<s>[INST] 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
$$
\begin{pmatrix}
        2 & 3 \\
        5 & 7 \\
        \end{pmatrix}
=
\begin{pmatrix}
        1 & 2 \\
        \end{pmatrix}
\otimes
\begin{pmatrix}
        1 \\
        2 \\
        \end{pmatrix}
+
\begin{pmatrix}
        1 & 1 \\
        \end{pmatrix}
\otimes
\begin{pmatrix}
        1 \\
        3 \\
        \end{pmatrix}
$$
 [/INST] 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.
</s>