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
$$\frac{d}{dt} e^{At} = A e^{At}$$

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

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 $$s \in \mathbb{R}$$. 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.