Chapter 6 - Linear Transformations - 6.4 Additional Properties of Linear Transformation - True-False Review - Page 417: k

False

We have $T_2$ is onto. Suppose $T_2(x,y)=(x,y)$ Define: $T_1(x,y)=(0,0)$ We know $ T_1 : V_1 → V_2$ and $T_2 : V_2 → V_3$ are linear transformations obtain $(T_2T_1))(x,y)=T_2(T_1(x,y))=T_2(0,0)=(0,0)\\ \rightarrow Rng(T)=(0,0) \ne R^2$ Hence, $T_2T_1$ is not onto.