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 $(1, 0) \mapsto (1,0)$).

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.
