Question

Question: Assume that$\left[V:F\right]=n$and that the following conditions are equivalent :

(i)$\left\{v_1,.....v_n\right\}$spans Vover F.

(ii)$\mathbf{\{}{\mathbf{v}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{v}}_{\mathbf{n}}\mathbf{\}}$is linearly independent V over F.

(iii)$\mathbf{\{}{\mathbf{v}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{v}}_{\mathbf{n}}\mathbf{\}}$is a basis of Vover F

Short answer

Answer

(i)$\left\{v_1,.....v_n\right\}$spans Vover F.

(ii)$\left\{v_1,.....v_n\right\}$is linearly independent V over F.

(iii)$\left\{v_1,.....v_n\right\}$is a basis of Vover F

Step-by-step solution

 (i)⇒(ii)

Given$\dim \left(v\left(F\right)\right)=n=\left[v:F\right)$

Let $\left\{v_1,v_2,.........v_n\right\}$span over .

Suppose$\left\{v_1,v_2,.........v_n\right\}$ is not linearly independent, and we know that minimum spaning and maximum linearly independent set from a basis.

.$\therefore \dim \left(v\left(F\right)\right)<n$

Which is a contradiction.

Hence$\left\{v_1,v_2,.........v_n\right\}$ is linearly independent.

Hence (i) to (ii) is proved.

(ii)⇒(iii)

Let $\left\{v_1,v_2,.........v_n\right\}$is linearly independent.

Suppose$\left\{v_1,v_2,.........v_n\right\}$ does not span V.

$\dim \left(v\left(F\right)\right)=\left[v:F\right)>n$

This is again a contradiction.

Hence $\left\{v_1,v_2,.........v_n\right\}$span V

it Is proved.

Step 3:(iii)⇒(i)

Let $\left\{v_1,v_2,.........v_n\right\}$is a basis for.

$\left\{v_1,v_2,.........v_n\right\}$span and linearly independent also.

$\left(iii\right)\Rightarrow \left(i\right)$

Hence All are equivalent.