Representing a linear operator on $V$ with an element of $V \otimes V^*$ I got interested by the first sentence of this wikipedia subsection. It claims that any linear operator $f:V\to V$ can be represented by an element of $V\otimes V^*$ in a very concrete way: the element $w\otimes h$ maps a vector $v$ to $h(v)w$. 
Clearly such an object is a linear operator, but given a basis I wanted to "see" the relationship between the matrix representing $f$ and the coefficients of $h$ and $w$. So I came up with an arbitrary matrix:
$$
        M=\begin{pmatrix}
        2 & 3  \\
        5 & 7 \\
        \end{pmatrix}
$$
and tried to determine $w$ and $h$ but found I couldn't.
I think I understand my mistake: this particular linear operator in this basis is represented by an element of $V\otimes V^*$ that isn't a pure tensor.
My question: Given that, is there an algorithmic way to find the right element of $V\otimes V^*$?

EDIT: Marko's answer gives the general solution but in the interim I found a solution for this particular matrix. I thought it might help someone in the future to see the grisly working-out, so here it is...
I noticed that the cases where the answer is a pure tensor are those where the matrix's columns are linearly dependent. So I wrote it like this:
$$
        \begin{pmatrix}
        2 & 3  \\
        5 & 7 \\
        \end{pmatrix}
=
\begin{pmatrix}
        a & \lambda a  \\
        b & \lambda b \\
        \end{pmatrix}
+
\begin{pmatrix}
        c & \gamma c  \\
        d & \gamma d \\
        \end{pmatrix}
$$
This was easy to solve by inspection:
$$
        \begin{pmatrix}
        2 & 3  \\
        5 & 7 \\
        \end{pmatrix}
=
\begin{pmatrix}
        1 & 2 \\
        2 & 4 \\
        \end{pmatrix}
+
\begin{pmatrix}
        1 & 1 \\
        3 & 3 \\
        \end{pmatrix}
$$
I could then express each of these as a pure tensor:
$$
\begin{pmatrix}
        1 & 2 \\
        2 & 4 \\
        \end{pmatrix}
=\begin{pmatrix}
        1 & 2 \\
        \end{pmatrix}
\otimes
\begin{pmatrix}
        1 \\
        2 \\
        \end{pmatrix}
$$
and
$$
\begin{pmatrix}
        1 & 1 \\
        3 & 3 \\
        \end{pmatrix}
=\begin{pmatrix}
        1 & 1 \\
        \end{pmatrix}
\otimes
\begin{pmatrix}
        1 \\
        3 \\
        \end{pmatrix}
$$
So the answer I was looking for was
$$
\begin{pmatrix}
        2 & 3 \\
        5 & 7 \\
        \end{pmatrix}
=
\begin{pmatrix}
        1 & 2 \\
        \end{pmatrix}
\otimes
\begin{pmatrix}
        1 \\
        2 \\
        \end{pmatrix}
+
\begin{pmatrix}
        1 & 1 \\
        \end{pmatrix}
\otimes
\begin{pmatrix}
        1 \\
        3 \\
        \end{pmatrix}
$$
 A: Let $\mathcal B=(e_1,\ldots,e_n)$ be a basis of $V$ and $\mathcal B^*=(e_1^*,\ldots, e_n^*)$ the associated dual basis of $V^*$.
The map $$V\otimes V^* \rightarrow End(V)$$ sends $e_i\otimes e_j^*$ to the linear operator : $$v=(v_1,\ldots,v_n)\mapsto e_j^*(v)e_i=\sum_{k=1}^nv_je_i.$$
Hence, the matrix representation of $e_i\otimes e_j^*$ in the basis $\mathcal B$ is the standard matrix $$E_{ij}=\begin{pmatrix}0 & 0 & \ldots & 0 \\ \vdots & \ddots & 1 & 0 \\ \vdots & \cdots & \cdots & \vdots \\ 0& \cdots & \cdots & \cdots \end{pmatrix}$$ with only one non-zero coefficient at row $i$ and column $j$.
Thus, the induced map $\varphi: V\otimes V^* \rightarrow M_n(K)$ sends $$e_i\otimes e_j^* \mapsto E_{ij}.$$
Conversely, there is a natural linear map $$\begin{array}{rccc}\psi:& M_n(K)&\rightarrow & V\otimes V^* \\ & A=(a_{ij})  & \mapsto & \displaystyle \sum_{k,l=1}^{n} a_{ij}e_i\otimes e_j^*\end{array}$$
One can easily check that $\psi$ is invertible and it is the inverse to $\varphi$.
Hence, if $A=\begin{pmatrix}2&5 \\ 3 & 7\end{pmatrix}$ represents a linear operator $f$ then it is represented in $V\otimes V^*$ as $$2e_1\otimes e_1^* + 5e_1\otimes e_2^* + 3e_2\otimes e_1^* + 7e_2\otimes e_2^*$$ which is very natural.
A: In your case you can write 
\begin{align*}
M&=(2e_1+5e_2)\otimes e_1+(3e_1+7e_2)\otimes e_2\\
&=2e_1\otimes e_1+5e_2\otimes e_1+3e_1\otimes e_2+7e_2\otimes e_2.
\end{align*}
The idea is the following: choose a basis $\{x_1,\ldots,x_k\}$ of the range of the operator $T.$ Then write 
$$Tx=\lambda_1(x) x_1+\cdots+\lambda_k(x) x_k$$ for some scalars $\lambda_1(x),\ldots,\lambda_k(x)$ (they do depend on $x$). Then this coefficients $\lambda_1,\ldots,\lambda_k$ represents linear functionals on $V$. To prove that they are really linear functionals, write
$$T(x+y)=Tx+Ty$$ and apply linear independence of the vectors $x_1,\ldots,x_k$ to get additivity of functions $\lambda_1,\ldots,\lambda_k$. That $\lambda_1,\ldots,\lambda_k$ are homogeneous it can be proved simiarly. 
A: We see this in quantum mechanics. We can expand linear operator in terms of bra and ket.
$$ A =     \sum |x><x|A|y><y| $$
Then  $\sum |x><x|=1 $ is the identity matrix and  $<x|A|y> $ are the matrix elements.
Then again. Physics is know for suggestive notations which may not work 100% of the time.
