Are all subspaces of equal dimension (of a vector space) the same? I haven't quite gotten my head around dimension, bases, and subspaces. It seems intuitively true, but are all subspaces of equal dimension of the same vector space the same?
If so, does it follow from the definitions of dimension, subspace, and vector space, or does it need to be proven?
Thanks
 A: No. Consider the two subspaces of $\Bbb  R^2$ generated respectively by $(1,0)$ and $(0,1)$. The first is the set $\{(a,0) \mid a \in \Bbb R\}$ and the other $\{(0,b) \mid b \in \Bbb R\}$, it's clear that they have the same dimension but are not the same. You will see later that they are isomorphic, indeed, any two subspaces of the same dimension (over the same field) are isomorphic.
A: Your statement can be made true by making a little improvement. Let $X$ be a vector space and $V$ and $U$ are subspaces of $X$ with $U\subseteq V$. Then, $U=V$ if and only if $\mbox{dim }V=\mbox{dim }U$.
A: While not the same, any two subspaces of the same dimension are similar in quite a strong sense. If you define a vector space (over$~\Bbb R$) of dimension$~d\in\Bbb N$ to be one that has a basis$~B$ consisting of $d~$vectors, then such a space$~V$ is isomorphic to $\Bbb R^d$ via the map $V\to\Bbb R^d$ that takes the coordinates with respect to$~B$ of vectors $v\in V$; since being isomorphic is symmetric and transitive this implies that all vector spaces of dimension$~d$ are isomorphic. There is very little to this proof; it does not even depend on the fact that dimension is uniquely defined (although of course it is). So in particular subspaces of the same dimension are automatically isomorphic.
But this is not the best one can say for subspaces. Not only are two subspaces $A,B$ of$~V$ that have the same dimension isomorphic, there is in fact an isomorphism $\phi:V\to V$ (this is called an automorphism of$~V$) such that restricted to$~A$ it gives an isomorphism $\phi|_A:A\to B$. This fact does require a non-trivial proof. It can be shown by choosing bases of the same number$~d$ of vectors for each one of the subspaces $A,B$, and then extending each of those bases to a basis of$~V$ by choosing additional vectors (it is a theorem that this is possible). Due to the well-definedness of dimension, the two extended basis will have the same number of elements. Then sending the elements of the first basis in order to those of the second basis defines an automorphism of$~V$ with the required property.
To see that this is a non-trivial strengthening of the earlier result, consider Abelian groups, where the corresponding statement is not true. An abelian group is like a vector space but it lacks the operation of scalar multiplication by the elements of a field; $\Bbb Z$ (with addition as only operation) is an example of an Abelian group. The even numbers ($2\Bbb Z$) and the multiples of $3$ ($3\Bbb Z$) are examples of subgroups of$~\Bbb Z$ that are isomorphic (an isomorphism is given by $2n\mapsto3n$) but there is no automorphism of$~\Bbb Z$ that restricts to an isomorphism $2\Bbb Z\to3\Bbb Z$. In fact $\Bbb Z$ has only two automorphisms, the identity and $n\mapsto-n$, neither of which works.
A: Here are some more intuitive "definitions".
Dimension - Number of degrees of freedom of movement. One-dimensional implies only one direction of movement: up and down a line. Two-dimensional means two distinct directions of movement, spanning a plane, etc.
Basis - The distinct directions of movement. One-dimensional movement only says we are moving along a line, but it does not specify which line. The basis tells us the directions we can move in. 
Can you see why the dimension of a subspace is not enough information to uniquely identify it? Why do we need to know the basis as well?
