Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 4 - Vector Spaces - 4.6 Bases and Dimension - Problems - Page 308: 2



Work Step by Step

The vectors are clearly linearly independent. (because there are $2$ and they are not parallel) Say $(a,b)=k(1,1)+l(-1,1)$, then $a=k-l$ and $b=k+l$, thus $k=0.5(a+b)$ and $l=0.5(b-a)$, thus every $2$ dimensional vector is expressable, thus it is a basis.
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