Answer
True
Work Step by Step
We can see that $Ker (T_1)$ and $Ker(T_2T_1)$ are subspaces of $V_1$
Obtain: $T_1(v_1)=0\\
\rightarrow (T_2T_1)v_1=T_2(T_1(v_1))=T_2(v_2)=0$
Hence, $Ker (T_1)$ is a subspace of $Ker (T_2T_1)$
The statement is true.
Differential Equations and Linear Algebra (4th Edition)
by Goode, Stephen W.; Annin, Scott A.
Published by
Pearson
ISBN 10:
0-32196-467-5
ISBN 13:
978-0-32196-467-0
Chapter 6 - Linear Transformations - 6.4 Additional Properties of Linear Transformation - True-False Review - Page 416: c
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