Question

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

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 section 1.4 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 columns than rows. 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, it does not span thevector space \({\mathbb{R}^n}\).

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