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 - Problems - Page 273: 20

Answer

See below

Work Step by Step

Given $V=P_2(R)$ We can write set $S$ as $S=\{p \in P_2(R):p(x)=ax^2+1\}$. We can notice that $0x^2+0x+0\ne 0$. Hence, $S$ is not a subspace of $P_2(R)$

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