91Ó°ÊÓ

Consider that V be a finitely generated linear space.

Let the set S as follows.

$\mathrm{S}=({\mathrm{g}}_1,{\mathrm{g}}_2,...,{\mathrm{g}}_{\mathrm{m}})\mathrm{spansV}$

Now let T be a subset of S such that T spans V and T is such minimal subset of S.

Suppose that$T=\left\{v_1,v_2,...,v_n\right\}$

Let$\underset{\mathrm{i}=1}{\overset{\mathrm{n}}{∑}}{\mathrm{a}}_{\mathrm{a}}{\mathrm{v}}_{\mathrm{i}}=0$ where a are scalars.

Suppose that$a_i≠0$ for some i.

In particular take $a_i≠0$.

$$\begin{gathered} a_1{v}_1+a_2{v}_2+...+a_n{v}_n=0 \\ {a}_1^{-1}\left(a_1{v}_1+a_2{v}_2+...+a_n{v}_n\right)=0 \\ {v}_1=\left(-a_1^{-1}{a}_2\right)v_2+\left(-a_1^{-1}{a}_3\right)v_3+...+\left(-a_1^{-1}{a}_n\right)v_n \\ ={\mathrm{β}}_2{\mathrm{v}}_2+{\mathrm{β}}_3{\mathrm{v}}_3+...+{\mathrm{β}}_{\mathrm{n}}{\mathrm{v}}_{\mathrm{n}} \end{gathered}$$

Where${\mathrm{β}}_i=a_1^{-1}{a}_i$ are scalars.

If$v∈V$be any element, then since T spans V.

Therefore, the equation as follows.

$$\begin{gathered} v={\mathrm{γ}}_1{v}_1+{\mathrm{γ}}_2{v}_2+...{\mathrm{γ}}_n{v}_n \\ ={\mathrm{γ}}_1\left({\mathrm{β}}_2{v}_2+{\mathrm{β}}_3{v}_3+...+{\mathrm{β}}_n{v}_n\right)+{\mathrm{γ}}_2{v}_2+...+.{\mathrm{γ}}_n{v}_n \end{gathered}$$

This gives that an element of V is a linear combination of $v_2+...+v_n$.

This contradicts the choice of T as T is minimal.

Hence, $a_1=0$.

That is$a_1=0$ for all i.

Hence, T is linearly independent set that spans V.

Thus, T is a basis of V.

Hence, V is a finite dimensional and has dimension n.

Therefore, $dimV≤m$.

Thus, V is a finite dimensional.