Proving Cyclic Decompositions: Let T be a Linear Operator on V

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In summary, the conversation discusses the concept of a linear operator T on a finite dimensional space V and its range R. The first part of the conversation states that R has a complementary T-invariant subspace if and only if it is independent of the null space N of T. The second part then discusses the case where R and N are independent, and proves that N is the unique T-invariant subspace complementary to R. The conversation also mentions the concept of T-admissibility, which refers to a subspace W that is invariant under T and satisfies a certain property involving the function f(T). The conversation ends with a request for a definition of T-admissibility and a hint for finding it.
  • #1
Bhatia
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Let T be a linear operator on the the finite dimensional space V, and let R be the range of T.

(a) Prove that R has a complementary T-invariant subspace iff R is independent of the null space N of T.

(b) If R and N are independent, prove that, N is the unique T-invariant subspace complementary to R.

I supposed R has a complementary T - invariant subspace, say, W. Then , R should be T- admissible. I assumed to the contrary, that R intersection T is not equal to {0}. I took a point in the intersection but could not proceed further. Please suggest.
 
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  • #2
Can you define "T-admisible"?
 
  • #3
arkajad said:
Can you define "T-admisible"?

Thanks for your reply.

Given a linear operator T on a vector space V then a subspace W is T- admissible if

i) W is invariant under T
(ii) if f(T) β belongs to W, there exists a vector η in W such that f(T)β = f(T) η.
 
  • #4
What is f(T) in (ii)?
 
  • #5
Hint: Fix a basis for the range of T and extend it to a basis for V. What can you say about these extra basis vectors?
 

FAQ: Proving Cyclic Decompositions: Let T be a Linear Operator on V

What is a cyclic decomposition?

A cyclic decomposition is a way of breaking down a vector space into a direct sum of smaller subspaces, each of which is spanned by a single vector. This decomposition is useful in understanding the behavior of linear operators on vector spaces.

How do you prove a cyclic decomposition for a linear operator?

To prove a cyclic decomposition, you must show that the vector space can be written as a direct sum of subspaces, each of which is spanned by a single vector. This can be done by finding a basis for the vector space that consists of vectors that are cyclic under the linear operator.

What is a cyclic vector?

A cyclic vector is a vector that generates a subspace under a linear operator. In other words, the subspace can be obtained by repeatedly applying the linear operator to the cyclic vector.

How do you find a cyclic vector for a linear operator?

To find a cyclic vector, you can start by choosing a vector in the vector space and applying the linear operator to it. If the resulting vector is not in the span of the original vector, then it is a cyclic vector. If it is in the span, you can repeat this process with a different vector until you find a cyclic vector.

Why is proving cyclic decompositions important?

Proving cyclic decompositions is important because it helps us understand the behavior of linear operators on vector spaces. It also allows us to simplify complicated vector spaces into smaller, more manageable subspaces. Additionally, cyclic decompositions have many applications in different areas of mathematics, such as in the study of matrices and differential equations.

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