Question

Suppose there was a basis for and a certain number of dimensions for subspace `W` in `RR^4`. Why is the number of dimensions `2`?

`W = {<< 4s - t, s, t, s >> | s,t in RR}`

For instance, apparently,

`{<< 0,1,4,1 >>,<< 1,1,3,1 >>}`

is a valid set, and it happens to be of dimension `2` in `RR^4`. Does a basis for `RR^n` have to have `n` vectors?

Answer 1

4 dimensions minus 2 constraints = 2 dimensions

Explanation:

The 3rd and the 4th coordinates are the only independent ones. The first two can be expressed in terms of the last two.

Answer 2

The dimension of a subspace is decided by its bases, and not by the dimension of any vector space it is a subspace of.

Explanation:

The dimension of a vector space is defined by the number of vectors in a basis of that space (for infinite dimensional spaces, it is defined by the cardinality of a basis). Note that this definition is consistent as we can prove that any basis of a vector space will have the same number of vectors as any other basis.

In the case of `RR^n` we know that `dim(RR^n) = n` as
`{(1,0,0,...0),(0,1,0,...,0),...,(0,0,...,0,1)}`
is a basis for `RR^n` and has `n` elements.

In the case of `W = {(4s-t, s, t, s)|s, t in RR}` we can write any element in `W` as `svec(u) + tvec(v)` where `vec(u) = (4,1,0,1)` and `vec(v) = (-1,0,1,0)`.

From this, we have that `{vec(u), vec(v)}` is a spanning set for `W`. Because `vec(u)` and `vec(v)` are clearly not scalar multiples of each other (note the positions of the `0`s), that means that `{vec(u), vec(v)}` is a linearly independent spanning set for `W`, that is, a basis. Because `W` has a basis with `2` elements, we say that `dim(W) = 2`.

Note that the dimension of a vector space is not dependent on the whether its vectors may exist in other vector spaces of larger dimension. The only relation is that if `W` is a subspace of `V` then `dim(W) <= dim(V)` and `dim(W) = dim(V) <=> W = V`