How to find a basis of a subspace V = {(x1; x2;….; xn) | Σni=1 xi=0}

[ahadmir]

Given V = {(x1; x2;….; xn) | Σni=1 xi=0}
(sum of vectors is equal to zero) be a subspace of Rn. How can we find a basis of V such that for each vector {(x1; x2;….; xn) in the basis Σi=1n x2i=1 ( i.e. sum of squares is equal to 1).

 

[HallsofIvy]

For problems like this start by looking at the easy cases. If n= 2, we are looking at vectors <x1, x2> such that x1+ x2= 0 and are looking for a basis. That is the single equation x2= -x1 so <x1, -x1>= x1<1, -1>. That is, a basis consists of simply the single vector <1, -1>. (Generally, one equation reduces the dimension of the space by 1. Since the over all space, R², has dimension 2, one equation reduces the dimension of the the subspace to 2- 1= 1.)
Since the length of a vector is "square root of sum of squares", the condition that the sum of the squares be 1 is exactly the same as requiring that the length of the vectors be 1. <1, -1> has length \sqrt{2} so dividing that vector by \sqrt{2} gives a unit vecotor: \{1/\sqrt{2}, -1/\sqrt{2}\} is a basis.

Now try n= 3. Any vector in R³ can be written in the form <x1, x2, x3> and we are requiring that x1+ x2+ x3= 0. Again this is a single equation so our subspace has dimension 3- 1= 2. We can solve for x3, say, in terms of the other two: x3= -x1- x2 so any vector is of the form <x1, x2, -x1-x2>= <x1, 0, -x1)+ <0 x2, -x2>= x1<1, 0, -1>+ x2<0, 1, -1>. A basis is {<1, 0, -1>, <0, 1, -1>}. Again, those both have length \sqrt{2} so you must divide by \sqrt{2}.

In R⁴, n= 4, we have x1+ x2+ x3+ x4= 0 or x4= -x1-x2-x3 so you should be able to see that a basis is {<1, 0, 0, -1>, <0, 1, 0, -1>, <0, 0, 1, -1>}. Again you will have to divide each by its length, \sqrt{2}.

Do you see the pattern now?