Solving a Basis Problem in R^2

[Chadlee88]

Hey, I've done this problem but i don't know what it all means? What does it mean to form a basis?

Problem: show that the vectors (1,1) , (1, -1) form a basis for R^2

(x,y) = a1(1, 1) + b1(1, -1)
x = a1 + b1 => a1 = x - b1
y= a1- b1 => b1 = a1-y

a1 = x - a1 + y
2a1 = x + y
a1 = 1/2(x + y) <----- What does this represent?

b1 = (x-b1)-y
2b1 = x - y
b1 = 1/2(x-y) <----- What does this represent?

 

[neurocomp2003]

Do you know the definition of basis/bases and span? Do you know what scalar/magnitude are?

 

[quasar987]

Ok. A "basis for R²" is a set B of vectors of R² that...

(1) ...are linearly independent and...

(2) ...such that any vector (x,y) of R² can be expressed as a linear combination of the vectors of the set B. (A set of vector that have property (2) is said to span R².)

It can be shown that any set of n vectors which span R^n are automatically linearly independent. So you only have to verify property (2) since it implies (1).

What you've done (without realizing it, perhaps), is you've said, if {(1,1), (1,-1)} spans R², then it must be that for any vector (x,y) of R², we can write it as a linear combination of (1,1) and (1,-1). In other words, there exists two constants a1 and b1 such that (x,y)=a1(1,1)+b1(1,-1). You've then manipulated this expression to express a1 and b1 in terms of x and y. By doing this, you proved that indeed, given a vector (x,y), there are constants a1 and b1 such that (x,y)=a1(1,1)+b1(1,-1) and they are given by a1 = (x + y)/2, b1 = (x-y)/2.

It takes practice and perseverance to integrate all the subtleties of mathematical proofs, but I hope my explanation was helpful.

 

[nocturnal]

It can be shown that any set of n vectors which span R^n are automatically linearly independent. So you only have to verify property (2) since it implies (1).

This is actually an if and only if as the converse holds as well. ie: any set of n vectors in R^n which is linearly independent spans R^n.

So as an alternative to what quasar suggested, you could just show that (1, 1), (1, -1) are linearly independent which amounts to showing that one is not a scalar multiple of the other since there are only two vectors.

 

[HallsofIvy]

Another way of looking at "basis" is that any vector can be written as a linear combination of the basis vectors in a unique way.

The fact that any vector can be written as a linear combination of the basis vectors is due to their spanning[/b] the space. The fact that that linear combination is unique is due to their being independent.

a1 = 1/2(x + y) <----- What does this represent?
b1 = 1/2(x-y) <----- What does this represent?

You started by asserting that an arbitrary vector, (x, y), could be written as a linear combination of the vectors (1, 1) and (1, -1):
(x,y)= a1(1, 1)+ b1(1,-1) for some real numbers a1 and b1. You have succeeded in showing how to calculate those numbers thus showing
(1) that the numbers exist, that (x,y) can be written as such a linear combination.
(2) that the numbers are unique, since the formulas you show are functions and have only one result, thus showing that the linear combination is unique.