Question 40518
Hi,

For the vectors to form a basis of *[tex \mathbb{R}^n], there must be n of them and they must be linearly independent. Clearly there are n of them as the matrix is an nxn matrix. So the proof reduces to showing that the rows of an invertable matrix are linearly independent.

Lets assume that an invertable matrix exists which has linearly dependent rows. Because the rows are linearly dependent we can perform elementary row operations to make a row entirely of zeroes. (Note elementary row operations don't affect the determinant) Now evaluating the determinant along this row of zeroes gives zero.

An invertable matrix must have non-zero determinant, so by contradiction there can be no invertable matrix with linearly dependent rows. Hence every invertable matrix has linearly independent rows which is what we needed.

Hope that helps,
Kev