Question

Exercises 29 and 30 show that every basis for must contain exactly n vectors.\({\mathbb{R}^n}\)

Let\(S = \left\{ {{{\bf{v}}_{\bf{1}}},....,{{\bf{v}}_k}} \right\}\)be a set of k vectors in\({\mathbb{R}^n}\), with\(k > n\). Use a theorem from chapter 1 to explain why S cannot be a basis for\({\mathbb{R}^n}\).

Short answer

Does not span \({\mathbb{R}^n}\)

Step-by-step solution

Set up a matrix with the vectors in S

A matrix of the order\(n \times k\) has the column vector \(\left\{ {{v_1},....,{v_k}} \right\}\). In the matrix, there are fewer rows than columns. Therefore, there cannot be a pivot element in each row.

Check for the span of S

As the matrix (order\(n \times k\)) cannot be a pivot in each row, the vectors in S do not span the vector space \({\mathbb{R}^n}\).

So, the given set does not span \({\mathbb{R}^n}\).