Matrix Representation of Nilpotent Linear Operator

In summary: Theorem 5.9.To find k, we need to find the index of nilpotency of \eta. Looking at the matrix representation, we can see that the operator becomes 0 when raised to the 4-th power (since all entries in the 4-th row and column are 0). Therefore, the index of nilpotency of \eta is p_{1} = 4.Next, we need to find the vectors \xi_{1}, ..., \xi_{k} that generate cyclic subspaces of index 4. Looking at the matrix representation, we can see that the first and fifth columns have non-zero entries only in
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Homework Statement



(5.5)[tex]V = Z_{\xi_{1}} \oplus Z_{\xi_{2}} \oplus ... \oplus Z_{\xi_{k}},[/tex] the basis for V is:

[tex]\xi_{1}, \eta(\xi_{1}), ..., \eta(\xi_{1})^{p_{1} - 1}[/tex]
[tex]\xi_{2}, \eta(\xi_{2}), ..., \eta(\xi_{2})^{p_{2} - 1}[/tex]
.
.
.
[tex]\xi_{k}, \eta(\xi_{k}), ..., \eta_(\xi_{k})^{p_{k} - 1}[/tex]

relative to which [tex]\eta[/tex] is represented by [tex]Dg[J_{p1}(0), J_{p2}(0), ..., J_{pk}(0)].[/tex]

Theorem 5.9: If [tex]\eta \in L(V, V)[/tex] is nilpotent of index [tex]p_{1}[/tex], then there exists an integer k, k distinct vectors [tex]\xi_{1}, ..., \xi_{k}[/tex] and k integers [tex]p_{1} \geq p_{2} \geq ... \geq p_{k} [/tex]such that the vectors in (5.5) form a basis for V. Moreover, V is the direct sum of the [tex]\eta[/tex]-cyclic subspaces generated by the [tex]\xi_{i}: V = Z_{\xi_{1}} \oplus Z_{\xi_{2}} \oplus ... \oplus Z_{\xi_{k}}.[/tex] Relative to the basis in (5.5), [tex]\eta [/tex] is represented by the matrix [tex]Dg[J_{p1}(0), J_{p2}(0), ..., J_{pk}].[/tex]

Problem: Find integer k and the vectors [tex]\xi_{1}, ..., \xi_{k}[/tex] of Theorem 5.9 for the nilpotent operator [tex]\eta \in L(\Re_{5 x 1}, \Re_{5 x 1}) [/tex]if

[tex]Mtx_{\epsilon}(\eta) = \left[ \begin{array} {ccccc} 0 & 0 & 0 & 0 & 0\\
2 & 0 & 0 & 0 & 0\\
1 & 3 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 4\\
0 & 0 & 0 & 0 & 0 \end{array} \right][/tex]



Homework Equations





The Attempt at a Solution


The answer is supposed to be k = 2 and [tex]\xi_{1} = \epsilon_{1}[/tex] and [tex]\xi_{2} = \epsilon_{5},[/tex] with the [tex]\epsilon_{i}[/tex] being the standard basis. I don't even know how they arrive at this answer. There is no example in this book to help guide how to do this problem. I'm self-studying, so I'm not asking for anyone to do this work for me. I just really want to understand how to do this problem. Anyone know where to start on this?
 
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  • #2



Hello, thank you for your question. This is an interesting problem that involves the concept of nilpotent operators and their cyclic subspaces. Let's break down the problem and try to understand it step by step.

First, let's define some of the terms used in the problem:

- Nilpotent operator: A linear operator \eta is called nilpotent if there exists a positive integer p such that \eta^{p} = 0, where \eta^{p} denotes the p-th power of \eta.
- Index of nilpotency: The smallest positive integer p for which \eta^{p} = 0 is called the index of nilpotency of \eta.
- Cyclic subspace: A subspace V \subset W is called cyclic with respect to a linear operator \eta if there exists a vector \xi \in V such that V = \text{span}\{\xi, \eta(\xi), \eta^{2}(\xi), ... \}.
- Direct sum: The direct sum of two subspaces V and W, denoted by V \oplus W, is the subspace containing all vectors of the form v + w, where v \in V and w \in W.

Now, let's look at the theorem given in the problem statement. It states that for a nilpotent operator \eta with index of nilpotency p_{1}, there exists an integer k, k distinct vectors \xi_{1}, ..., \xi_{k}, and k integers p_{1} \geq p_{2} \geq ... \geq p_{k} such that the vectors in (5.5) form a basis for V. Moreover, V is the direct sum of the \eta-cyclic subspaces generated by the \xi_{i}.

In simpler terms, this means that for a nilpotent operator \eta, there exists a basis for its vector space V consisting of k distinct vectors \xi_{1}, ..., \xi_{k}, where each vector \xi_{i} generates a cyclic subspace of V of index p_{i}.

Now, let's apply this theorem to the given problem. We are given a nilpotent operator \eta \in L(\Re_{5 x 1}, \Re_{5 x 1}) and its matrix representation Mtx_{\epsilon}(\eta). We need to find the integer k and the vectors
 

What is a nilpotent linear operator?

A nilpotent linear operator is a type of linear transformation on a vector space that satisfies the property that applying the operator repeatedly will eventually result in the zero vector. In other words, there exists a positive integer n such that Tn = 0, where T is the nilpotent operator.

What is the matrix representation of a nilpotent linear operator?

The matrix representation of a nilpotent linear operator is a square matrix with all entries above the main diagonal equal to 0, and the entries on the main diagonal representing the coefficients of the linear transformation. The matrix will also have a last row of all 0's, as this corresponds to the zero vector.

Why is the matrix representation of a nilpotent linear operator useful?

The matrix representation of a nilpotent linear operator is useful because it can simplify calculations and provide a more visual representation of the linear transformation. It also allows for the use of matrix operations, such as multiplication and addition, to be applied to the linear transformation.

How can the matrix representation of a nilpotent linear operator be determined?

The matrix representation of a nilpotent linear operator can be determined by finding a basis for the vector space and expressing the linear transformation in terms of that basis. The coefficients of the basis vectors will then form the entries of the matrix representation.

What are some real-world applications of the matrix representation of a nilpotent linear operator?

The matrix representation of a nilpotent linear operator has various applications in fields such as quantum mechanics, computer graphics, and image processing. For example, in quantum mechanics, the matrix representation is used to represent the state of a quantum system, and in computer graphics, it is used to transform and manipulate objects in a 3D space.

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