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Q23E

Expert-verifiedFound in: Page 216

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**

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

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

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

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

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

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

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

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

**Theorem: Property of the orthogonal compliment.**

** **

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

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

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

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

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

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.

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

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

Hence, the values $V=({V}^{\perp}{)}^{\perp}$for subset *V* of ${\mathrm{\mathbb{R}}}^{\mathrm{n}}$.

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