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Let $T:V\to V$ be a linear operator on a finite dimensional vector space $V$. I need to prove that: If $T$ is irreducible then $T$ is cyclic

My definitions are:

$T$ is an irreducible linear operator iff $V$ and {$0$} are the only complementary invariant subspaces

T is a cyclic linear operator iff $V$ is a cyclic subspace (i.e. there is a vector $v\in V$ such that $V$ is generated by the set of vectors {$v, T(v),T^2(v),...$}

I don´t know where to start. Any comment, suggestion or hint would be highly appreciated

user128422
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  • Suppose $T$ is not cyclic. What then follows from analyzing the span of any such set ${v, T(v), T^2(v), \ldots}$? – Darrin Dec 31 '14 at 23:49
  • If $T$ is not cyclic then the span of {$v, T(v),T^2(v),...$} $\neq V$ also the span of this set has a complement (because every subspace has a complement) so I need to prove that this complement is invariant, is this correct? – user128422 Jan 01 '15 at 00:02
  • also a basis of the span of {$v, T(v),..$} is {$v, T(v),...T^{k-1}(v)$} (where $k<n$(dimension of $V$))hence every element of the span is a linear combination of this basis but does this implies that the elements of the complement of the span of {$v, T(v),..$} are linear combinations of {$w,T(w),...T^{r-1}(w)$} where $r<n$ , $k+r=n$and $w\notin$ span of {$v, T(v),...$}? – user128422 Jan 01 '15 at 00:20

2 Answers2

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I'm $11$ months late, but here is a solution anyway:

Assume $T$ is irreducible and let $0 \ne v \in V$. Then $\langle T,v \rangle = \{f(T)(v):f(x) \in \mathbb{F}[x]\}$ is a $T$-invariant subspace containing $v \ne 0$, so $T = \langle T,v \rangle$, proving that $T$ is cyclic.

St Vincent
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  • The question defines irreducible as no complementary invariant subspaces. Why is your proof valid? – Arrow Apr 10 '19 at 08:29
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It is shown in section 13.12 of Greub's Linear Algebra that for finite dimensional vector spaces one always has a direct sum decomposition into cyclic subspaces. The proof is intricate and relies on orthogonal complements.

Given the existence of such a decomposition, it's clear irreducible implies cyclic: take a cyclic decomposition and apply the definition of irreducible to conclude equality with one of the cyclic subspaces.

Arrow
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