Discussion
Let A be an n × n matrix with all entries equal to 1. Then :
Answer: D
A has exactly one non-zero eigenvalue, namely n.
Every row of A is the same, so A has rank 1 and its null space has dimension \(n - 1\): the eigenvalue 0 has multiplicity \(n - 1\). The trace is n, which equals the sum of the eigenvalues, so the remaining eigenvalue is n, with eigenvector \((1, 1, \ldots, 1)\).
So for \(n \ge 2\) the eigenvalues are 0 (repeated \(n - 1\) times) and n. That rules out 'all eigenvalues positive', '0 has multiplicity 1' (it has multiplicity \(n - 1\)), and 'at least three distinct eigenvalues' (there are only two).
No comment is present. Be the first to comment.
Loading…
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