Conjugation of $R_x(theta)$ with $CNOT$

Question

Section 2.5 (4.3) of the Qiskit textbook, see here, discusses the conjugation of $R_x(theta)$ by $CNOT$. The following expression is given:

$$CX_{j,k}(R_x(theta)otimes 1) CX_{j,k}=color{brown}{CX_{j,k}e^{ifrac{theta}{2}(Xotimes1)}CX_{j,k}=e^{ifrac{theta}{2}CX_{j,k}(Xotimes 1)CX_{j,k}}}=e^{ifrac{theta}{2}Xotimes X}$$
I am confused by the part highlighted in yellow. What exponentiation rules allow this? Could someone show me the intermediate steps or the relevant identities/rules to take us from the LHS to RHS of the highlighted part?

Adam Zalcman · Accepted Answer

This is an application of the following identity
$$
Be^AB^{-1} = e^{BAB^{-1}}tag1
$$
where $A$ is any $ntimes n$ real or complex matrix and $B$ is any invertible $ntimes n$ real or complex matrix.
Proof of $(1)$. First, recall that the matrix exponential of $A$ is defined as
$$
e^A = sum_{k=0}^infty frac{1}{k!}A^k.
$$
Next, note that for any integer $k$
$$
(BAB^{-1})^k = BAB^{-1}BAB^{-1}dots BAB^{-1} = BAAdots AB^{-1} = BA^kB^{-1}tag2.
$$
Finally, calculate
$$
e^{BAB^{-1}} = sum_{k=0}^infty frac{1}{k!}(BAB^{-1})^k = sum_{k=0}^infty frac{1}{k!}BA^kB^{-1} = B left(sum_{k=0}^infty frac{1}{k!}A^kright) B^{-1} = Be^AB^{-1}
$$
where we used $(2)$ in the second step. $square$

Conjugation of $R_x(theta)$ with $CNOT$

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