Question

Vectors - Span `RR^2`?

Any help here is much appreciated

Answer 1

A requirement for any two vectors to span `RR^2` is that the vectors are linearly independent .

For convenience we normally use a natural basis for vectors based on a standard cartesian coordinate system. Thus we normally use standard vectors `bb(ul hat i)` and `bb(ul hat j)`, or in column format the basis:

`bb(B_1) = { bb(ul hat i), bb(ul hat j) } = { ((1),(0)), ((0),(1)) }`

Using this basis `bb(B_1)` we can can represent any vector by just using its standard cartesian coordinates in the column vector, so that the coordinate `(3,2)` is trivially represented as:

`((3),(2)) = 3bb(ul hat i) + 2bb(ul hat j) = 3((1),(0)) + 2((0),(1))`

However we can readily show that the vectors:

`bb(ul u) = ((1),(1)) \ \` and `bb(ul v)= ((-3),(2))`

are also linearly independent, and as such can also be used as a basis:

`bb(B_2) = { bb(ul u), bb(ul v)} = { ((1),(1)), ((-3),(2)) }`

And now to represent the coordinate `(3,2)`, we seek `lamda,mu in RR` st:

`((3),(2)) = lamda bb(ul u) + mu bb(ul v)`
`\ \ \ \ \ \ \ \ = lamda ((1),(1)) + mu ((-3),(2))`
`\ \ \ \ \ \ \ \ = ((1,-3),(1,2)) ((lamda),(mu))`

And solving this system, yields the solution:

`lamda= 12/5 \ \`, and `mu=-1/5`

Thus we can write the coordinate using this spanning basis `bb(B_2)` as:

`((3),(2)) = 12/5 ((1),(1)) -1/5 ((-3),(2))`