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Let T be a nilpotent operator on an n-dimensional vector space V . Then T^n = 0, where 0 ∈ L(V ) is the zero map.

(a) First, show that since T is nilpotent, dimkerT > 0.

(b) Next, show that either T is the zero map or dimker T 2 > dimkerT.

(c) Show in general that for k ∈ {2, 3, . . .} either T k−1 is the zero map or dimker T k > dimker T k−1 .

(d) Using parts (a)-(c), why must T n = 0?

I am beyond confused with how to solve this proof. Any help would be appreciated!

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    Solution for $a)$ : Since $T$ is nilpotent, the only eigenvalue is $0$. Hence, $T$ is not invertible, implying $kern(T)\ne$ {$0$} – 2017-02-12

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