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 $T^n = 0$


1 Answer 1


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).

  • $\begingroup$ 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, $T^{n+1} = 0$ $\endgroup$ Sep 11, 2021 at 8:06
  • $\begingroup$ I slightly misread your question - let me fix my answer. $\endgroup$
    – JackT
    Sep 11, 2021 at 8:14

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