Chapter 12: Chapter 12

Problem 293

Let the linear operator \(\mathrm{T}\) be nilpotent of degree 4 on \(\mathrm{C}^{6}\), \(\mathrm{T} \neq 0^{\rightarrow}\). Find the Jordan canonical form of \(\mathrm{T}\).

Short Answer

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The Jordan canonical form of the nilpotent linear operator T is: \[ J_T = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}. \]

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Step 1: Determine the minimal polynomial of T

A linear operator T is nilpotent of degree 4 if T⁴ = 0, but T³ ≠ 0. This means that the minimal polynomial of T divides x⁴ (since T⁴ = 0), but it doesn't divide x³ (because T³ ≠ 0). Hence, the minimal polynomial of T is \(m_T(x) = x^4\).

Step 2: Determine the eigenvalues of T

As T is a nilpotent operator with minimal polynomial \(m_T(x) = x^4\), it has only one eigenvalue, which is λ = 0.

Step 3: Determine the Jordan blocks associated with the eigenvalue λ = 0

To determine the Jordan blocks associated with the single eigenvalue λ = 0, we need to find the size of each Jordan block. As T is nilpotent of degree 4, it is clear that the largest Jordan block must be of size 4, otherwise, T⁴ would be nonzero. Since the vector space has dimension 6, and we have determined that there must be a Jordan block of size 4, there are 2 dimensions left for the remaining Jordan blocks. The remaining dimensions can be distributed among Jordan blocks of size 1 or size 2. Now let's consider the dimension of the null space of T³: \(\text{dim } \ker(T^3)\). If there are no other blocks with size bigger than 1 for the remaining 2 dimensions, then we have 1 block of size 4, and 2 blocks of size 1. In this case, \(\text{dim } \ker(T^3) = 1 + 1 + 3 = 5\). However, since T³ is nonzero, the kernel of T³ is not the entire vector space and, hence, \(\text{dim } \ker(T^3) < 6\). This means the previous case with two blocks of size 1 is not possible. Therefore, we must have 1 Jordan block of size 4 and 1 Jordan block of size 2. In this case, \(\text{dim } \ker(T^3) = 1 + 2 = 3\), which satisfies the condition that the kernel is not the entire vector space.

Step 4: Assemble the Jordan canonical form from the Jordan blocks

The Jordan canonical form of T can be assembled by arranging the Jordan blocks we found along the diagonal of a 6x6 matrix. The Jordan block corresponding to λ = 0 and size 4 is: \[ J_4 = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \] and the Jordan block of size 2 is: \[ J_2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. \] The Jordan canonical form for T is: \[ J_T = \begin{pmatrix} J_4 & 0 \\ 0 & J_2 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}. \]

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Chapter 12: Chapter 12

Let the linear operator \(\mathrm{T}\) be nilpotent of degree 4 on \(\mathrm{C}^{6}\), \(\mathrm{T} \neq 0^{\rightarrow}\). Find the Jordan canonical form of \(\mathrm{T}\).

Short Answer

Step by step solution

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Step 1: Determine the minimal polynomial of T

Step 2: Determine the eigenvalues of T

Step 3: Determine the Jordan blocks associated with the eigenvalue λ = 0

Step 4: Assemble the Jordan canonical form from the Jordan blocks

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Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

Chapter 13

Chapter 14

Chapter 15

Chapter 16

Chapter 17

Chapter 18

Chapter 19

Chapter 20

Chapter 21

Chapter 22

Chapter 23

Chapter 24

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