91Ó°ÊÓ

We can find such a vector of minimal length by using Cauchy- Schwarz inequality$\mathbf{|}\stackrel{\mathbf{âŠ\yen }}{\mathbf{v}}\mathbf{.}\stackrel{\mathbf{âŠ\yen }}{\mathbf{w}}\mathbf{|}\mathbf{≤}\mathbf{||}\stackrel{\mathbf{âŠ\yen }}{\mathbf{v}}\mathbf{||}\mathbf{.}\mathbf{||}\stackrel{\mathbf{âŠ\yen }}{\mathbf{w}}\mathbf{||}$

Take, $\stackrel{r}{w}=\left[\begin{array}{c}1\\ 1\\ .\\ .\\ 1\end{array}\right]$

By the above formula observe that$\stackrel{âŠ\yen }{v.}\stackrel{âŠ\yen }{w}=v_1+v_2+...+v_n$.

$$\begin{gathered} ⇒\left|{}_{j-1}{}^{n}V_j\right|≤\sqrt{n} \\ ⇒-\sqrt{n}≤{}_{j-1}{}^{n}V_j≤\sqrt{n} \\ ⇒{}_{j-1}{}^{n}V_j≤\sqrt{n} \end{gathered}$$

Also note that equality holds only and only if $\stackrel{âŠ\yen }{u}$and $\stackrel{âŠ\yen }{w}$are linearly dependent.Thus,

$\stackrel{r}{v}\left[\begin{array}{c}\mathrm{λ}\\ \mathrm{λ}\\ .\\ .\\ \mathrm{λ}\end{array}\right],$

$$\begin{gathered} {v^2}_1+{v^2}_2+...+{v^2}_n=1 \\ ⇒\mathrm{λ}=\frac{1}{\sqrt{n}} \end{gathered}$$

For $n=2$, $V_1+V_2=1I\sqrt{2}$.

Draw the graph for the equation$V_1+V_2=1I\sqrt{2}$ .

From the above figure it is clear that the vector of minimal length will be the vector which is perpendicular to the length to the tangent line$V_1+V_2=1I\sqrt{2}$. In the above figure the dotted area shows that all possible vectors but for them$V_1+V_2$is not minimal.

Hence, the vector with the minimal length will be $\stackrel{r}{v}\left[\begin{array}{c}1/\sqrt{n}\\ 1/\sqrt{n}\\ .\\ .\\ 1/\sqrt{n}\end{array}\right]$.