Correct answer: (B) S is neither compact nor connected.
Explanation
Let v=(1,−3,2)ᵀ. Then $A^2=(w^Tv)A$. If wᵀv ≠ 0, the minimal polynomial divides t(t−wᵀv), with distinct roots, so A is diagonalizable. If wᵀv = 0, A is nonzero nilpotent and not diagonalizable. Thus S is the unit sphere minus the great circle wᵀv=0: two disjoint open hemispheres. It is disconnected and not closed/bounded-closed compact. **B**.
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