Properties of Matrix Exponential

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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}$


Proofs

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:

\(\ds \map {\Phi'} t\) \(=\) \(\ds \mathbf A e^{\mathbf A t} e^{\mathbf A s} - \mathbf A e^{\mathbf A \paren {t + s} }\)
\(\ds \) \(=\) \(\ds \mathbf A \paren {e^{\mathbf A t} e^{\mathbf A s} - e^{\mathbf A \paren {t + s} } }\)
\(\ds \) \(=\) \(\ds \mathbf A \map \Phi t\)


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.

$\blacksquare$