CNOT quantum gate using tensor products

Question

In textbook, it states that the CNOT gate with the X gate applied on second qubit is \begin{array}{cccc} 1&0&0&0\\ 0&1&0&0\\ 0&0&0&1\\ 0&0&1&0 \end{array} However, By tensor products, wouldn't it be $I$ $\otimes$ $U_{x}$ , which gives \begin{array}{cccc} 0&1&0&0\\ 1&0&0&0\\ 0&0&0&1\\ 0&0&1&0 \end{array}

Answer 1

The way that we can write controlled-not operation explicitly algebraically with the control on the first qubit is as follows: $$ \mathit{CNOT} \;=\; \def\ket#1{\lvert #1 \rangle}\def\bra#1{\langle #1 \rvert}\ket{0}\!\bra{0} \!\otimes\! \mathbf 1 \,+\, \ket{1}\!\bra{1} \!\otimes\! \: \sigma_x$$ where $\mathbf 1$ is an identity matrix of $2\times2$ order.

You can now see why it should give the desired answer.