If \(A\) and \(B\) are any two square matrices of the same rank and \(C \equiv[B, A]\), show that \(e^{A+B}=e^{A} e^{B} e^{\frac{1}{2} C}\) provided that \([C, A]=[C, B]=0\)

The matrix exponential is a significant concept in linear algebra and is especially important when dealing with differential equations and quantum mechanics. It extends the idea of the exponential function to matrices. For a square matrix \(M\), the exponential \(e^M\) is defined by the series:

\[ e^M = I + M + \frac{M^2}{2!} + \frac{M^3}{3!} + .... \]

Here, \(I\) is the identity matrix. This series is infinite but converges for any square matrix \(M\). The matrix exponential has properties similar to the exponential function for real numbers and is used to solve linear systems of differential equations. If you understand the concept of an exponential function from calculus, the matrix exponential builds on that intuition. In practical terms, it transforms a matrix in such a way that solutions to certain matrix equations become manageable.

In matrix algebra, the commutator is a measure of how much two matrices differ in their multiplication order. For any two square matrices \(A\) and \(B\), the commutator \([A, B]\) is defined as:

\[ [A, B] = AB - BA \]

If the commutator is zero, i.e., \([A, B] = 0\), then the matrices are said to commute. In this context, commuting matrices have a special relationship that often simplifies complex matrix expressions. For this exercise, it is given that the commutator \(C = [B, A]\), and you are also given the conditions \([C, A] = 0\) and \([C, B] = 0\). This means that not only do \(A\) and \(B\) commute with each other when subtracted but also, \(C\) commutes with both \(A\) and \(B\). These properties are crucial for applying the Baker-Campbell-Hausdorff formula correctly.

Matrix commutation relations describe how a set of matrices interact when multiplied in different orders. These relations have significant implications, especially in quantum mechanics, where operators that commute have simultaneous eigenstates. In this problem, we deal with the Baker-Campbell-Hausdorff formula, which translates the addition of matrices into the product of their exponentials. The formula is given by:

\[ e^{A+B} = e^{A} e^{B} e^{-\frac{1}{2}[A,B]} \]

Given that \([C, A] = 0\) and \([C, B] = 0\), we understand that \(C\) can be combined with the exponents without affecting the order. So, the formula simplifies to:

\[ e^{A+B} = e^{A} e^{B} e^{\frac{1}{2}C} \]

This result highlights how the commutation (or lack thereof) influences how matrix exponentials are calculated and combined. In essence, knowing the commutation relations among matrices allows us to manipulate exponential expressions efficiently and accurately.