Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Prove Theorem 5.1.8d. ${(V^{⊥})}^{⊥} = V$ for any subspace V of $ℝ^{n}$ .

It is proved that ${V = (V^{⊥})}^{⊥}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Show that V⊆(V⊥)⊥.

Consider a subspace V of $ℝ^{n}$ such that $\vec{x} \in V$ and $\vec{y} \in V^{⊥}$ .

If the vector $\vec{x}$ is perpendicular $\vec{y}$ then $\vec{x} - \vec{y} = 0$ .

By the definition, the dot product of the vectors $\vec{x}$ and $\vec{y}$ is defined as follows.

$\vec{x} - \vec{y} = 0$

Similarly, for role="math" localid="1659438267836" $\vec{x} \in {(V^{⊥})}^{⊥}$ and $\vec{y} \in V^{⊥}$ , the dot product of the vectors $\vec{x}$ and $\vec{y}$ is defined as follows.

$\vec{x} - \vec{y} = 0$

Therefore, the set V is subset of ${(V^{⊥})}^{⊥}$ .

Step 2: Show that V=(V⊥)⊥.

Theorem: Property of the orthogonal compliment.

Consider a subspace V of $R^{n}$ .

a. The orthogonal complement role="math" localid="1659438558378" $V^{⊥}$ of is a subspace of $R^{n}$ .

b. The intersection of V and $V^{⊥}$ consist of the zero vector: $V \cap V^{⊥} = {\vec{0}}$ .

c. $\dim (V) + \dim (V^{⊥}) = n$

d. $(V^{⊥})^{⊥} = V$

By the theorem, the equation for the set V is defined as follows.

$\dim V + \dim V^{⊥} = n$

By the theorem, the equation for the set $V^{⊥}$ is defined as follows.

$\dim V^{⊥} + \dim (V^{⊥})^{⊥} = n$

Compare the equations ${dimV}^{1} + \dim (V^{1}) = n$ and $\dim V + \dim V^{⊥} = n$ as follows.

$\dim V + {dimV}^{⊥} = {dimV}^{⊥} + \dim {(V^{⊥})}^{⊥} \dim V = \dim {(V^{⊥})}^{⊥}$

As $\dim V = \dim (V^{⊥})^{⊥}$ and $V \subset (V^{⊥})^{⊥}$ implies $V = (V^{⊥})^{⊥}$ .

Hence, the values $V = (V^{⊥})^{⊥}$ for subset V of $ℝ^{n}$ .

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Linear Algebra With Applications

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Short Answer

Prove Theorem 5.1.8d. ${(V^{⊥})}^{⊥} = V$ for any subspace V of $ℝ^{n}$ .

Step by Step Solution

Step 1: Show that V⊆(V⊥)⊥.

Step 2: Show that V=(V⊥)⊥.

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Linear Algebra With Applications

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Prove Theorem 5.1.8d.(V⊥)⊥=V for any subspace V of ℝn.

Step by Step Solution

Step 1: Show that V⊆(V⊥)⊥.

Step 2: Show that V=(V⊥)⊥.

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Prove Theorem 5.1.8d. ${(V^{⊥})}^{⊥} = V$ for any subspace V of $ℝ^{n}$ .