Question

Let be a finite set in a vector spaceVwith the proper\(S\)ty that every x in \(V\) has a unique representation as a linear combination of elements of \(S\). Show that \(S\) is a basis of \(V\).

Short answer

It is proved that \(S\) is a basis of \(V\).

Step-by-step solution

State the condition for a basis

Let \(H\) be a subspace of vector space \(V\). An indexed set of vectors \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _p}} \right\}\) in \(V\) is a basis for \(H\) if

a. \(B\) is a linearly independent set, and

b. the subspace spanned by \(B\) coincides with \(H\), that is, \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _p}} \right\}\).

Show that \(S\) is a basis of \(V\)

Set \(S\) spans \(V\) since every \({\mathop{\rm x}\nolimits} \) in \(V\) can be represented as a linear combination of elements in \(S\). To demonstrate linear independence, assume that \(S = \left\{ {{{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _n}} \right\}\) and \({c_1}{{\mathop{\rm v}\nolimits} _1} + \cdots + {c_n}{{\mathop{\rm v}\nolimits} _n} = 0\) for any scalar \({c_1},...,{c_n}\) in which \({c_1} = .... = {c_n} = 0\) is one possibility. According to the hypothesis, this is the unique possible way to represent the zero vector as a linear combination of the elements in \(S\). Here, \(S\) is linearly independent and thus is a basis for \(V\).