Properties of Matrix Exponential

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

The matrix exponential $e^{At}$ has the following properties:

Derivative

 * $\dfrac{d}{dt} e^{At} = A e^{At}$

Nonvanishing Determinant

 * $\det e^{At} \ne 0$

Same-Matrix Product

 * $e^{At} e^{As} = e^{A(t+s)}$

Inverse

 * $(e^{At})^{-1} = e^{-At}$

Commutative Product (1)

 * $AB = BA \implies e^{At} B = B e^{At}$

Commutative Product (2)

 * $AB = BA \implies e^{At} e^{Bt} = e^{(A+B)t}$

Series Expansion

 * $\displaystyle e^{At} = \sum_{n=0}^{\infty} \frac{t^n}{n!} A^n$

Decomposition

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

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

Nonvanishing Determinant
The linear system $x' = Ax$ 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, it follows it is unique. By linear independence of its columns, $\det e^{At} \ne 0$. The nonzero determinant property also follows as a corollary to Liouville's Theorem (Differential Equations).

Same-Matrix Product
Let
 * $\Phi(t) = e^{At} e^{As} - e^{A(t+s)}$

for some fixed $s \in \R$. Then
 * $\Phi'(t) = A e^{At} e^{As} - Ae^{A(t+s)}$
 * $ = A( e^{At} e^{As} - e^{A(t+s)} ) = A \Phi(t)$.

Since $\Phi(0) = e^{As} - e^{As} = 0$, it follows $\Phi(t) = e^{At} \Phi(0) = 0$ independent of $s$, hence the result holds.

Inverse
Using the Same-Matrix Product property,
 * $e^{At} e^{-At} = e^{-At} e^{At} = e^{0} = I$,

hence $e^{At}$ and $e^{-At}$ are inverses of each other.

Commutative Product (1) & (2)
The proofs let $\Phi_1(t) = e^{At} B - B e^{At}$ and $\Phi_2(t) = e^{At} e^{Bt} - e^{(A+B)t}$, 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
Since $ (P B P^{-1})^n = P B^n P^{-1}$ by induction, the result follows from plugging in the matrices and factoring $P$ and $P^{-1}$ to their respective sides.