Discussion
Let \(T : \mathbf{R}^{n} \to \mathbf{R}^{n}\) be defined as \(T(\bar{e}_{1}) = \bar{0}\), \(T(\bar{e}_{j}) = \bar{e}_{j - 1}, j = 2, ...., n\) where \(\{\bar{e}_{1}, \bar{e}_{2}, ........, \bar{e}_{n}\}\) is the standard basis of \(\mathbf{R}^{n}\). Then :
Answer: D
T is nilpotent.
T is the backward shift: \(e_{n} \mapsto e_{n-1} \mapsto \cdots \mapsto e_{1} \mapsto 0\). Applying it \(n\) times sends every basis vector to 0, so \(T^{n} = 0\).
It is linear by construction, since it is defined on a basis and extended linearly. It is not idempotent: \(T^{2}e_{3} = e_{1}\) while \(Te_{3} = e_{2}\) (for \(n \ge 3\); for \(n = 2\), \(T^{2} = 0 \ne T\)). Its kernel contains \(e_{1}\), so Ker T is not {0}.
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