Does matrix has a underlying basis? Can I say a matrix (M) as a liner transformation and it operates on a vector? The vector must have a basis and the matrix M gave us a new vector. Now is there any basis associated with the matrix. This is elementary linear algebra but still I am not clear. Would someone please explain?
 A: To clear up some confusion the following. A matrix $A$ is just an object given by two dimensions $m,n$ (giving its size), and by $mn$ scalars $A_{i,j}$, indexed by pairs $(i,j)$ with $0<i\leq m$ and $0<j\leq n$. On the other hand a vector (unless one is talking specifically about a column-vector or a row-vector) is an element of a vector space, which is an abstract notion that (satisfies certain axioms but) does not stipulate what its elements are; they could be functions or forces or formal power series or sequences of numbers or whatever. A vector does not need a basis to exist. However, to represent a vector $v\in V$ by a sequence of coordinates (commonly taken to constitute a column-vector, which is in fact a particular kind of matrix), one needs to specify an (ordered) basis of$~V$. Similarly to represent a linear map $f$ between (finite dimensional) vector spaces $V,W$ by a matrix, one needs to specify (ordered) bases $B_V$ of $V$ and $B_W$ of $W$. This is done in such a way that this matrix of $f$ multiplied by the column vector of coordinates of $v$ with respect to $B_V$ results in a column-vector that contains coordinates of $f(v)\in W$ with respect to $B_W$.
To answer your question: there is no basis associated to a matrix in itself. However a matrix may be associated to a linear map, and this association depends on two bases, one in each of the vector spaces concerned. It is useful to note that if one identifies the set of column-vectors of size$~k$ with $\Bbb R^k$, then multiplication by a given $m\times n$ matrix $A$ defines a linear map $\Bbb R^n\to\Bbb R^m$, and the matrix of this linear map with respect to the standard bases of $\Bbb R^n$ and of $\Bbb R^m$ is precisely$~A$; since standard bases are fixed and require no choice, one could say that $A$ is canonically associated with this linear map. But this does not prevent $A$ from also being associated, for some bases, with other linear maps.
A: First of all, a $m\times n$ matrix $M$ can be associated with a linear transformation $L_M: \mathbb{R}^n \rightarrow \mathbb{R}^m$ defined by
$$L_M(\mathbf{x}) = M\mathbf{x} \text{ for } \mathbf{x} \in \mathbb{R}^n$$
On the other hand, the collection $\mathbb{M}_{m \times n}$ of all $m \times n$ matrix $M$ forms a vector space with the vector addition as the usual matrix addition and the scalar-vector multiplication as the usual scalar-matrix multiplication. The vector space $\mathbb{M}_{m \times n}$ has dimension $mn$ and it has a basis $\beta = \{M^{11}, M^{12}, \ldots M^{1n}, M^{21}, M^{22}, \ldots M^{2n}, \ldots, M^{mn}\}$ where $M^{kl}$ is a $m \times n$ matrix with
$$M^{kl}_{ij} = \delta_{ik}\delta{jl} = \begin{cases} 1 & \text{ when } i = k \text{ and } j = l\\ 0 & \text{ otherwise}\end{cases}$$
For example,
$$M^{12} = \begin{pmatrix}
0 & 1 & 0 & \cdots & 0\\
0 & 0 & 0 & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \cdots & 0
\end{pmatrix}$$
A: Consider a basis for a vector space $\mathbb{V}$ as $\{v_1,..v_n\}$, and a basis for a matrix space $\mathbb{M}$ as $\{M_{11},...,M_{mn}\}$. We can define the set of linear transformations by considering each operation $\mathbb{M}\otimes\mathbb{V}$.
As you can see, the basis' for the $\mathbb{V}$ and $\mathbb{M}$ are not compatible, having different dimensions. Any vector $\vec v$ (with the correct number of dimensions) can be created from $\mathbb{V}$, and similarly any matrix $M$ with the correct dimensions, can be created from $\mathbb{M}$.
There are numerous basis' for both $\mathbb{V}$ and $\mathbb{M}$ - neither an individual vector nor an individual matrix creates a basis.
