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