An arbitrary vector x in a space can be written as a linear combination of the base vectors...?

Question

If a vector space has a basis {e1, e2, ..., en} and x is some vector in that space, then for the equation:

x = x1e1 + x2e2 + ... + xnen
where x1, x2, ..., xn are the components of x

to hold, do the base vectors e1, e2, ..., en have to form a standard basis (i.e. the basis must be {(1,0,...,0), (0,1,...,0), ..., (0,0,...,1)})?

Thanks for the answers, things are much clearer now. I just thought the "components of x" were always written relative to the standard basis: x = (x1, x2, ..., xn). Apparently this is not the case

Answer 1

no
they just have to be n vectors linearly independent:
n=2
e1=(1,2) and e2= ( -2,1) are linearly independent, and they form a basis for R^2 (the plane).

Answer 2

Yes the standard basis will do the job and so will any other set of independent unit vectors. Just think of your standard set being rotated into another set. That can also be made into a linear combination describing the vector.

Answer 3

No. There are many bases other than the standard basis. Every vector can be written as a linear combination of the vectors in any basis. The components of a vector depend on the basis used. They are simply the coefficients of the basis vectors.

For example. One basis for R^3 is the standard one:
{e_1, e_2, e_3}={(1,0,0), (0,1,0), (0,0,1)}

Another basis for R^2 could be
{b_1, b_2, b_3}={(1,0,0), (1,1,0), (1,1,1)}.

Now, the vector (1,2,3) can be written as
1*e_1 +2*e_2 +3*e_3, so
1,2,and 3 are the components of that vector in the first basis.

But (1,2,3) can also be written as
(-1)*b_1 +(-1)*b_2 +3*b_3,
so -1,-1, 3 are the components of that same vector in the second basis.

The components depend on the basis used.