## Playing with bases

Here is one last push before the main result. Exercise 1 is one of the basic facts about bases.

**Exercise 1**. In any system of linearly independent vectors can be completed to form a basis.

**Proof**. Let be a basis in If each of belongs to then by the lemma we would have contradicting the assumption Hence, among there is at least on vector that does not belong to We can add it to denoting it

Suppose Since is independent of the other vectors, we have but then because of independence of all other coefficients are zero. Thus, are linearly independent.

If we can repeat the addition process, until we obtain linearly independent vectors By construction, belong to Since span do too and therefore form a basis.

**Definition 1**. Let be two subspaces. Vectors are called **linearly independent relative to** if any nontrivial linear combination does not belong to For the purposes of this definition, it is convenient to denote by a generic element of plays the role of zero and the definition looks similar to usual linear independence: for any nonzero vector Rejecting this definition, we can say that are called **linearly dependent relative to** if for some nonzero vector

**Definition 2**. Let be two subspaces. Vectors are called a **basis relative to** if they are linearly independent and can be completed by some basis from to form a basis in

**Exercise 2**. Show existence of a relative basis in

**Proof**. Take any basis in (say, ) and, using Exercise 1, complete it by some vectors (say, ) to get a basis in Then, obviously, form a basis in relative to Besides, none of belongs to

**Exercise 3**. Any system of vectors linearly independent relative to can be completed to form a relative basis in

**Proof**. Take a basis in (say, ) and add it to The resulting system is linearly independent. Indeed, if then

By assumption of relative linear independence but then the remaining coefficients are also zero.

By Exercise 1 we can find such that is a basis in Now the system is a relative basis because these vectors are linearly independent and together with form a basis in