Answer to Question #219070 in Abstract Algebra for Komal

Question

If dim V < infinity and T is a linear transformation such that for all v belongs to V , there exist a positive integer k such that(T)^k(v) =0, then prove that T is nilpotent.

Answer 1

Let $\text{dim} \ V=n$ and let $\ T:\ V\rightarrow V$ be a linear transformation such that for all $v\in V,$ there exists integer $k>0$ such that $T^k v=0$ .

Suppose $\beta =\{\beta_1,\ \beta_2,\ …\ , \ \beta_n\}$ is a basis for $V$ .

For each vector $\beta _i$ there exists $k_i$ such that $T^{k_i}\beta_i=0$ .

Let $k=\max \{ k_1,\ k_2,\ … , \ k_n\}$ . In this case we have that $T^k\beta _i=T^{k-k_i} \left( T^{k_i}\beta_i \right)=T^{k-k_i}(0)=0$ .

Since every vector $v\in V$ is a linear combination of $\beta_i$ , we can consider $v=\sum\limits_{i=1}^nc_i\beta_i$ for some $c_i$ .

$T^kv=T^k \left(\sum\limits_{i=1}^nc_i\beta_i\right)= \sum\limits_{i=1}^nc_i T^k(\beta_i)=\sum \limits_{i=1}^nc_i\cdot 0=0$ .

So, $T^k=0$ and it proves that $T$ is nilpotent.