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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 $$\mathrm{T}$$.

Expert verified
The Jordan canonical form of the nilpotent linear operator T is: $J_T = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}.$
See the step by step solution

## 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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