### Vector spaces: Spans

### Finding bases

We return to the problem of finding efficient spanning sets of vectors for a given vector space. This now appears to come down to finding a *basis* for the vector space. The following results can be put to good use for finite-dimensional subspaces.

Growth criterion for independence

Let #n# be a natural number and #V# a vector space. If the set of vectors #\vec{u}_1, \ldots ,\vec{u}_n# of #V# satisfies

\[

\vec{u}_1 \neq \vec{0} ,\, \vec{u}_2 \not\in \linspan{\vec{u}_1 } ,\,

\vec{u}_3\not\in \linspan{ \vec{u}_1 , \vec{u}_2 } ,\, \ldots , \vec{u}_n \not\in \linspan{\vec{u}_1 ,\ldots , \vec{u}_{n-1}}

\]

then it is independent.

By choosing a first vector arbitrarily, and then again and again selecting a vector outside the span of all selected vectors, we find a basis. If a spanning set of vectors for the vector space is given, another method is more suitable.

Two ways to find a basis

Below are two ways to find a basis for a given vector space #V#.

- If #V# is described by means of a spanning set, we find a basis by
*thinning*this set. - If we do not have a spanning set for #V#, we must produce vectors by ourselves:
- Start with a vector #\vec{u}_1 \neq\vec{0}# (if possible, otherwise we are ready).
- Then choose (if possible) a vector #\vec{u}_2 \not\in \linspan{\vec{u}_1}#. The vectors #\vec{u}_1, \vec{u}_2# are independent due to the theorem above.
- Then choose (if possible) a vector #\vec{u}_3 \not\in \linspan{\vec{u}_1, \vec{u}_2}#. The vectors #\vec{u}_1, \vec{u}_2, \vec{u}_3# are independent due to the theorem above.
- Continue this way.

The rule gives two general possibilities; sometimes ad hoc methods are faster.

Ad 1. If we have a finite set of vectors in #\mathbb{R}^n# then the row reduction techniques of section *Systems of linear equations and matrices* come in handy. This more systematic and faster method for finding a basis from a given spanning set will be discussed *later*. As we will also see *later*, the same techniques can be applied to a finite set of vectors in vector spaces other than #\mathbb{R}^n# by use of *coordinates*.

Ad 2. If the dimension of #V# is finite, then the process stops at #n=\dim{V}# because no vectors outside #\linspan{\vec{u}_1,\ldots,\vec{u}_n}# can be found. If there is no end to the process, then the dimension of #V# is infinite. A good example is the choice #\vec{u}_i=x^{i-1}# for all natural numbers #i# in the vector space of all polynomials in #x#.

To this end we choose a second vector outside of #\linspan{\rv{-5,7,0}}#, that is, it is independent of #\rv{-5,7,0}#. The standard basis vector #\vec{u}_2 =\rv{1,0,0}# will do. Because #\dim{\mathbb{R}^3}=3# we are still looking for a third vector outside of #\linspan{\rv{-5,7,0},\rv{1,0,0}}#. The standard basis vector #\vec{u}_3 = \rv{0,0,1}# will do.

**Pass Your Math**independent of your university. See pricing and more.

Or visit omptest.org if jou are taking an OMPT exam.