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: 8


The set $S=\{(-1,2),(1,-2),(2,4)\}$ is not a basis for $R^2$.

Work Step by Step

The set $S=\{(-1,2),(1,-2),(2,4)\}$ is not a basis for $R^2$ because the vectors in $S$ are not linearly independent. Indeed, considering the linear combination $$a (-1,2)+b(1,-2)+c(2,4)=(0,0),$$ then we have the system $$\begin{align*} -a+b+2c&=0\\ 2a-2b+4c&=0 \end{align*}$$ It is easy to see that the above system has the solution $a=t, b=t , c=0$ . Hence, the vectors of $S$ are not linearly independent. Hence, the set $S=\{(-1,2),(1,-2),(2,4)\}$ is not a basis for $R^2$
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