Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 4 - Vector Spaces - 4.1 Exercises - Page 198: 7

Answer

not a subspace of $\mathbb{P}_{n}$

Work Step by Step

For $ H=\{p(t)| p(t)=a+bt+ct^{2}, a,b,c\in \mathbb{Z}\}$ following the definition of a subspace on p.195, (a) if a=b=c=0$\in \mathbb{Z}$, the zero vector is in H. (b) ($a_{1}+b_{1}t+c{}_{1}t^{2})+(a_{2}+b_{2}t+c_{2}t^{2})=$ $=(a_{1}+a_{2})+(b_{1}+b_{2})t+(c_{1}+c_{2})t^{2}$ belongs to H, because the set of integers is closed under addition. (c) Let $a+bt+ct^{2}, a,b,c\in \mathbb{Z}$, and, at least one of a,b,c is not zero. Also, let $d\in \mathbb{R}, d\not\in \mathbb{Z}$. Then $da+dbt+dct^{2}\not\in H$, because at least one coefficient is not an integer. (H is not closed under multiplication by scalars) So, $H$ is not a subspace of $\mathbb{P}_{n}$
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