Is "Kernel Matrix" a bad choice of wording? I'm reading an article, in it, a solution to an homogeneous rank deficient system of linear equations is presented. Before presenting the solution, it says: "the corresponding kernel matrix is given as".
I'm a bit shocked by the choice of wording, or my understanding of kernel.
1- Isn't the "kernel matrix" presented only an instance of all the matrices that are contained in the subspace of the kernel ?
2- Under which conditions is there a "kernel matrix" i.e. a unique representation of the kernel ?
 A: This is actually excellent wording!
One of the big lessons of modern math is that it's helpful to think in terms of morphisms instead of thinking in terms of objects alone; this unites a lot of otherwise-disparate concepts.  So, instead of thinking of a subobject of an object, we think of the inclusion map from a subobject to an object.
That is, if we have a linear map $T : V \to W$ and $U \subset V$ maximal such that $T|_U = 0$, then instead of saying that the kernel of $T$ is $U$ as we would classically we say that the kernel of $T$ is the inclusion map $i : U \hookrightarrow V$.  This can be characterized by a universal property, which means that it makes sense even in contexts where the are no underlying sets or anything.
The "kernel matrix," therefore, is just the matrix representation of the kernel!
Of course a matrix representation of a map $i : U \hookrightarrow V$ depends on a choice of basis for both $V$ and $U$, and in particular the choice of basis on $U$ isn't determined by the choice of basis on $V$ (since $U$ probably isn't spanned by elements of the chosen basis for $V$).  Different choices of basis for $U$ correspond to the different matrices mentioned by Brain Fitzpatrick in his answer.
A: Let $A$ be an $m\times n$ matrix of rank $r$. Then the null space of $A$ has dimension $n-r$. Any $n\times(n-r)$ matrix $K$ whose columns are a basis for the null space of $A$ can be referred to as a kernel matrix of $A$. 
Since bases of vector spaces are not unique, $A$ can have many kernel matrices. 
It's likely that the term "the corresponding kernel matrix" was used because the result was independent of choice of basis for the null space of $A$. 
Note that many computer algebra systems accept the command "kernel". Here's a snippet from macaulay2 in which we compute "the kernel matrix" of an integral matrix.
i1 : A = matrix{{1,2,3,4,5},{6,7,8,9,10}}

o1 = | 1 2 3 4 5  |
     | 6 7 8 9 10 |

              2        5
o1 : Matrix ZZ  <--- ZZ

i2 : K = gens kernel A

o2 = | -1 -2 -2 |
     | 2  3  2  |
     | -1 0  1  |
     | 0  -1 0  |
     | 0  0  -1 |

              5        3
o2 : Matrix ZZ  <--- ZZ

i3 : A * K

o3 = 0

              2        3
o3 : Matrix ZZ  <--- ZZ

Note that multiplying $K$ on the right by any unimodular matrix defines another "kernel matrix" for $A$.
i4 : P = matrix{{-5,-18,34},{1,4,-7},{-2,-7,13}}

o4 = | -5 -18 34 |
     | 1  4   -7 |
     | -2 -7  13 |

              3        3
o4 : Matrix ZZ  <--- ZZ

i5 : L = K*P

o5  = | 7   24  -46 |
      | -11 -38 73  |
      | 3   11  -21 |
      | -1  -4  7   |
      | 2   7   -13 |

               5        3
o5 : Matrix ZZ  <--- ZZ


i6 : rank(L)

o6 : 3

i7 : A * L

o7 = 0

              2        3
o7 : Matrix ZZ  <--- ZZ

