Question

Consider some perpendicular unit vectors ${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{..}{\overrightarrow{\mathbf{v}}}_{\mathbf{m}}$in${\mathit{ℝ}}^{{}^{\mathbf{n}}}$. Show that these vectors are necessarily linearly independent. Hint: Form the dot product of${\overrightarrow{\mathbf{v}}}_{\mathbf{i}}$and both sides of the equation

${\mathit{c}}_{\mathbf{1}}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}{\mathit{c}}_{\mathbf{i}}{\overrightarrow{\mathbf{v}}}_{\mathbf{i}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}{\mathit{c}}_{\mathbf{m}}{\overrightarrow{\mathbf{v}}}_{\mathbf{m}}\mathbf{=}\overrightarrow{\mathbf{0}}\mathbf{.}$

Short answer

It has been proved that the vectors are linearly independent.

Step-by-step solution

Given that

Given are some unit vectors${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,\mathrm{...}{\overrightarrow{v}}_m$ in${ℝ}^{{}^n}$ .

Find dot product as per hint

As per the hint, Form the dot product of ${\overrightarrow{v}}_i$and both sides of the equation

$c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2+\mathrm{...}+c_i{\overrightarrow{v}}_i+\mathrm{...}+c_m{\overrightarrow{v}}_m=\overrightarrow{0}$

So

$\begin{array}{l}{\overrightarrow{v}}_i(c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2+\mathrm{...}+c_i{\overrightarrow{v}}_i+\mathrm{...}+c_m{\overrightarrow{v}}_m)={\overrightarrow{v}}_{i}0\\ c_1{\overrightarrow{v}}_1\cdot {\overrightarrow{v}}_i+c_2{\overrightarrow{v}}_2\cdot {\overrightarrow{v}}_i+\mathrm{...}+c_i{\overrightarrow{v}}_i\cdot {\overrightarrow{v}}_i+\mathrm{...}+c_m{\overrightarrow{v}}_m\cdot {\overrightarrow{v}}_i=0\end{array}$

Prove that the vectors are linearly independent

It is given that the vectors are perpendicular so all the products will be zero except ${\overrightarrow{v}}_i\cdot {\overrightarrow{v}}_i$which is equal to 1 as these are unit vectors.

So,

$\begin{array}{l}{c}_{1}0+\mathrm{...}+c_{i}1+\mathrm{...}+c_{m}0=0\\ c_i=0\end{array}$

This process can be generalized and the only solution is

$c_1=c_2=\mathrm{...}{c}_i=\mathrm{...}{c}_m=0$

Since,$c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2+\mathrm{...}+c_i{\overrightarrow{v}}_i+\mathrm{...}+c_m{\overrightarrow{v}}_m=\overrightarrow{0}$ has a unique solution ,$c_1=c_2=\mathrm{...}{c}_i=\mathrm{...}{c}_m=0$ therefore,

The vectors are linearly independent.