Discussion
Let \(A\) be a \(2 \times 2\) real symmetric matrix such that the trace of \(A\) is \(6\) and the determinant of \(A\) is \(5\). Which one of the following statements is true?
Answer: D
The true statement is \(A^{2} - 6A\) is negative definite.
A real symmetric matrix is described by its eigenvalues. Here \(\lambda_{1} + \lambda_{2} = 6\) and \(\lambda_{1}\lambda_{2} = 5\), so the eigenvalues are the roots of \(\lambda^{2} - 6\lambda + 5 = 0\), namely \(1\) and \(5\). A polynomial \(p(A)\) has eigenvalues \(p(1)\) and \(p(5)\):
- \(A - I_{2}\): \(0\) and \(4\). Positive semi-definite only, because of the zero eigenvalue.
- \(A - 3I_{2}\): \(-2\) and \(2\). Indefinite.
- \(A^{2} + A - 3I_{2}\): \(-1\) and \(27\). Indefinite.
- \(A^{2} - 6A\): \(1 - 6 = -5\) and \(25 - 30 = -5\). Both negative, so negative definite.
In fact \(A^{2} - 6A + 5I_{2} = 0\) by the Cayley-Hamilton theorem, so \(A^{2} - 6A = -5I_{2}\).
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