Answer to Question #105717 in Algebra for Sourav Mondal

Question

Let V be a vector space which is generated
by a finite set of vectors {v1, v2, ..., vn.}. Prove
that any linearly independent set of vectors
in V is finite and contains not more than n
elements.

Answer 1

$\begin{Bmatrix} v_1,v_2,...v_n \\ \end{Bmatrix}$ is spanning set of the vector space V. Let linearly independent vectors $u_1,u_2,...u_m$ in vector space V and there exist $c_{ij}\isin F$ such that,

$u_1=c_{11}v_1+c_{21}v_2+....+c_{n1}v_n\\ u_2=c_{12}v_1+c_{22}v_2+....+c_{n2}v_n\\ .\\ .\\ u_m=c_{1m}v_1+c_{2m}v_2+....+c_{nm}v_n\\$

Let $a_1,a_2,....a_m \isin F$ be such that,

$a_1u_1+a_2u_2+....a_mu_m=0$

Therefore, this produced a homogeneous system,

$\begin{bmatrix} c_{11} & c_{21}...& c_{n1}\\ c_{12} & c_{22}...&c_{n2}\\ .\\ c_{1m} & c_{2m}...&c_{nm} \end{bmatrix} \begin{bmatrix} a_1\\ a_2\\ .\\ a_m \end{bmatrix} =\begin{bmatrix} 0\\ 0\\ .\\ 0 \end{bmatrix}.$

If m>n ,

since this $Ca=0$ system is a under determined system,it has a non-trivial solution $\hat{a_1},\hat{a_2},...\hat{a_2}$ such that $a_1u_1+a_2u_2+....a_mu_m=0$ and $\hat{a_1},\hat{a_2},...\hat{a_2}$ are not all zeros. Therefore $u_1,u_2,...u_m$ are linearly dependent and initial linearly independent criterion is contradictory.

$\therefore$ any linear independent set of vector in V is finite and not more than n.