about a basis of a vector space

Question

If I want to prove that ${v_1=(0,1,1,), v_2=(1,0,1),v_3=(1,1,0)}$ is a basis for $R^3$, is it sufficient to show that the determinant of these vectors is not equal to zero? can we verify the linearly independence and the generating these vectors to $R^3$ through the value of the determinant? thanks

Answer 1

That's exactly right!

The determinant of a matrix gives the volume spanned by its rows, and also the volume spanned by its columns.

In three dimensions, three vectors can span a parallelepiped. If that volume is non-zero, i.e. the determinant is non-zero, then the row/columns are linearly independent. If the row/columns are linearly dependent then they collapse to span a parallelogram, or a line segment, which all have zero volume.

Answer 2

A standard Linear Algebra theorem asserts that a set of $n$ vectors of $\mathbb{R}^n$ is a basis if and only if the determinant of the matrix whse columns are those vectors is distinct from $0$.