Question

Among all the unit vectors in${\mathbf{R}}^{\mathbf{n}}$, find the one for which the sum of the components is maximal. In the case $\mathbf{n}\mathbf{=}\mathbf{2}$, explain your answer geometrically, in terms of the unit circle and the level curves of the function${\mathit{X}}_{\mathbf{1}}\mathbf{+}{\mathit{X}}_{\mathbf{2}}$.

Short answer

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

Geometrical representation is,

Step-by-step solution

Use Cauchy-Schwarz inequality

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

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

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

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

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

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

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

Geometrical representation

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 $\stackrel{r}{v}\left[\begin{array}{c}1/\sqrt{n}\\ 1/\sqrt{n}\\ .\\ .\\ 1/\sqrt{n}\end{array}\right]$.