Discussion
Let V be a finite dimensional vector space over \(\mathbf{R}\) and \(T \in L(V)\). If \(\lambda\) is an eigenvalue of T, then which of the following statements is false ?
Answer: B
The false statement is \(\ker(T - \lambda I) = \{0\}\).
\(\lambda\) is an eigenvalue exactly when there is a non-zero \(v\) with \(Tv = \lambda v\), that is, a non-zero vector in \(\ker(T - \lambda I)\). So the kernel is not \(\{0\}\), which makes (A) true and (B) false.
On a finite dimensional space, a non-trivial kernel means \(T - \lambda I\) is not injective, so by rank-nullity its rank is less than \(\dim V\) (C), and a singular operator has determinant 0 (D). Those three are equivalent forms of the same fact.
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