0
$\begingroup$

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 $T^n = 0$

$\endgroup$

1 Answer 1

4
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.