Solution of Supplementary Problem For Vector Spaces

 Show that the vectors (1, 1) and (-1, 2) form a basis of R ².

Solution: 

This requires showing; a) the vectors are linearly independent, and b) they generate R².

As before (earlier problems), we set out the condition via expression for linear independence:

a(1, 1) + b(-1, 2) = (0, 0)

-> a - b = 0 and a + 2b = 0

a(1, 1) + b(-1, 2) = (0, 0)

-> a - b = 0 and a + 2b = 0

solve simultaneously by subtracting the 2nd from the 1st:

a - b = 0
a + 2b = 0
----------
0 - 3b = 0 so that b = 0 and a = 0

Thus the vectors are linearly independent.

(b) To show generation of R², let (a,b) be an arbitrary element of R² and write out:

x (1, 1) + y(-1, 2) = (a, b)

which leads to the pair of simultaneous equations:

x - y = a and x + 2y = b

As before, subtracting the 2nd from the 1st eqn.

x - y = a
x + 2y = b
----------
0 - 3y = a - b or y = (b - a)/ 3

Therefore, (x,y) are the coordinates of (a,b) with respect to the basis {(1,1), (-1,2)}.