Question

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

Let$S=\left\{{\mathbf{v}}_{\mathbf{1}},....,{\mathbf{v}}_k\right\}$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 ${\mathbb{R}}^n$.