# Operators acting on a single subsystem within a combined system's state

I was reading over combined systems and operators acting on a single system within the combined system, and am confused by the math.

For example, we have individual qubit states for subsystems $$A$$ and $$B$$ that, as a combined system, produce the state:

\begin{align} A\otimes B &= \begin{bmatrix} a_{0}b_{0} \\ a_{0}b_{1} \\ a_{1}b_{0} \\ a_{1}b_{1} \end{bmatrix} \end{align}

We have an operator $$O$$ that acts on subsystem $$A$$. To express this on the combined state, it is simply (where $$I$$ is the identity matrix of the same dimension as $$A$$ and $$B$$) \begin{align} (O\otimes I)(A\otimes B) \end{align}

But why is this? I've been reading a text that says this combined operator changes only the coefficients $$a_0$$ and $$a_1$$, while leaving $$b_0$$ and $$b_1$$ unchanged. But I don't see that. I see this operation as changing whatever the value of the product of $$a_i$$ and $$b_j$$ is.

Welcome to SE! Another way of saying what your text claims is $$(O\otimes I)(A\otimes B)=(OA)\otimes B.$$ To prove this to yourself, you could write a general $$O$$, say $$O=\begin{pmatrix}w&x\\y&z\end{pmatrix}$$, and then just evaluate both sides of the above equation directly for $$A\equiv\begin{pmatrix}a_0\\a_1\end{pmatrix}$$ and $$B\equiv\begin{pmatrix}b_0\\b_1\end{pmatrix}$$ (i.e., for the LHS take the tensor product $$O\otimes I$$ first, then multiply $$A\otimes B$$ by this, and for the RHS multiply $$A$$ by $$O$$ first, then take the tensor product with $$B$$).
You will find that the equation indeed holds; this is the sense in which $$A$$ "just acts on" the state of the first qubit.