How to apply operators across parts of a tensor product. Question
I have a system which is the tensor product of two vectors from different vector spaces, I would like to now apply operations to both sub systems separately. The question is how can I do this numerically.
For example let $|a \rangle \in \mathbb{C}^2$ and $|b \rangle \in \mathbb{C}^3$, I have an element $|a \rangle \otimes|b \rangle$ of the total space $\mathbb{C}^6$ which I am representing as a $2 \times 3$ complex valued matrix.
What I would like to do is numerically determine $A|a \rangle \otimes B|b \rangle$, where A and B are operators that transform the vectors into another vector of the same form. I understand this means they are from the spaces $( \mathbb{C}^2 )^{*} \otimes \mathbb{C}^2 \simeq \mathbb{C}^4$ and $( \mathbb{C}^3 )^{*} \otimes \mathbb{C}^3 \simeq \mathbb{C}^9$ respectively.
Attempted solution
So far I have tried to do two separate partial traces of the total $2 \times 3$ complex valued matrix, then I apply the operations separately and take their tensor product. My result seems to be off by a factor of the sum of the original matrix which I think may be from the partial trace. However it is highly questionable if you can even do this since we are effectively only operating on a 5 dimensional space. How are we not losing information.
A method I have seen done is to take the Kronecker product of the operators as represented by square matrices then use matrix multiplication on the original $2 \times 3$ matrix listed out in a column vector. I have written out the basis elements of this method but I can't seem to make sense of it. If this is the only correct method could someone please try explain it or link to a decent reference explaining it.    
 A: If we view $|a\rangle \in \mathbb{C}^2$ and $|b\rangle \in \mathbb{C}^3$ in terms of their direct sum $(|a\rangle,|b\rangle) \in \mathbb{C}^2 \times \mathbb{C}^3 =\mathbb{C}^5$ where I am ignoring some parenthetical demarcations. Then $|a\rangle \mapsto A|a\rangle$ and $|b\rangle \mapsto B|b\rangle$ are lumped together as the direct sum (viewing $A,B$ as $2 \times 2$ and $3 \times 3$ complex matrices respective):
$$ (|a\rangle,|b\rangle) \mapsto \left[ \begin{array}{cc} A & 0 \\ 0 & B \end{array} \right] \left[ \begin{array}{c} |a\rangle \\ |b\rangle \end{array}\right]$$
here I make the convention that $(|a\rangle, |b\rangle) =\left[ \begin{array}{c} |a\rangle \\ |b\rangle \end{array}\right]$. 
However, let us discuss $\mathbb{C}^2 \otimes \mathbb{C}^3$. You wish to calculate
$A |a \rangle \otimes B|b\rangle$. I believe your question (doubt) is to the verity of the identity $A |a \rangle \otimes B|b\rangle=(A \otimes B)|a \rangle \otimes |b\rangle$. I'll begin by counting. $A \otimes B$ the Kronecker product is $6 \times 6$. Then $|a \rangle \otimes |b\rangle$ is naturally identified with a $6$-component vector. Let's be explicit $|b \rangle = (b_1,b_2,b_3)$ then 
$$ |a \rangle \otimes |b\rangle = (b_1|a \rangle,b_2|a \rangle,b_3|a \rangle)$$
On the other hand,
$$ A \otimes B = A \otimes \left[ \begin{array}{ccc} B_{11} & B_{12} & B_{13} \\
B_{21} & B_{22} & B_{23}\\
B_{31} & B_{32} & B_{33}
\end{array}\right] = \left[ \begin{array}{ccc} B_{11}A & B_{12}A & B_{13}A \\
B_{21}A & B_{22}A & B_{23}A\\
B_{31}A & B_{32}A & B_{33}A
\end{array}\right]$$
So, calculate:
\begin{align} 
(A \otimes B)|a \rangle \otimes |b\rangle 
&= \left[ \begin{array}{ccc} B_{11}A & B_{12}A & B_{13}A \\
B_{21}A & B_{22}A & B_{23}A\\
B_{31}A & B_{32}A & B_{33}A
\end{array}\right] \left[\begin{array}{c} b_1|a \rangle \\ b_2|a \rangle \\ b_3|a \rangle\end{array} \right] \\
&= \left[\begin{array}{c} (B_{11}b_1+B_{12}b_2+B_{13}b_3)A|a \rangle \\ 
(B_{21}b_1+B_{22}b_2+B_{23}b_3)A|a \rangle \\ (B_{31}b_1+B_{32}b_2+B_{33}b_3)A|a \rangle\end{array} \right] \\
&= \bigl((B_{11}b_1+B_{12}b_2+B_{13}b_3)A|a \rangle, \\
 &\qquad \qquad (B_{21}b_1+B_{22}b_2+B_{23}b_3)A|a \rangle,(B_{31}b_1+B_{32}b_2+B_{33}b_3)A|a \rangle \bigr) \\
&= A|a \rangle) \otimes \bigl((B_{11}b_1+B_{12}b_2+B_{13}b_3, \\ & \qquad \qquad B_{21}b_1+B_{22}b_2+B_{23}b_3,B_{31}b_1+B_{32}b_2+B_{33}b_3 \bigr) \\
&= A|a \rangle \otimes B|b \rangle.
\end{align}
Perhaps the detail I've shown will be helpful. I do think other conventions are possible, but this one ought to suffice for your problem.
