Find T cyclic operator that has exactly N distinct T-invariant subspaces

In summary: N = 8:If N = 8, then there are eight distinct T-invariant subspaces. This means that V can be decomposed into eight T-invariant subspaces, denoted as V = W1 ⊕ W2 ⊕ W3 ⊕ W4 ⊕ W5 ⊕ W6 ⊕ W7 ⊕ W8. In this case, the dimension of V must be at least 8, since each Wi has dimension at least 1. Similarly to the previous cases, we can construct eight T-invariant subspaces using the basis {v, T(v), T^2(v), ..., T^7(v)}. Therefore, an example of a cyclic operator T with N
  • #1
toni07
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Let T be a cyclic operator on $R^3$, and let N be the number of distinct T-invariant subspaces. Prove that either N = 4 or N = 6 or N = 8. For each possible value of N, give (with proof) an example of a cyclic operator T which has exactly N distinct T-invariant subspaces.

Am I supposed to assume that T is a linear operator on V, after assuming that V is a finite dimensional vector space? Also, I think I'm supposed to use the dimension of subspaces, but I don't understand how it's supposed to be used.
 
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Yes, you can assume that T is a linear operator on a finite dimensional vector space V. The dimension of subspaces can be used to prove the number of distinct T-invariant subspaces.

First, let's define a T-invariant subspace. A subspace W of V is called T-invariant if T(W) ⊆ W, meaning that the image of W under T is contained in W.

Now, let's consider the possible values of N.

1. N = 4:
If N = 4, then there are four distinct T-invariant subspaces. This means that V can be decomposed into four T-invariant subspaces, denoted as V = W1 ⊕ W2 ⊕ W3 ⊕ W4. In this case, the dimension of V must be at least 4, since each Wi has dimension at least 1. Since T is a cyclic operator, there exists a vector v in V such that {v, T(v), T^2(v)} forms a basis for V. This basis can be used to construct four T-invariant subspaces: span{v}, span{T(v)}, span{T^2(v)}, and span{v, T(v), T^2(v)}. Therefore, an example of a cyclic operator T with N = 4 is the left shift operator on R^3, defined as T(x,y,z) = (y,z,0). It is easy to verify that T is cyclic and has four distinct T-invariant subspaces.

2. N = 6:
If N = 6, then there are six distinct T-invariant subspaces. This means that V can be decomposed into six T-invariant subspaces, denoted as V = W1 ⊕ W2 ⊕ W3 ⊕ W4 ⊕ W5 ⊕ W6. In this case, the dimension of V must be at least 6, since each Wi has dimension at least 1. Similarly to the previous case, we can construct six T-invariant subspaces using the basis {v, T(v), T^2(v), T^3(v), T^4(v), T^5(v)}. Therefore, an example of a cyclic operator T with N = 6 is the rotation operator on R^3, defined as T(x,y,z) = (y,z,x). It is easy to verify that T is cyclic and has six distinct T-invariant
 

FAQ: Find T cyclic operator that has exactly N distinct T-invariant subspaces

What is a cyclic operator?

A cyclic operator is an operator on a vector space that has a single vector that generates the entire space when repeatedly acted upon by the operator. This vector is called a cyclic vector.

How do you find a cyclic operator with N distinct T-invariant subspaces?

To find a cyclic operator with N distinct T-invariant subspaces, you would need to first determine the dimension of the vector space and then construct a matrix representation of the operator using a basis for the space. Then, you can use the matrix to determine the minimal polynomial, which will give you the number of distinct T-invariant subspaces. You can then manipulate the matrix to find the desired number of subspaces.

What is the significance of having exactly N distinct T-invariant subspaces?

Having exactly N distinct T-invariant subspaces means that the operator has a very specific structure and set of properties. This can provide valuable insight into the behavior of the operator and its relationship to other operators on the same vector space.

Are there any practical applications for finding a cyclic operator with N distinct T-invariant subspaces?

Yes, there are several practical applications for finding a cyclic operator with N distinct T-invariant subspaces. For example, it can be useful in designing efficient algorithms for solving linear systems of equations or in understanding the behavior of dynamical systems.

Is it always possible to find a cyclic operator with N distinct T-invariant subspaces?

No, it is not always possible to find a cyclic operator with N distinct T-invariant subspaces. This is because the number of distinct T-invariant subspaces is dependent on the dimension of the vector space and the structure of the operator, so it may not always be possible to manipulate the matrix representation to achieve the desired number of subspaces.

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