Question

question 6

if a is nilpotent matrix, then --- is the only eigenvalue of a.

Explanation:

Define a nilpotent matrix

A square matrix \(A\) is defined as nilpotent if there exists a positive integer \(k\) such that:

$$A^k = 0$$

Relate eigenvalues to matrix powers

Let \(\lambda\) be an eigenvalue of \(A\), and \(\mathbf{v}
eq \mathbf{0}\) be its corresponding eigenvector:

$$A\mathbf{v} = \lambda\mathbf{v}$$

Multiplying both sides by \(A\) repeatedly yields:

$$A^k\mathbf{v} = \lambda^k\mathbf{v}$$

Solve for the eigenvalue

Since \(A\) is nilpotent, we substitute \(A^k = 0\):

$$0\mathbf{v} = \lambda^k\mathbf{v}$$

$$\mathbf{0} = \lambda^k\mathbf{v}$$

Since \(\mathbf{v}
eq \mathbf{0}\), we must have:

$$\lambda^k = 0 \implies \lambda = 0$$

Thus, \(0\) is the only eigenvalue of \(A\).

Answer:

If A is nilpotent matrix, then <blank>0</blank> is the only eigenvalue of A.