Properties of Matrix Exponential

Theorem
In the following:
 * $\mathbf A$ and $\mathbf B$ are constant square matrices
 * $\mathbf P$ is a nonsingular matrix
 * $t, s \in \R$

The matrix exponential $e^{\mathbf A t}$ has the following properties:

Derivative

 * $\dfrac \d {\d t} e^{\mathbf A t} = A e^{\mathbf A t}$

Nonvanishing Determinant

 * $\det e^{\mathbf A t} \ne 0$

Same-Matrix Product

 * $e^{\mathbf A t} e^{\mathbf A s} = e^{\mathbf A \paren {t + s} }$

Inverse

 * $\paren {e^{\mathbf A t} }^{-1} = e^{-\mathbf A t}$

Commutative Product (1)

 * $\mathbf A \mathbf B = \mathbf B \mathbf A \implies e^{\mathbf A t} \mathbf B = \mathbf B e^{\mathbf A t}$

Commutative Product (2)

 * $\mathbf A \mathbf B = \mathbf B \mathbf A \implies e^{\mathbf A t} e^{\mathbf B t} = e^{\paren {\mathbf A + \mathbf B} t}$

Series Expansion

 * $\ds e^{\mathbf A t} = \sum_{n \mathop = 0}^\infty \frac {t^n} {n!} \mathbf A^n$

Decomposition

 * $ e^{\mathbf P \mathbf B \mathbf P^{-1} } = \mathbf P e^{\mathbf B} \mathbf P^{-1}$

Derivative
The derivative rule follows from the definition of the matrix exponential.

Nonvanishing Determinant
The linear system $x' = \mathbf A x$ has $n$ linearly independent solutions.

Putting together these solutions as columns in a matrix creates a matrix solution to the differential equation, considering the initial conditions for the matrix exponential.

From Existence and Uniqueness Theorem for 1st Order IVPs, this solution is unique.

By linear independence of its columns:
 * $\det e^{\mathbf A t} \ne 0$

The nonzero determinant property also follows as a corollary to Liouville's Theorem (Differential Equations).

Same-Matrix Product
Let
 * $\map \Phi t = e^{\mathbf A t} e^{\mathbf A s} - e^{\mathbf A \paren {t + s} }$

for some fixed $s \in \R$.

Then:

Since $\map \Phi 0 = e^{\mathbf A s} - e^{\mathbf A s} = 0$, it follows that:
 * $\map \Phi t = e^{\mathbf A t} \map \Phi 0 = 0$

independent of $s$.

Hence the result.

Inverse
Using the Same-Matrix Product property,
 * $e^{\mathbf A t} e^{-\mathbf A t} = e^{-\mathbf A t} e^{\mathbf A t} = e^0 = I$

hence $e^{\mathbf A t}$ and $e^{-\mathbf A t}$ are inverses of each other.

Commutative Product (1) & (2)
Let:
 * $\map {\Phi_1} t = e^{\mathbf A t} \mathbf B - \mathbf B e^{\mathbf A t}$
 * $\map {\Phi_2} t = e^{\mathbf A t} e^{\mathbf B t} - e^{\paren {\mathbf A + \mathbf B} t}$

and then follows the same program outlined in the Same-Matrix Product proof.

Series Expansion
Differentiating the series term-by-term and evaluating at $t=0$ proves the series satisfies the same definition as the matrix exponential, and hence by uniqueness is equal.

Decomposition
$\paren {\mathbf P \mathbf B \mathbf P^{-1} }^n = \mathbf P \mathbf B^n \mathbf P^{-1}$ by induction.

The result follows from plugging in the matrices and factoring $\mathbf P$ and $\mathbf P^{-1}$ to their respective sides.