Discussion
Let
\[ \begin{aligned} S = \Big\{ w &= \begin{bmatrix} w_{1} \\ w_{2} \\ w_{3} \end{bmatrix} \in \mathbb{R}^{3} : \begin{bmatrix} 1 \\ -3 \\ 2 \end{bmatrix} \begin{bmatrix} w_{1} & w_{2} & w_{3} \end{bmatrix} \\ &\text{is diagonalizable and } \|w\| = 1 \Big\}, \end{aligned} \]
where \(\|w\| = (w_{1}^{2} + w_{2}^{2} + w_{3}^{2})^{1/2}\). Then, which one of the following is TRUE?
Answer: B
S is neither compact nor connected.
Write \(v = (1, -3, 2)^{T}\). The matrix \(vw^{T}\) has rank 1, and its only possibly non-zero eigenvalue is its trace, \(w^{T}v = w_{1} - 3w_{2} + 2w_{3}\). A rank-1 matrix is diagonalizable exactly when this trace is non-zero; if it is zero the matrix is nilpotent and not zero, so not diagonalizable.
So S is the unit sphere with the great circle \(w_{1} - 3w_{2} + 2w_{3} = 0\) removed. That leaves two open hemispheres. S is not closed, since points on the removed circle are limits of points of S, so it is not compact; and it falls into two disjoint open pieces, so it is not connected.
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