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Expert-verified Found in: Page 216 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Prove Theorem 5.1.8d.${\left({V}^{\perp }\right)}^{{\mathbf{\perp }}}{\mathbf{=}}{\mathbf{V}}$ for any subspace V of ${{\mathbf{ℝ}}}^{{\mathbf{n}}}$.

It is proved that ${\mathrm{V}=\left({\mathrm{V}}^{\perp }\right)}^{{\perp }}$.

See the step by step solution

## Step 1: Show that V⊆(V⊥)⊥.

Consider a subspace V of ${\mathrm{ℝ}}^{n}$ such that $\stackrel{\to }{x}\in V$and $\stackrel{\to }{y}\in V{}^{\perp }$.

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

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

$\stackrel{\to }{x}-\stackrel{\to }{y}=0$

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

$\stackrel{\to }{x}-\stackrel{\to }{y}=0$

Therefore, the set V is subset of ${\left({V}^{\perp }\right)}^{\perp }$.

## Step 2: Show that V=(V⊥)⊥.

Theorem: Property of the orthogonal compliment.

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

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

b. The intersection of V and ${{\mathbf{V}}}^{{\mathbf{\perp }}}$ consist of the zero vector: ${\mathbit{V}}{\mathbf{\cap }}{{\mathbit{V}}}^{{\mathbf{\perp }}}{\mathbf{=}}\left\{\stackrel{\to }{0}\right\}$.

c. ${\mathbf{dim}}{\mathbf{\left(}}{\mathbf{V}}{\mathbf{\right)}}{\mathbf{+}}{\mathbf{dim}}{\mathbf{\left(}}{{\mathbf{V}}}^{{\mathbf{\perp }}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{n}}$

d. ${\mathbf{\left(}}{{\mathbf{V}}}^{{\mathbf{\perp }}}{\mathbf{\right)}}^{\mathbf{\perp }}{\mathbf{=}}{\mathbf{V}}$

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

$\mathrm{dim}\mathrm{V}+\mathrm{dim}{\mathrm{V}}^{\perp }=\mathrm{n}$

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

$\mathrm{dim}{\mathrm{V}}^{\perp }+\mathrm{dim}\left({\mathrm{V}}^{\perp }{\right)}^{\perp }=\mathrm{n}$

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

$\mathrm{dim}\mathrm{V}+{\mathrm{dimV}}^{\perp }={\mathrm{dimV}}^{\perp }+\mathrm{dim}{\left({\mathrm{V}}^{\perp }\right)}^{\perp }\phantom{\rule{0ex}{0ex}}\mathrm{dim}\mathrm{V}=\mathrm{dim}{\left({\mathrm{V}}^{\perp }\right)}^{\perp }$

As $\mathrm{dim}\mathrm{V}=\mathrm{dim}\left({\mathrm{V}}^{\perp }{\right)}^{\perp }$ and $V\subset \left({V}^{\perp }{\right)}^{\perp }$ implies $V=\left({V}^{\perp }{\right)}^{\perp }$.

Hence, the values $V=\left({V}^{\perp }{\right)}^{\perp }$for subset V of ${\mathrm{ℝ}}^{\mathrm{n}}$. ### Want to see more solutions like these? 