Question-76

Use the following definitions to solve this question:

• A square matrix \(\displaystyle A\) of order \(\displaystyle n\) is called nilpotent if \(\displaystyle A^{k} =0\) for some positive integer \(\displaystyle k\). For example, \(\displaystyle \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}\) is nilpotent.

• A matrix is called diagonalizable if \(\displaystyle \mathbb{R}^{n}\) has a basis of eigenvectors of \(\displaystyle A\).

Which of the following statements are true?

• Zero is the only eigenvalue of a nilpotent matrix.

• A nilpotent matrix has at least one non-zero eigenvalue.

• If a matrix is nilpotent and diagonalizable, it has to be the zero matrix.

• All diagonal matrices are nilpotent.

Answer

• Zero is the only eigenvalue of a nilpotent matrix.

• A nilpotent matrix has at least one non-zero eigenvalue.

• If a matrix is nilpotent and diagonalizable, it has to be the zero matrix.

• All diagonal matrices are nilpotent.

Solution

Eigenvalues of a nilpotent matrix

We have:

\[ \begin{equation*} \begin{aligned} Av & =\lambda v \end{aligned} \end{equation*} \]

Pre-multiplying by \(\displaystyle A\) on both sides \(\displaystyle k-1\) times, we have:

\[ \begin{equation*} A^{k} v=\lambda ^{k} v\Longrightarrow \lambda ^{k} v=0 \end{equation*} \]

Since \(\displaystyle v\) is an eigenvector, \(\displaystyle v\neq 0\). This implies that \(\displaystyle \lambda =0\). Therefore, zero is the only eigenvalue of a nilpotent matrix.

Nilpotent and diagonalizable

Since \(\displaystyle A\) is nilpotent, we have \(\displaystyle A^{k} =0\) for some \(\displaystyle k\). Since \(\displaystyle A\) is diagonalizable, there is a basis \(\displaystyle B=\{v_{1} ,\cdots ,v_{n}\}\) for \(\displaystyle \mathbb{R}^{n}\), where \(\displaystyle v_{i}\) is an eigenvector of \(\displaystyle A\). Since zero is the only eigenvalue, we have \(\displaystyle Av_{i} =0\) for every eigenvector. Since \(\displaystyle A\) maps every basis vector to the zero vector, \(\displaystyle A\) has to be the zero matrix.