Question

Consider a basis${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}{\overrightarrow{\mathbf{v}}}_{\mathbf{m}}$ of a subspaceVofrole="math" localid="1659434380505" ${\mathbf{ℝ}}^{\mathit{n}}$. Show that a vector$\overrightarrow{\mathbf{x}}$ inrole="math" localid="1659434402220" ${\mathbf{ℝ}}^{\mathit{n}}$is orthogonal toV if and only if$\overrightarrow{\mathbf{x}}$ is orthogonal to all vectors${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}{\overrightarrow{\mathbf{v}}}_{\mathbf{m}}$.

Short answer

The vector $\overrightarrow{x}$in ${\mathrm{ℝ}}^n$is orthogonal to Vif and only if $\overrightarrow{x}$is orthogonal to all vectors ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,...{\overrightarrow{v}}_m$.

Step-by-step solution

Orthogonal condition x→⊥v→i

Consider a subspace V of ${\mathrm{ℝ}}^n$such that ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,...,{\overrightarrow{v}}_m$is basis of V and $\overrightarrow{x}\in {\mathrm{ℝ}}^n$ is orthogonal to V.

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

As$\overrightarrow{x}\in {\mathrm{ℝ}}^n$is orthogonal to V, by the definition the dot product of the vectors$\overrightarrow{x}$ and$\overrightarrow{v}$ is defined as follows:

$\overrightarrow{x}.\overrightarrow{v}=0$

For any$\overrightarrow{v}\in V$ the value $\overrightarrow{x}.\overrightarrow{v}=0$, assume$\overrightarrow{v}={\overrightarrow{v}}_i$ then if$\overrightarrow{x}.\overrightarrow{v}=0$ implies $\overrightarrow{x}.{\overrightarrow{v}}_i=0$.

Show that x→⊥v→ .

Assume that is orthogonal to all ${\overrightarrow{v}}_1.{\overrightarrow{v}}_2,...,{\overrightarrow{v}}_m$.

As${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,...{\overrightarrow{v}}_m$ is basis ofV then for any$\overrightarrow{v}\in V$there exist$a_i\in R$such that $\overrightarrow{V}=\sum _{i=1}^m{a}_i{\overrightarrow{v}}_i$.

Take dot product with respect to both side in the equation$\overrightarrow{V}=\sum _{i=1}^m{a}_i\overrightarrow{v}$as follows.

$$\begin{gathered} \overrightarrow{V}=\sum _{i=1}^m{a}_i\overrightarrow{v} \\ \overrightarrow{x}.\overrightarrow{v}=\overrightarrow{x}.\left\{\sum _{i=1}^m{a}_i\overrightarrow{v}\right\} \\ \overrightarrow{x}.\overrightarrow{v}=\sum _{i=1}^m{a}_i\left\{\overrightarrow{x}.\overrightarrow{v_i}\right\} \end{gathered}$$

As$\overrightarrow{x}\in {\mathrm{ℝ}}^n$is orthogonal to all${\overrightarrow{v}}_1{\overrightarrow{v}}_2,...{\overrightarrow{v}}_m$means$\overrightarrow{x}.{\overrightarrow{v}}_i=0$for all ${\overrightarrow{v}}_i$.

Substitute the value 0 for $\overrightarrow{x}.{\overrightarrow{v}}_i$in the equation $\overrightarrow{x}.\overrightarrow{v}=\sum _{i=1}^m{a}_i\left\{\overrightarrow{x}.{\overrightarrow{v}}_i\right\}$as follows.

$$\begin{gathered} \overrightarrow{x}.\overrightarrow{v}=\sum _{i=1}^m{a}_i\left\{\overrightarrow{x}.{\overrightarrow{v}}_i\right\} \\ \overrightarrow{x}.\overrightarrow{v}=\sum _{i=1}^m{a}_i\left\{0_i\right\} \\ \overrightarrow{x}.\overrightarrow{v}=0 \end{gathered}$$

By the definition of orthogonal, the point$\overrightarrow{x}\in {\mathrm{ℝ}}^n$is orthogonal to V.

Hence, the vector $\overrightarrow{x}$in ${\mathrm{ℝ}}^n$is orthogonal to Vif and only if $\overrightarrow{x}$is orthogonal to all vectors ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,...,{\overrightarrow{v}}_m$.