Basis, Span and Equivalence of Vector Spaces This question here, forms the basis (no pun intended) of the question I'm asking below : If $n$ vectors are linearly independent, is their span $\mathbb{R}^n$?

Firstly if I have
$$A = \begin{bmatrix}a \\ a\end{bmatrix} \ \text{and} \ \  A' = \begin{bmatrix}a \\ a\\ a\end{bmatrix}$$
Does $span(A) = span(A') = \mathbb{R^1}$? According to the question I've linked above it does.
Secondly can we define a vector space in the following way?
$$V := span\{b_1, b_2, b_3, ...,b_n\}$$
where $b_i$ are the basis vectors for $V$. 

If both of the above premises are true, does that not imply that the vector space created by the basis vector $A$ and the vector space created by the vector $A'$ are equivalent?
To show this, we let
$$V = \left\{\lambda\begin{bmatrix}a\\a\end{bmatrix}  \Bigg\vert \ \lambda \in \mathbb{R}\right\}$$
$$V' = \left\{\lambda\begin{bmatrix}a\\a\\a\end{bmatrix}  \Bigg\vert \ \lambda \in \mathbb{R}\right\}$$
Now $\dim(V) = 1$ and $\dim(V')=1$, they are both one-dimensional vector spaces. But $V$ is a one-dimensional vector space in $\mathbb{R^2}$ and $V'$ is a one-dimensional vector space in $\mathbb{R^3}$, intuitively they shouldn't be equivalent. 
But according to our definition of vector spaces $V = V' = \mathbb{R^1}$ as $span(V) = span(V') = \mathbb{R^1}$
According to this definition of a vector space, we could then define two different planes (two-dimensional vector spaces where $\dim(W) = 2$ and $\dim(W') = 2$), with different orientations in $\mathbb{R^3}$ and they would be equivalent vector spaces. 
We could generalize and take it even further and define two different hyperplanes, with completely different orientations, where $\dim(H) = n$ and $\dim(H') = n$ in $\mathbb{R^{n+1}}$ and $H = H'$ by our definition, which obviously cannot be true.

So my question boils down to this, how can we relate the concepts of span and basis in terms of the definition of a vector space, and how can we relate them so that contradictions like I've shown in examples above do not occur, when we try to equate two vector spaces.
If you can identify any misconceptions I may have, please let me know!
 A: You must note that $A$ is a vector in $\Bbb R^2$ and $A'$ is a vector in $\Bbb R^3$. That being said, the span of $A$ is not the same as the span of $A'$ because they live in different spaces!
However, there is some sense in which these spans are the "same." Consider that for any vector $v \in \operatorname{span}A$, $v = \lambda(a, a)$ for some $\lambda \in \Bbb R$. Similarly, for any vector $u \in \operatorname{span}A'$, $u = \mu(a, a, a)$ for some $\mu \in \Bbb R$. Both vectors, though they live in different spaces, effectively encode one piece of data, namely the number $\lambda a$ or $\mu a$.
More precisely, we say there is an isomorphism, or a relabeling of these spaces with the set of real numbers, $\Bbb R$. That is, for any real number $r$, there is a vector in $\operatorname{span} A$ and $\operatorname{span} A'$ such that the components of that vector are each the number $r$. Similarly, given a vector in $\operatorname{span} A$ or $\operatorname{span} A'$, we can find a real number that is equal to that vector's single piece of data.
It is not the case that the span of any vector in $\Bbb R^2$ or $\Bbb R^3$ is equal to $\Bbb R$, but it is the case that they can be isomorphic to $\Bbb R$.
In symbols, we write
$$
\operatorname{span}A \cong \operatorname{span}A' \cong \Bbb R.
$$
To address your main question, the definition of a vector space is a set closed under an addition $+$ and scalar multiplication $\cdot$ such that various other conditions hold. The glaring issue with your saying that $\operatorname{span}A$ and $\operatorname{span}A'$ are equal is that we cannot add vectors in $\operatorname{span}A$ and $\operatorname{span}A'$, so this does not respect closure under addition.
Since not both of your premises are true, the conclusion that $\operatorname{span}A$ and $\operatorname{span}A'$ are equal is invalid. The conclusion that they can be isomorphic is valid, as I have described.
A: It seems that your confusion comes from the improper use of the terms ''equivalent'', ''equals'' and ''isomorphic'', referred to a vector space.
Equals means that the two vector spaces are the same space. So, a plane, as a subspace of $\mathbb{R}^3$, is equal only to itself.
Equivalent is not a defined word in this context, in the sense that to say that two vector space are equivalent has not a defined meaning, also if I can understand the you are using this word in the sense of ''isomorphic'', that is:
Isomorphic means that there is a function betwee the two spaces $V$ and $W$ that is one-to-one and onto and preserve the structure of the space $V$ in the sense that 
$$
f(x+y)=f(x)+f(y) \quad \mbox{and} \quad f(kx)=kf(x)
$$
Note that this imlpies that the two vector spaces must to be defined over the same fileld $\mathbb{F}$  such that $kx$ and $kf(x)$ ($k\in \mathbb{F}$) are well defined.
Given this, the kay fact is that:

two vector space that have the same finite dimension $n$ are isomorphic. And this means that they are spanned by basis that have the same number of vectors.

This does not means that they are the same space in a concrete way, but this means that, in an abstract way, they are essentially the ''same space'' also if they are not equals. This is the case of two subspaces of $\mathbb{R}^n$ of the same dimension  or also of the vector space $\mathbb{R}^2$ ( the plane $x-y$) and a subspace of dimension two in any vector space  $\mathbb{R}^n$.
Finally note that two vector spaces can be isomorphic also if the element of the spaces are not $n-$ples of real numbers. E.g. The set of complex numbers $\mathbb{C}$, as a vector space over $\mathbb{R}$, is isomorphic to $\mathbb{R}^2$ and to the vector space of the polynomial of degree at most $1$  in $\mathbb{R}$ ( the polynomial $ax+b$ with $a,b \in \mathbb{R}$).
So the correct statement is:

If $n$ vectors are linearly independent,  their span is isomorphic to $\mathbb{R}^n$!

