Question 1.12.3: In R3, consider the set B = {e1, e2, e3}, where e1 = [1 0 0]...

First, we observe that while the formal definition of a basis refers to the basis of a subspace H of a vector space V, since every vector space is a subspace of itself, it follows that we can also discuss a basis for a vector space.

Considering the set B = {e_{1}, e_{2}, e_{3}}, we observe that the vectors in this set are the columns of the 3×3 identity matrix. By the Invertible Matrix Theorem,it follows that the set B is linearly independent because I_{3} has a pivot in every column. Likewise, the set B spans R_{3} since I_{3}has a pivot in every row. As a linearly independent spanning set in R_{3}, B is indeed a basis.

For the set S whose elements are the columns of the matrix

A=\begin{bmatrix} 1 & -1 &0 \\ 2 & 1 &3 \\ -1 &3 &1 \end{bmatrix}

we again use the Invertible Matrix Theorem to determine whether or not S is a basis for R_{3} . Row-reducing A, it is straightforward to see that A is row equivalent to the identity matrix, and therefore is invertible. In particular, A has a pivot in every column and every row, and thus the columns of A are linearly independent and span R_{3}. It follows that S is also a basis for R_{3}.