Irreducible linear operator is cyclic

[david34]

I´m having a hard time proving the next result:
Let T:V→V be a linear operator on a finite dimensional vector space V . If T is irreducible then T cyclic.
My definitions are: T is an irreducible linear operator iff V and {{\vec 0}} are the only complementary invariant subspaces.
T is cyclic linear operator iff V is a cyclic vector space (i.e. there is a vector \vec v ∈ V such that V is generated by the set of vectors {{ \vec v, T(\vec v), T^{2}(\vec v),...}}

I was trying to do it by contradiction: suppose T is not a cyclic linar operator then ∀ \vec v ∈ V the generated space by the set {{ \vec v, T(\vec v), T^{2}(\vec v),...}} is not equal to V also if \vec v \neq \vec 0 then span { \vec v, T(\vec v), T^{2}(\vec v),... } is not equal to {\vec 0}.

Moreover span {\vec v, T(\vec v), T^{2}(\vec v),... } is invariant (I´ve already proven it); also I know that every subspace has a complement that is : ∃ W subspace of V such that W ⊕ span{\vec v, T(\vec v), T^{2}(\vec v),...} = V

Then I think a just need to prove that the complementary subspace W is invariant that is : T(W)⊆ W but this is the part that I´m having trouble.

Any comment, suggestion, hint would be highly appreciated

 

[lavinia]

If the span of a non-zero vector,v, is not the whole space then there is a vector ,w ,which is not in the span of v. The span of w is a second cyclic subspace. Both of these subspaces are irreducible and invariant. So what is their intersection?

 

[david34]

But how can we know that T(w) is not in span{v,T(v),...} ?

 

[lavinia]

But how can we know that T(w) is not in span{v,T(v),...} ?

By assumption, w is not in the span of v.