Homepage › Solution manuals › Stephen Friedberg › Linear Algebra › Exercise 6.2.7

Exercise 6.2.7

Answers

The necessity comes from the definition of orthogonal complement, since every element in $β$ is an element in $W$ . For the sufficiency, assume that $⟨ z, v ⟩ = 0$ for all $v \in β$ . Since $β$ is a basis, every element in $W$ could be written as

\sum_{i = 1}^{k} a_{i} v_{i},

where $a_{i}$ is some scalar and $v_{i}$ is element in $β$ . So we have

⟨ z, \sum_{i = 1}^{k} a_{i} v_{i} ⟩ = \sum_{i = 1}^{k} \bar{a_{i}} ⟨ z, v_{i} ⟩ = 0 .

Hence $z$ is an element in $W^{⊥}$ .

jephianlin

2011-06-27 00:00

Exercise 6.2.7

Answers

Comments

Add answer