# 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

Related Linear and Abstract Algebra News on Phys.org

#### lavinia

Gold Member
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),...} ?