Does the linear transformation that a matrix encodes depend on a choice of basis? Context
Let $M$ be an $m \times n$ matrix of real numbers. Let $\mathbf{x}$ be column vector of length $n$ with elements $x_1, \ldots , x_n \in \mathbb{R}$. Let $\vec{x} = (x_1, \ldots , x_n) \in \mathbb{R}^n$ be its analogue in $\mathbb{R}^n$.
Let $\vec{\mathbf{e}}^t$ denote the standard basis in $\mathbb{R}^n$ so that
$$
\vec{\mathbf{e}} = 
\left[
\begin{array}
.\vec{e_1} \\
\vec{e_2} \\
\vdots \\
\vec{e_n}
\end{array}
\right]
$$
and let $\vec{\mathbf{b}}^t$ denote a non-standard basis in $\mathbb{R}^n$ with
$$
\vec{\mathbf{b}} = 
\left[
\begin{array}
.\vec{b_1} \\
\vec{b_2} \\
\vdots \\
\vec{b_n}
\end{array}
\right]
$$
Fact from Linear Algebra: $M$ encodes a linear transformation $T:  \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that 
$$
M \mathbf{x} = \mathbf{y}
$$
where $\mathbf{\vec{e}}^t M \mathbf{x} = T(\vec{x}) = \vec{y}$ is some vector in $\mathbf{R}^m$.

Question
Does the linear transformation $T$ that $M$ corresponds to depend on our choice of basis? That is, if instead of working with  $\mathbf{\vec{e}}^t$  we instead worked with $\mathbf{\vec{b}}^t$, would $M$ encode a different linear transformation? For example, if
$$
\mathbf{\vec{e}}^t \mathbf{x} = \vec{x} = \mathbf{\vec{b}}^t \mathbf{x'}
$$
then do we have that
$$
\mathbf{\vec{e}}^t M \mathbf{x} = T(\vec{x}) = \mathbf{\vec{b}}^t M \mathbf{x'}?
$$
EDIT
Ok. Upon reflection, it is obvious that our choice of basis matters significantly. For example, let us work in $\mathbb{R}^2$ and let our non-standard basis be $(1,1)$ and $(1, -1)$. 
Then the identity matrix
$$
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
$$
will now send $(1,0)$ to $(1,1)$ since
$$
\begin{bmatrix} (1, 1) & (1, -1) \end{bmatrix} \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} \left[
\begin{array}
& 1 \\
0
\end{array}
\right] = (1,1)
$$
which is clearly a different mapping than the identity mapping that would result were we to use the standard basis (where $(1, 0) \mapsto (1,0)$).
 A: A matrix, in the context you are looking at, is a representation of a linear map with respect to a choice of basis, both in source and target. 
If you want the same map but want to use another basis, you have to transform the matrix, too. Textbooks on linear algebra will tell you how. 
Put from the other point of view: if you fix the matrix $M$ but change the basis, you'll get a different map.
A: Going strictly by the definitions, yes: the matrix $M$ does not necessarily encode a transformation with respect to any one basis, and as such it can encode several different transformations depending on one's choice of basis.
That being said: when a basis isn't specified, the tacit assumption is that $M$ encodes the transformation $T:\Bbb R^n \to \Bbb R^m$ given by $T(x) = Mx$, with respect to the standard basis.  So, if you mean something other than this specific transformation, you should say so.
A: Yes, the choice of basis makes a difference. For this reason, some books will write something like the following: If $T$ is a linear transformation, then the matrix representation of $T$ relative to a basis $\beta = \{\beta_1, \beta_2,...,\beta_n\}$ will be written 
$$[T]_\beta$$
For example, let the matrix
$$A = \begin{pmatrix} 6&4&-6\\4&6&-6\\ 6&6&-7 \end{pmatrix}$$
Typically when we write $A$ in this way without any subscripts, we are implying that $A$ is being represented relative to the standard basis, 
$$\{ \begin{pmatrix} 1\\0\\0 \end{pmatrix}, \begin{pmatrix} 0\\1\\0 \end{pmatrix} , \begin{pmatrix} 0\\0\\1 \end{pmatrix} \}$$
We can see that the columns of $A$ are just the action of the matrix $A$ on the usual basis vectors. In other words,
$$A \vec{e}_1 = (6,4,6)$$
$$A \vec{e}_2 = (4,6,6)$$
$$A \vec{e}_3 = (-6,-6,-7)$$
The matrix seems to "scramble up" the usual basis vectors. However, with some work (namely, the eigenvalue problem), we can find a different basis
$$\beta = \{\beta_1, \beta_2, \beta_3\} = \{ \begin{pmatrix} -1\\-2\\-2 \end{pmatrix}, \begin{pmatrix} -2\\-1\\-2 \end{pmatrix} , \begin{pmatrix} 2\\2\\3 \end{pmatrix} \}$$
where
$$A \beta_1 = \begin{pmatrix} 6&4&-6\\4&6&-6\\ 6&6&-7 \end{pmatrix} \begin{pmatrix} -1\\-2\\-2 \end{pmatrix} = \begin{pmatrix} -2\\-2\\-4 \end{pmatrix} = 2 \beta_1$$
Similarly,
$$A \beta_2 = 2 \beta_2 \quad \text{and} \quad A \beta_3 = \beta_3$$
So, the matrix $A$ (which represents some linear transformation) has the representation relative to the basis $\beta$ above
$$[A]_\beta = \begin{pmatrix} 2&0&0\\ 0&2&0\\ 0&0&1 \end{pmatrix}$$
A nice, diagonal matrix! In this case, we can see that $A$ acts by simply lengthening the $\beta$-basis vectors. We may also say that $A$ dilates its eigenspaces.
Hope this helps.
