Elementary Linear Algebra 7th Edition

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

Chapter 5 - Inner Product Spaces - 5.2 Inner Product Spaces - 5.2 Exercises - Page 245: 9


see the counterexample below.

Work Step by Step

Since $\langle (u_1,u_2), (v_1, v_2 )\rangle= u_1v_1 $, let $u=(0,1)$, then $\langle u.u\rangle=\langle (0,1).(0,1)\rangle=0$. But $u\neq (0,0)$, so the function does not define an inner product.
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