Question

Let \(B = \left\{ {{v_1},...,{v_n}} \right\}\) be a linearly independent set in \({\mathbb{R}^{\bf{n}}}\). Explain why Bmust be a basis for \({\mathbb{R}^{\bf{n}}}\).

Short answer

Set B is a basis for \({\mathbb{R}^n}\).

Step-by-step solution

State the invertible matrix theorem

Recall the invertible matrix theorem, which states that if the square matrix is invertible, then the columns are linearly independent, and the columns form a basis for \({\mathbb{R}^n}\).

Apply the invertible matrix theorem and the basis theorem

It is given that\(\left\{ {{v_1},...,{v_n}} \right\}\)are a set of vectors. It forms a square matrix of the order\(n \times n\), as shown below:

\(A = \left[ {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}& \cdots &{{{\bf{v}}_n}}\end{array}} \right]\)

It is given that\(B = \left\{ {{{\bf{v}}_1},...,{{\bf{v}}_n}} \right\}\)is a linearly independent set in\({\mathbb{R}^n}\). By the invertible matrix theorem, the columns of B span\({\mathbb{R}^n}\).

According to the basis theorem, if the columns of B span\({\mathbb{R}^n}\)and the columns arelinearly independent, then B must be a basis for \({\mathbb{R}^n}\).