Measure two qubits, one from entangled pair, in Bell basis?

Question

In quantum teleportation we require both sender (Alice) and receiver (Bob) to share an entangled pair of qubits in the $|\beta_{00}\rangle$ Bell state. Then, if Alice wants to teleport a quibt $|v\rangle$ to Bob, she needs to measure $|v\rangle$ and her qubit from an entangled pair in the Bell basis. Such measurment requires her to perform a CNOT on the two qubits with $|v\rangle$ being the control qubit and then apply Hadamard transformation to $|v\rangle$.

I can't understand how can she perform the above. If we represent quantum gates as matrices, what vector would represent the input to the CNOT matrix (the pair $|v\rangle$ and Alice's qubit from an entangled pair?) In other words, how can we represent a pair that Alice wants to apply CNOT to (when performing measurement in the Bell Basis?)

I know that Bell measurement does not require CNOT in general, but this is one of the ways it could be implemented (assuming we just want to get the classical 2-bit value).

My intuition is that because we need to operate on a tensor product of $|v\rangle$ and the entangled pair, we need to create a matrix representing an operation performing CNOT on $v\rangle$ and first qubit from the pair, and I on the second qubit from the pair, so it would be like $CNOT \otimes I$. Is this correct?

Answer 1

I figured out that as we are operating on 3 qubits ($|v\rangle$ and $\beta_{00}\rangle$ pair) and we want to perform CNOT on the first two (if we enumerate like 1- $|v\rangle$, 2- first from pair, 3 - second from pair) it is just like we want to perform $CNOT \otimes I$ on $|v\rangle \otimes |\beta_{00}\rangle$.

After performing the CNOT described above we need to perform H on the $|v\rangle$ qubit, which is now entangled (as CNOT is an entangling operation), so it is like we want to perform $H \otimes I$ on the three (where $I$ is 3 x 3 identity matrix).

Beaing able to perform both steps described above we are able to perform a measurement of qubit $|v\rangle$ and the first qubit from the entangled pair $|\beta_{00}\rangle$ in the Bell basis.