Question

Prove Theorem 5.1.8d.${\left(V^{\perp }\right)}^{\mathbf{\perp }}\mathbf{=}\mathbf{V}$for any subspaceV of${\mathbf{ℝ}}^{\mathbf{n}}$.

Short answer

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

Step-by-step solution

Show that V⊆(V⊥)⊥.

Consider a subspace V of ${\mathrm{ℝ}}^n$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, forrole="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 setV is subset of ${\left(V^{\perp }\right)}^{\perp }$.

Show that V=(V⊥)⊥.

Theorem: Property of the orthogonal compliment.

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

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 Vand ${\mathbf{V}}^{\mathbf{\perp }}$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.

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

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

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

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

$$\begin{gathered} \dim \mathrm{V}+{\mathrm{dimV}}^{\perp }={\mathrm{dimV}}^{\perp }+\dim {\left({\mathrm{V}}^{\perp }\right)}^{\perp } \\ \dim \mathrm{V}=\dim {\left({\mathrm{V}}^{\perp }\right)}^{\perp } \end{gathered}$$

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

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