What is the connection between the fundamental subspaces and finite-dimensional vector spaces being isomorphic to $\mathbb R^n$ I know that if $V$ is a finite dimensional vector space of dim $n$, then $V$ is isomorphic to $\mathbb{R}^n$. We know this is true because if dim $V = n$, there is some basis $\mathbf{v}_1, \mathbf{v}_2,...,\mathbf{v}_n \in V$, and we can define the invertible linear transformation $T: V \rightarrow \mathbb{R}^n$ by $A\mathbf{v}_k = \mathbf{e}_k$ for $k = 1,2,...,n$. A corollary to various invertibility theorems asserts that a matrix $A$ is invertible only if its columns form a basis in $\mathbb{R}^n$. So the basis vectors $\mathbf{v}_1, \mathbf{v}_2,...,\mathbf{v}_n \in V$ are $n$-tuples (have $n$ entries/$n$ coordinates)?, so $T$ is an $n \times n$ matrix and everything works out and makes sense. Now as to where I get confused...
All subspaces are vector spaces. So shouldn't the four fundamental subspaces, ran $A$, ker $A$, etc, be isomorphic to $\mathbb{R}^n$? To look at a simple example, say we have the matrix 
\begin{align*}
A = \begin{bmatrix}1&1&2\\2&2&4\\2&3&5\end{bmatrix} \rightarrow \begin{bmatrix}1&0&1\\0&1&1\\0&0&0\end{bmatrix}.
\end{align*}
Since there is a pivot in the first and second columns of the echelon form of $A$, we know that the first two column vectors of $A: (1,2,2)^T$ and $(1,2,3)^T$ give us a basis in the Ran $A$. 
The dimension of Ran $A$ is $2$, so referring back to our first statement, it must be true that Ran $A$ is isomorphic to $\mathbb{R}^2$. So there must be some invertible matrix $T: \text{Ran}\; A \to \mathbb{R}^2$. The Ran $A$ consists of column vectors with $3$ entries, and all invertible matrices must be square, so $T$ must be a $3\times 3$ matrix. Thus, the output vectors in the codomain are elements in $\mathbb{R}^3$. But this contradicts the fact that the codomain should be $\mathbb{R}^2$. 
I would appreciate if someone could tell me where I'm getting confused. 
 A: The confusion you make comes from a space and linear subspaces. For example, $\text{Ran}A$ is a linear subspace of dimension $2$ of $\mathbb R^3$. Indeed, there is an invertible linear transformation between $\text{Ran}A$ and $\mathbb R^2$. This doesn’t imply that $\text{Ran}A$ is isomorphic to $\mathbb R^3$.
A: @mathcounterexamples.net @Niki Di Giano.
Thank you two for the help! If I try to reiterate what you two are saying to make sure i'm understanding correctly: Let $T$ represent the matrix of the transformation. The input to the transformation matrix that is the isomorphism $T$: Ran $A \rightarrow \mathbb{R}^2$ is a vector $\mathbf{v} = (x_1,x_2)^T \in$ Ran $A$ which is a two dimensional subspace of $\mathbb{R}^3$. So my mistake was incorrectly assuming that the input was a vector with 3 coordinates. Let $\mathcal{B}$ be the basis for Ran $A$. Then the input vector is the coordinate vector of $\mathbf{v}$ relative to $\mathcal{B}$, or $[\mathbf{v}]_\mathcal{B}$ (by defn).  So it was also incorrect of me to state in the first part that the basis vectors of an $n$-dimensional vector space are $n$-tuples. The dimension of a vector space does not have to equal the number of coordinates in the basis vectors for that space, since two dimensional subspaces in $\mathbb{R}^3$ exist (they are planes in $\mathbb{R}^3$). Furthermore, by defining $T$ for each vector in $\mathcal{B}$ such that $T([1,2,2]^T) = \mathbf{e_1}$ and $T([1,2,3]^T) = \mathbf{e_2}$, we get that the matrix representation of $T$ is the identity $I_2$. In my textbook the mapping $T: V \rightarrow \mathbb{F}^n$ is correlated to $\mathbf{v} \rightarrow [\mathbf{v}]_\mathcal{B}$, given $\mathcal{B}$ is the basis for $V$. But we just showed that the input $\mathbf{v}$ is $[\mathbf{v}]_\mathcal{B}$ which only happens if $T$ is the identity. In other words, we transformed $\mathcal{B}$ into the standard basis, which we can do for any finite dimensional vector space.
