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!
Best Answer
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.