Discover Linear Algebra: A first course in linear algebra

One way to obtain a basis for a space (and hence to determine its dimension) is to assign parameters — then each independent parameter corresponds to a basis vector.

For example, in \(\R^2\) we have natural parameters associated to the \(x\)- and \(y\)-coordinates: \(\uvec{x} = (x,y)\text{.}\) Expanding this into a linear combination, we get \(\uvec{x} = x(1,0) + y(0,1)\text{.}\) Parameter \(x\) corresponds to vector \((1,0)\) and parameter \(y\) corresponds to vector \((0,1)\text{,}\) and together the two corresponding vectors form a basis \(\{(1,0),(0,1)\}\) for \(\R^2\text{.}\) (In fact, the standard basis for \(\R^2\text{!}\)). Since there were two independent parameters required to described an arbitrary vector in the space, this led to two basis vectors, and so the dimension of \(\R^2\) is (surprise!) \(2\text{.}\)

A vector space is called finite-dimensional if it can be spanned by a finite set; otherwise, it is called infinite-dimensional. For example, \(\R^n\) is finite-dimensional for each value of \(n\text{,}\) because it can be spanned by the finite set of standard basis vectors \(\{\uvec{e}_1,\uvec{e}_2,\dotsc,\uvec{e}_n\}\text{.}\)

We’ve already seen that a linearly dependent spanning set can be reduced to a basis (Proposition 18.5.1). Working the other way, we will use Proposition 17.5.6 to argue in Subsection 19.5.2 that a linearly independent set that is not a spanning set can be built up to a basis by including additional vectors (Proposition 19.5.4). Proposition 17.5.6 tells us exactly how to do this: to ensure linear independence at each step, the new vector to be included should not be in the span of the old (i.e. the new should not be any linear combination of the old).