# 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#?

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#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?

For instance, apparently,

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

is a valid set, and it happens to be of dimension

##### 2 Answers

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.

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

is a basis for

In the case of

From this, we have that

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