i}$.  
2) I'm not sure whether you understand the definition of a decomposable endomorphism.  It just means that you can write $V_F$ nontrivially as a direct sum of invariant subspaces, i.e., $V_F = W_1 \oplus W_2$ in which $W_1$ and $W_2$ are both proper and nonzero.  It would be logically equivalent to say that both are proper; it would also be logically equivalent to say that both are nonzero.  Let's also assume that $V_F$ has positive dimension to exclude trivialities.  
From this definition it is immediate that -- since $V_F$ is finite-dimensional -- $V$ can be written as a finite direct sum of indecomposable invariant subspaces.  
You seem to be asking about the converse: namely, if $k = 1$ must $V$ be indecomposable?  The answer is no.  In fact $V$ is indecomposable iff $V$ is primary -- i.e., $k = 1$ -- and $V$ is cyclic, i.e., spanned by the powers of $T$ acting on a single vector $v$ . To see this in general requires another big theorem, the Cyclic Decomposition Theorem, which indeed says (in particular) that any $V$ is a direct sum of cyclic, primary invariant subspaces and that the number of factors in this decomposition is unique.  However, you can get an easy example just by taking $T = 0$: then the minimal polynomial is $T$, which is a prime power, so $k = 1$.  But as long as $\operatorname{dim} V > 1$ there are plenty of decompositions into direct sums of invariant subspaces, since every subspace is invariant in this case.
3) Your third assertion is true and is really a tautology: if $W_i$ is the kernel of a certain linear operator, then indeed that operator does not send any vector outside of $W_i$ to $0$.  
