Answer
True
Work Step by Step
Since, $S$ is closed for a scalar multiple. For any vector $v$ in $S$, we have $v \in S$ and for a scalar $k$, we have $kv \in S$.
This suggests that $0 \in S \subseteq V$
Therefore, the given 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 4 - Vector Spaces - 4.3 Subspaces - True-False Review - Page 272: e
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