Homepage › Solution manuals › Stephen Friedberg › Linear Algebra › Exercise 6.2.10

Exercise 6.2.10

Answers

By Theorem 6.6, we know that $V = W \oplus W^{⊥}$ since $W \cap W^{⊥} = {0}$ by definition. So there’s a nature projection $T$ on $W$ along $W^{⊥}$ . That is, we know tht every element $x$ in $V$ could be writen as $u + v$ such that $u \in W$ and $v \in W^{⊥}$ and we define $T (x) = u$ . Naturally, the null space $N (T)$ is $W^{⊥}$ . And since $u$ and $v$ is always orthogonal, we have

∥ x ∥^{2} = ∥ u ∥^{2} + ∥ v ∥^{2} \geq ∥ u ∥^{2} = ∥ T (x) ∥^{2}

by Exercise 6.1.10. And so we have

∥ T (x) ∥ \leq ∥ x ∥ .

jephianlin

2011-06-27 00:00

Exercise 6.2.10

Answers

Comments

Add answer