Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 4 - Vector Spaces - 4.5 Basis and Dimension - 4.5 Exercises - Page 187: 43


$S$ is not a basis for $P_3$.

Work Step by Step

The set $S=\left\{4-t, t^{3}, 6 t^{2}, t^{3}+3 t, 4 t-1\right\}$ is not a basis for $P_3$. Indeed, since $P_3$ has dimension $4$, then any basis for $P_3$ must contain $4$ vectors. But, here $S$ contains $ 5$ vectors and hence $S$ is not a basis for $P_3$.
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