📂Matrix Algebra

Lie Product Formula

Theorem¹

matrix $X, Y \in M_{n \times n}(\mathbb{C})$ satisfies the following.

$$ e^{X + Y} = \lim\limits_{m \to \infty} \left( e^{\frac{X}{m}} e^{\frac{Y}{m}} \right)^{m} $$

Here $e^{X}$ is the matrix exponential.

Proof

matrix exponential

$$ e^{X} = \sum\limits_{m=0}^{\infty} \dfrac{X^{m}}{m!} $$

By the definition of the matrix exponential, the product of $e^{\frac{X}{m}}$ and $e^{\frac{Y}{m}}$ can be arranged as follows.

$$ e^{{\frac{X}{m}}} e^{{\frac{Y}{m}}} = I + \dfrac{X}{m} + \dfrac{Y}{m} + O \left( \dfrac{1}{m^{2}} \right) $$

Here $O$ denotes Big-O notation. Therefore $e^{\frac{X}{m}} e^{\frac{Y}{m}} \to I$ $\text{as}$ $m \to \infty$, and for sufficiently large $m$ and sufficiently small $\epsilon$, $\| e^{\frac{X}{m}} e^{\frac{Y}{m}} - I \| \lt \epsilon$ holds.

properties of the matrix logarithm

$$ e^{\log A} = A, \quad \text{for } \| A - I\| \lt 1 $$

$$ \log(I + B) = B + O(\| B^{2} \|), \quad \text{for } \| B \| \lt 1/2 $$

From the properties of the matrix logarithm we obtain

$$ \begin{align*} \log(e^{\frac{X}{m}} e^{\frac{Y}{m}}) &= \log \left( I + \dfrac{X}{m} + \dfrac{Y}{m} + O \left( \dfrac{1}{m^{2}} \right) \right) \\ &= \dfrac{X}{m} + \dfrac{Y}{m} + O \left( \left\| \dfrac{X}{m} + \dfrac{Y}{m} + O \left( \dfrac{1}{m^{2}} \right) \right\|^{2} \right) \\ &= \dfrac{X}{m} + \dfrac{Y}{m} + O \left( \dfrac{1}{m^{2}} \right) \end{align*} $$

Taking the exponential of both sides (note that this is possible because the above conditions are satisfied, unlike the exponential function defined only on the real numbers),

$$ e^{\frac{X}{m}} e^{\frac{Y}{m}} = \exp \left( \dfrac{X}{m} + \dfrac{Y}{m} + O \left( \dfrac{1}{m^{2}} \right) \right) $$

$$ \implies (e^{\frac{X}{m}} e^{\frac{Y}{m}})^{m} = \exp \left(X + Y + O \left( \dfrac{1}{m} \right) \right) $$

Now, taking the limit $m \to \infty$, and since $\exp$ is continuous,

$$ \lim\limits_{m \to \infty} (e^{{\frac{X}{m}}} e^{{\frac{Y}{m}}})^{m} = \exp \left(X + Y\right) = e^{X+ Y} $$

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