# Prove that $T^n$ is a zero map.

Let $$n$$ be a positive integer and $$V$$ be a $$(n+1)$$ dimensional vector space over $$\Bbb R$$. If $$\{e_1,e_2,...,e_{n+1}\}$$ be a basis of $$V$$ and $$T: V \to V$$ is a linear transformation satisfying $$T(e_i) = e_{i+1}, i = 1,2,...,n$$ and $$T(e_{n+1})=0$$ Then

$$T^n = T•T•T•...•T(n-times)$$

is a zero map.

How can we prove it ?

My try:

Let $$T = \begin{pmatrix} 0&1&0&0...0\\0&0&1&0...0\\0&0&0&1...0\\ ...\\ 0&0&0&0...1\\ 0&0&0&0...0 \end{pmatrix}$$

Here we see that rank of $$T = n$$ and nullity of $$T = 1$$ and Trace of $$T = 0$$. But I have no idea how to prove $$T^n = 0$$

It is not true that $$T^n=0$$. Indeed, $$T^n e_1 = e_{n+1} \neq 0$$.
However, it is true that $$T^{n+1}=0$$. It is enough to prove that $$T^{n+1}e_i$$=0 for all $$i=1,\dots, n+1$$. By definition, $$Te_{n+1} =0$$, so $$T^{n+1}e_{n+1}=0$$. Let $$1 \leqslant k \leqslant n$$. Then $$Te_k = e_{k+1},\quad T^2e_k = e_{k+2}, \dots, \quad T^{n+1-k} e_{k} = e_{n+1}.$$ Therefore, $$T^{n+2-k} e_k = 0$$ for all $$k=1,\dots ,n$$. Finally, note that $$n+2-k \leqslant n+1$$, so $$T^{n+1} e_k =T^{k-1} T^{n+2-k} e_k=T^{k-1} 0=0$$(with the understanding that $$T^0$$ is the identity).
• I couldn't understand your logic. Here $T$ is nilpotent matrix and all eigen values of nilpotent matrix are zero therefore characteristic polynomial is $Ch_T(x) = x^{n+1}$ . So it is trivially, $T^{n+1} = 0$ Sep 11, 2021 at 8:06