if $z$ in $Ker(A)$, then $zA = 0$? It seems that every where I look, the definition for a vector being in the kernel of a matrix i.e.  $z \in Ker(A)$ 
is that $Az = 0$, then $z^TA^T =0$
Is it possible to say $z \in Ker(A)$ then $zA=0$? where $z$ is a row vector?
 A: EDIT: Your notation is perfectly well defined, and it could make sense, but the issue is that you'd have to switch the meanings of other notation in order to make it make sense. This is because if $z$ is a row vector, then $zA = 0$ isn't the same as $Az^T = 0$. This is easy to see, just compute some examples. This implies that if $T_A$ is the linear transformation usually represented by $A$, and $v$ the vector representing $z$ then $zA = 0$ doesn't mean $T(v) = 0$ (instead, $zA^T = 0$ would mean $T(v) = 0$). Of course, if you reinterpret your matrix $A$ to represent to the linear transformation which is usually represented by $A^T$, then you'd be okay, but you'd also have to switch the order of matrix multiplication. 
The following has some technical issues: (But I'm leaving it there anyway. Feel free to figure out the issue)
Here's the problem with your notation.
Usually if $S,T$ are linear transformations represented by matrices $A_S,A_T$, then the usual notation for matrices is set up nicely so that the matrix representing $S\circ T$ is $A_SA_T$. Similarly, if $v$ is the vector represented by $z$, then expressions like $(S\circ T)(v)$ naturally become $A_SA_Tz$. Your suggestion would have you write $z(A_SA_T) = 0$ to mean $(S\circ T)(v) = 0$.
What's worse, it's not even associative, since $z(A_SA_T) = 0$ in your notation would mean that $(S\circ T)(v) = S(T(v)) = 0$, but $(zA_S)A_T = 0$ in your notation would mean that $T(S(v)) = 0$.
In general unless the objects you're considering "commute", you can't freely rearrange them.
