Verified answer

Yes they can, the easiest way to test this is as follows:

Make the 3 x 3 matirix with entries u, v and w as its rows:

1st row: 1 0 2

2nd row: 0 3 -1

3rd row: 1 1 2

Work out the determinant of this matrix (you should get 1)

The determinant is the volume of the (3d) parallelepiped with u, v and w as its edges. If the volume is zero, the three vectors are co-planar and so cannot possibly represent all the vectors in ℝ³ - if, as in your example, the volume is non-zero, then the three vectors must have components in all three dimensions, so they can be used as a basis for all of 3d real space.

Let's see if linear combinations of these vectors can create three obviously independent vectors.

7u + 2v - 6w = 1 0 0 = i

w - u = 0 1 0 = j

- 3u - v + 3w = 0 0 1 = k

The vectors i, j, and k are linear combinations of u, v, w and obviously independent and span R³. Therefore u, v, and w also span R³ and can be considered a base.

Let's define new vectors:

x = w - u = (0, 1, 0)

y = 3x - v = (0 ,3 , 0) - (0, 3, -1) = (0, 0, 1)

z = u - 2y = (1, 0,2 ) - (0, 0, 2) = (1, 0, 0)

Hence, u,v,w are three linearly independant vectors in R3, or in other words they are a base.