Question

Assume that V is finite dimensional over F and S is a linearly independent subset of V. Prove that S is contained in a basis of V.

Short answer

S contain a basis of V.

Step-by-step solution

Defintion 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:Showing that S contain a basis of V.

Let V be a finite dimensional vector space over F and is a linearly independent subset of V.

Let $S=\left\{u_{1,}{u}_2,.....u_{\mathrm{m}}\right\}⊑V$be the set linearly independent vectors in V. Now if $V=\mathrm{span}\left\{u_{1,}{u}_2,.....u_{\mathrm{m}}\right\}$ . Then S is contain in a basis of V.

So let$V\ne \mathrm{span}\left\{{\mathrm{u}}_{1,}{\mathrm{u}}_2,.....{\mathrm{u}}_{\mathrm{m}}\right\}$

Since a linearly independent set of vector can be extended. So extend the list of vectors by a vector ${\mathrm{u}}_{\mathrm{m}+1}\in \mathrm{V}-\mathrm{span}\left\{{\mathrm{u}}_{1,}{\mathrm{u}}_2,.....{\mathrm{u}}_{\mathrm{m}}\right\}$

Again if

$\mathrm{V}-\mathrm{span}\left\{{\mathrm{u}}_{1,}{\mathrm{u}}_2,.....{\mathrm{u}}_{\mathrm{m}}\right\}$

Then S is contained in a basis of V. Proceeding the same way from a set of vector $\left\{{\mathrm{u}}_{1,}{\mathrm{u}}_2,.....{\mathrm{u}}_{\mathrm{m}}\right\}$with ${\mathrm{u}}_{\mathrm{j}}\notin \mathrm{span}\left\{{\mathrm{u}}_{1,}{\mathrm{u}}_2,.....{\mathrm{u}}_{\mathrm{m}}\right\}\mathrm{for}2\le \mathrm{j}\le \mathrm{n}$for .

The set of vector is also linearly independent. This process must stop after finite spanning set of V provides an upper bound to the length of the independent set of vectors of V.

Therefore S is contain in a basis of V.