Irreducible linear operator is cyclic

In summary: Therefore, T(w) cannot be written as a linear combination of v, T(v), T^2(v), etc. This means that T(w) is not in the span of {v, T(v), T^2(v), ...} and is therefore not in the intersection of the two cyclic subspaces. In summary, the conversation discusses the concept of irreducible and cyclic linear operators on a finite dimensional vector space. It is proposed to prove that an irreducible operator is also cyclic by contradiction. The conversation mentions the existence of complementary invariant subspaces and the need to prove that the complementary subspace is also invariant. It is then suggested that the intersection of two cyclic subspaces is not equal to the span of their generators.
  • #1
david34
3
0
I´m having a hard time proving the next result:
Let [itex] T:V→V [/itex] be a linear operator on a finite dimensional vector space [itex] V [/itex] . If [itex] T [/itex] is irreducible then [itex] T [/itex] cyclic.
My definitions are: [itex] T [/itex] is an irreducible linear operator iff [itex] V [/itex] and {[itex] {\vec 0} [/itex]} are the only complementary invariant subspaces.
T is cyclic linear operator iff V is a cyclic vector space (i.e. there is a vector [itex] \vec v ∈ V [/itex] such that [itex] V [/itex] is generated by the set of vectors {[itex] { \vec v, T(\vec v), T^{2}(\vec v),...} [/itex]}

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

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

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

Any comment, suggestion, hint would be highly appreciated
 
Physics news on Phys.org
  • #2
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?
 
  • #3
But how can we know that T(w) is not in span{v,T(v),...} ?
 
  • #4
david34 said:
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.
 
  • #5
.Your approach is on the right track. To prove that the complementary subspace W is invariant, we can use the fact that T is irreducible. Since W is a complementary subspace to the span of {v, T(v), T^2(v), ...}, we know that V = W ⊕ span{v, T(v), T^2(v), ...}.

Now, suppose there exists a vector w in W such that T(w) is not in W. Since T is irreducible, this means that W and T(w) are the only complementary invariant subspaces of V. But this contradicts the fact that W is a complementary subspace to the span of {v, T(v), T^2(v), ...}, as T(w) is not in W. Therefore, T(w) must be in W for all w in W, and W is invariant under T.

This shows that T is a cyclic linear operator, as the span of {v, T(v), T^2(v), ...} is equal to V.
 

Related to Irreducible linear operator is cyclic

1. What is an irreducible linear operator?

An irreducible linear operator is a type of linear operator that cannot be decomposed into smaller, nontrivial subspaces. This means that there is no proper subspace of the original space that is invariant under the operator.

2. How is an irreducible linear operator different from a reducible linear operator?

A reducible linear operator can be decomposed into smaller subspaces, whereas an irreducible linear operator cannot. This means that a reducible linear operator has at least one proper invariant subspace, while an irreducible linear operator does not have any proper invariant subspaces.

3. What is a cyclic vector?

A cyclic vector is a vector in a vector space that, when acted upon by a linear operator, generates the entire vector space. In other words, the span of the vector and all its images under the linear operator is the entire vector space.

4. How does an irreducible linear operator relate to a cyclic vector?

An irreducible linear operator is cyclic if and only if it has a cyclic vector. This means that the span of the cyclic vector and all its images under the linear operator is the entire vector space, and there are no proper invariant subspaces.

5. What are some applications of irreducible linear operators?

Irreducible linear operators have many applications in mathematics and physics. They are used in the study of dynamical systems, quantum mechanics, and group representation theory. They also have practical applications in fields such as signal processing, control systems, and data analysis.

Similar threads

  • Linear and Abstract Algebra
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
3
Views
497
  • Linear and Abstract Algebra
Replies
8
Views
1K
  • Linear and Abstract Algebra
Replies
3
Views
3K
  • Linear and Abstract Algebra
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
7
Views
2K
Replies
4
Views
960
  • Linear and Abstract Algebra
Replies
10
Views
585
Replies
15
Views
2K
  • Linear and Abstract Algebra
Replies
9
Views
2K
Back
Top