Question

Among all the vectors in ${\mathbf{R}}^{\mathbf{n}}$whose components add up to 1, find the vector of minimal length. In the case$\mathbf{n}\mathbf{=}\mathbf{2}$ , explain your solution geometrically.

Short answer

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

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$\left|\stackrel{\perp }{v}.\stackrel{\perp }{w}\right|\mathbf{\le }||\stackrel{\perp }{v}||\mathbf{.}||\stackrel{\perp }{w}||$.

Take,

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

By the above formula observe that $\stackrel{\perp }{v.}\stackrel{\perp }{w}=v_1+v_2+...+v_n=1$.

$$\begin{gathered} \Rightarrow 1\le ||\stackrel{r}{v}||\sqrt{n} \\ \Rightarrow ||\stackrel{r}{v}||\ge \frac{1}{\sqrt{n}} \end{gathered}$$

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

Thus,

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

$$\begin{gathered} v_1+v_2+...+v_n=1 \\ \Rightarrow n\lambda \Rightarrow \lambda =\frac{1}{n} \end{gathered}$$

consider the diagram

For $n=2$, $V_1+V_2=1$.

Now, draw the graph for equation$V_1+V_2=1$.

From the above figure it is clear that the vector of minimal length will be the vector which is perpendicular to the line$V_1+V_2=1$ .

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