Discussion
Eigen values of a nilpotent matrix A, i.e., a matrix satisfying \(A^{K} = 0\) for some integer K are :
Answer: B
The eigenvalues of a nilpotent matrix are all zero.
Let \(A\vec{v} = \lambda\vec{v}\) with \(\vec{v} \neq 0\). Applying A repeatedly gives \(A^{K}\vec{v} = \lambda^{K}\vec{v}\). Since \(A^{K} = 0\), this means \(\lambda^{K}\vec{v} = 0\), and because \(\vec{v}\) is not the zero vector, \(\lambda^{K} = 0\), so \(\lambda = 0\).
Thus every eigenvalue is 0; equivalently the characteristic polynomial is \(\lambda^{n}\). They cannot be distinct (all are equal), and they are neither purely imaginary nor negative. A simple example is the matrix with a single 1 above the diagonal, whose square is zero.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development