Question

Assume that the subset $\left\{v_1,v_2,........v_n\right\}$of Vis linearly independent over F and that $\mathbf{w}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{v}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{c}}_{\mathbf{n}}{\mathbf{v}}_{\mathbf{n}}$with $\mathbf{ci}\mathbf{\in }\mathbf{F}$. Prove that the set $\left\{w-v_1,w-v_2,....w-v_n\right\}$ is linearly independent over F if and only if ${\mathbf{c}}_{\mathbf{1}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{c}\mathbf{\ne }{\mathbf{1}}_{\mathbf{F}}$.

Short answer

$\left\{\mathrm{w}-{\mathrm{v}}_1,\mathrm{w}-{\mathrm{v}}_2,.....\mathrm{w}-{\mathrm{v}}_{\mathrm{n}}\right\}$is linearly independent over F.

Step-by-step solution

Definition of vector space.

In a vector space a set of vector is said to be linearly independent If one of the vector in the set can be written as a linear combination of other, and if no vector can be written as a linear combination of others then the vector is said to be linearly independent.

Step 2:

Let the subset $\left\{{\mathrm{v}}_1,{\mathrm{v}}_2,......{\mathrm{v}}_{\mathrm{n}}\right\}$of V is linearly independent over F and $\mathrm{w}={\mathrm{c}}_1{\mathrm{v}}_1+{\mathrm{c}}_2{\mathrm{v}}_2+....+{\mathrm{c}}_{\mathrm{n}}{\mathrm{v}}_{\mathrm{n}}$with ${\mathrm{c}}_{\mathrm{i}}\in \mathrm{F}$.

Let $\left\{\mathrm{w}-{\mathrm{v}}_1,\mathrm{w}-{\mathrm{v}}_2,....\mathrm{w}-{\mathrm{v}}_{\mathrm{n}}\right\}$ is linearly independent over F. This implies

${\mathrm{\alpha }}_1\left(\mathrm{w}-{\mathrm{v}}_1\right)+{\mathrm{\alpha }}_2\left(\mathrm{w}-{\mathrm{v}}_2\right)+..............+{\mathrm{\alpha }}_{\mathrm{n}}\left(\mathrm{w}-{\mathrm{v}}_{\mathrm{n}}\right)=0$

Simplify further,

localid="1659075987458" $\left({\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2+...+{\mathrm{\alpha }}_{\mathrm{n}}\right)\mathrm{w}-\left({\mathrm{\alpha }}_1{\mathrm{v}}_1+{\mathrm{\alpha }}_2{\mathrm{v}}_2+...+{\mathrm{\alpha }}_{\mathrm{n}}{\mathrm{v}}_{\mathrm{n}}\right)=0$

Put the values of w in the above equation

$\left({\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2+...+{\mathrm{\alpha }}_{\mathrm{n}}\right)\left({\mathrm{c}}_1{\mathrm{v}}_1+{\mathrm{c}}_2{\mathrm{v}}_2+....+{\mathrm{c}}_{\mathrm{n}}{\mathrm{v}}_{\mathrm{n}}\right)-\left({\mathrm{\alpha }}_1{\mathrm{v}}_1+{\mathrm{\alpha }}_2{\mathrm{v}}_2+...+{\mathrm{\alpha }}_{\mathrm{n}}{\mathrm{v}}_{\mathrm{n}}\right)=0$

This gives

localid="1659076239143" $\left(\left({\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2+...+{\mathrm{\alpha }}_{\mathrm{n}}\right){\mathrm{c}}_1-{\mathrm{\alpha }}_1){\mathrm{v}}_1+(\left({\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2+...+{\mathrm{\alpha }}_{\mathrm{n}}\right){\mathrm{c}}_2-{\mathrm{\alpha }}_2){\mathrm{v}}_2+....+(\left({\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2+...+{\mathrm{\alpha }}_{\mathrm{n}}\right){\mathrm{c}}_{\mathrm{n}}-{\mathrm{\alpha }}_{\mathrm{n}}\right)v_n=0$

Since $\left\{{\mathrm{v}}_1,{\mathrm{v}}_2,...,{\mathrm{v}}_{\mathrm{n}}\right\}$of V is linearly independent over F.

Therefore $\left(\left({\alpha }_1+{\alpha }_2+....{\alpha }_n\right)c_i-{\alpha }_i\right)=0$for all $\mathrm{i}=1,2,3....\mathrm{n}$

This implies $\left({\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2+....{\mathrm{\alpha }}_{\mathrm{n}}\right){\mathrm{c}}_{\mathrm{i}}={\mathrm{\alpha }}_{\mathrm{i}}$for all $i=1,2,3....n$$i=1,2,3....n$

$$\begin{gathered} \sum _{\mathrm{i}=1}^{\mathrm{n}}\left(\left({\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2+...{\mathrm{\alpha }}_{\mathrm{n}}\right){\mathrm{c}}_{\mathrm{i}}\right)=\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{\alpha }}_{\mathrm{i}} \\ \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{\alpha }}_{\mathrm{i}}\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{i}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{\alpha }}_{\mathrm{i}} \end{gathered}$$

Simplify further

$$\begin{gathered} \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{\alpha }}_{\mathrm{i}}\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{i}}-\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{\alpha }}_{\mathrm{i}}=0 \\ \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{\alpha }}_{\mathrm{i}}\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{i}}-1_{\mathrm{F}}=0 \end{gathered}$$

This gives

$\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{\alpha }}_{\mathrm{i}}=0\mathrm{or}\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{i}}=1_{\mathrm{F}}$

Now the set $\left\{\mathrm{w}-{\mathrm{v}}_1,\mathrm{w}-{\mathrm{v}}_2,.....\mathrm{w}-{\mathrm{v}}_{\mathrm{n}}\right\}$ is linearly independent if and only if all the ${\alpha }_i=0$. So,

$\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{i}}\ne 1_{\mathrm{F}}$

Or

${\mathrm{c}}_1+{\mathrm{c}}_2+{\mathrm{c}}_3....{\mathrm{c}}_{\mathrm{n}}\ne 1_{\mathrm{F}}$

Hence proved.