How to find a vector basis for this impossible form?

[AngryHippo]

Find a basis for the subspace S of R^3 consisting of all vectors of the form
(a, 2a-b, b)^T, where a and b are real numbers

Relevant equations would really just be the determinant of the system.

I have tried so many 3x3 matrix combinations of the given form but no matter what the determinant always equals zero preventing me from saying it is a basis. I am guessing it is impossible to do but how would I prove that it is impossible?

 

[losiu99]

Hint: what is the dimension of that subspace? Can this explain why you fail to find 3 linearly independent vectors of that form?

 

[AngryHippo]

Well if the question says that S is in R^3 wouldn't S have to have the dimension 3?

 

[Mark44]

Well if the question says that S is in R^3 wouldn't S have to have the dimension 3?

No. There are many subspaces of R³. One of them consists only of the zero vector, so has dimension zero. There are an infinite number of lines through the origin, each of which is a subspace of R³, and each of which has dimension one.

There are an infinite number of planes that include the origin, each of which is a subspace of R³, and each of which has dimension two.

A subspace of a space with dimension n can be of any dimension from 0 up to and including n.

 

[AngryHippo]

But it has to be a basis so it has to span and be linear independent. Doesn't have to be in dimension 3 to be able to span R^3?

 

[Mark44]

A basis for a subspace has to span the subspace and must be linearly independent, yes. Each vector in the basis has three coordinates, as do all vectors in R³, but the dimension of a basis is how many vectors are in the basis, not how many coordinates there are in each vector.

 

[Mark44]

For example, the set S = {(x, y, z) | x = y, z = 0} is a line in R³ and happens to be a subspace of R³. There are an infinite number of points on this line, a couple of which are (2, 2, 0) and (-1, -1, 0). You can also think of these as vectors, <2, 2, 0> and <-1, -1, 0>.

A basis for this subspace is {<1, 1, 0>}. Since there is only one vector in the basis, the dimension of the subspace is one.