Since the kernel must grow with each power for a nilpotent operator the the $\Leftarrow$ direction is easy. Why is the other direction true?
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Think about the Jordan normal form. – Angina Seng May 04 '17 at 19:39
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\iff gives $\iff$ – Fly by Night May 04 '17 at 19:43
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Is $n$ the dimension of the space? – lhf May 04 '17 at 20:02
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See the answers here. – Dietrich Burde May 04 '17 at 20:32
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To prove the "$\Rightarrow$" direction use contraposition: assume $\dim(\ker(T))<2$ and show that $T^{n-1}\neq 0$.
If $\dim(\ker(T))=0$ then we are trivially done.
Assume $\dim(\ker(T))=1$, then $T$ has a one dimensional eigenspace associated with the zero eigenvalue and eigen vector $p_0$. Let $P$ be the matrix formed by the generalized eigenvectors of $T$ with $p_0$ as its first column. We then have $$T = PJP^{-1}$$ where $J$ is of the form $$J = \begin{bmatrix} 0 & 0 & \dots & 0 \\ 0 & J_1 & \;& \; \\ \vdots & \; & \ddots& \; \\ 0 & \; & \; & J_p\end{bmatrix},$$ and each $J_l$ is a Jordan block.
We then have $$T^{n-1}=PJ^{n-1}P^{-1}$$ where $$J^{n-1}=\begin{bmatrix} 0 & 0 & \dots & 0 \\ 0 & J_1^{n-1} & \;& \; \\ \vdots & \; & \ddots& \; \\ 0 & \; & \; & J_p^{n-1}\end{bmatrix}.$$ $J^{n-1}$, and therefore $T^{n-1}$, clearly has a one dimensional kernel. We conclude by contraposition that $T^{n-1}=0$ implies $\dim(\ker(T))\geq 2$.
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