# 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?

## $W = \left\{\left\langle4 s - t , s , t , s\right\rangle | s , t \in \mathbb{R}\right\}$ For instance, apparently, $\left\{\left\langle0 , 1 , 4 , 1\right\rangle , \left\langle1 , 1 , 3 , 1\right\rangle\right\}$ is a valid set, and it happens to be of dimension $2$ in ${\mathbb{R}}^{4}$. Does a basis for ${\mathbb{R}}^{n}$ have to have $n$ vectors?

Mar 16, 2016

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.

Mar 16, 2016

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 ${\mathbb{R}}^{n}$ we know that $\dim \left({\mathbb{R}}^{n}\right) = n$ as
$\left\{\left(1 , 0 , 0 , \ldots 0\right) , \left(0 , 1 , 0 , \ldots , 0\right) , \ldots , \left(0 , 0 , \ldots , 0 , 1\right)\right\}$
is a basis for ${\mathbb{R}}^{n}$ and has $n$ elements.

In the case of $W = \left\{\left(4 s - t , s , t , s\right) | s , t \in \mathbb{R}\right\}$ we can write any element in $W$ as $s \vec{u} + t \vec{v}$ where $\vec{u} = \left(4 , 1 , 0 , 1\right)$ and $\vec{v} = \left(- 1 , 0 , 1 , 0\right)$.

From this, we have that $\left\{\vec{u} , \vec{v}\right\}$ 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 $\left\{\vec{u} , \vec{v}\right\}$ is a linearly independent spanning set for $W$, that is, a basis. Because $W$ has a basis with $2$ elements, we say that $\dim \left(W\right) = 2$.

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 $\dim \left(W\right) \le \dim \left(V\right)$ and $\dim \left(W\right) = \dim \left(V\right) \iff W = V$