91Ó°ÊÓ

We want to demonstrate that range \(T^{k}=\) range \(T^{m}\) for all \(k > m\). Let \(v \in V\) be an arbitrary element in the vector space V.

Given the condition that range \(T^m =\) range \(T^{m+1}\), we can find \(v' \in V\), such that \(T^{m+1}(v') = T^m(v)\)

We want to show that for all \(k > m\), range \(T^{k}=\) range \(T^{m}\). We will use induction. 1. Base Case (k=m+1): As given, the condition holds for k = m+1 by definition. Now we assume the condition holds for some \(k = p\) where \(p > m\) and show that it will also hold for \(k = p+1\). 2. Inductive Hypothesis: Assume that range \(T^p =\) range \(T^m\). 3. Induction step: We need to show that range \(T^{p+1} =\) range \(T^m\). Let \(T^m(v_p) = T^p(v')\) for some \(v_p, v' \in V\). Now, we apply the operator T to both sides of the equation: \(T^{m+1}(v_p) = T^{p+1}(v')\) Since we have assumed the condition to be true for \(k = p\), we know that range \(T^p =\) range \(T^m\). Therefore, there exists a vector \(v'' \in V\) such that \(T^m(v'') = T^{m+1}(v_p)\). By substitution, we get \(T^m(v'') = T^{p+1}(v')\) Which means range \(T^{p+1} =\) range \(T^m\). Thus, our induction step is completed. By induction, we can conclude that range \(T^{k}=\) range \(T^{m}\) for all \(k > m\).