2.2 Exponential of matrices

The IVP in one-dimensional space

$$x' = ax, \quad x(0) = x_{0}, \quad x_{0} \in R,$$

has a unique solution x(t) = x₀ e^at, $t \geq 0$. It is natural to ask whether we can define the ``exponential of a matrix'' so that the IVP in $\Real^{n}$

$$x' = Ax, \quad x(0) = x_{0}, \quad x_{0} \in \Real^{n}$$ (2.3)

has a unique solution x(t) = e^At x₀, $t \geq 0$. Before answering this question, we need to define the norm of a linear operator and the exponential of matrices. Let $T : R^{n} \raro R^{n}$ be a linear operator. We define the norm of T by

$$\parallel T \parallel = \max_{\mid x \mid \, \leq 1} \mid T (x) \mid \, ,$$

where $\mid \cdot \mid$ denotes the Euclidean norm. Let the linear operator T on Rⁿ be represented by the $n \times n$ matrix A with respect to a basis for Rⁿ. Then the norm of A is defined by

$$\parallel A \parallel = \parallel T \parallel \, .$$

Let L(Rⁿ) be the space of all linear operators on Rⁿ. Then for $S, \, T \in L(R^{n})$, $x \in R^{n}$,

(1)

$\parallel T \parallel \geq 0$ and $\parallel T \parallel = 0$ iff T=0;

(2)

$\parallel kT \parallel = \mid k \mid \; \parallel T \parallel \quad \forall \,\, k \in R$;

(3)

$\parallel S + T \parallel \, \leq \, \parallel S \parallel \, + \, \parallel T \parallel$;

(4)

$\mid T (x) \mid \, \leq \, \parallel T \parallel \; \mid x \mid$ $\forall \, x \in \Real^{n}$;

(5)

$\parallel TS \parallel \, \leq \, \parallel T \parallel \; \parallel S \parallel$;

(6)

$\parallel T^{k} \parallel \, \leq \, \parallel T \parallel^{k}$ $k = 0, 1,2, \ldots$ .

Proof: (omitted)

Lemma 2.2.1:     Let A be an $n \times n$ matrix and $r \in R_{+}$. Then the series

$$\sum_{k =0}^{\infty} \frac{A^{k}t^{k}}{k!}$$

is absolutely and uniformly convergent for $\mid t \mid \, \leq \, r$.

Proof: (omitted)

Definition 2.2.1:     Let A be an $n \times n$ matrix. Then $\forall \; t \in R$, we define

$$e^{At} = \sum_{k=0}^{\infty} \frac{A^{k}t^{k}}{k!} \, .$$

Let $A, \, B$ and P be $n \times n$ matrices where P is nonsingular. Then

(1)

e⁰ = I, 0 being $n \times n$ zero matrix;

(2)

e^A+B = e^A e^B, provided AB = BA;

(3)

e^A is invertible and (e^A)^-1 = e^-A;

(4)

P^-1 AP = B implies e^B = P^-1 e^A P.

Remark:     e^At is an $n \times n$ matrix whose entries are limits of the corresponding n² infinite series.

Theorem 2.2.1:     Let A be an $n \times n$ matrix. Then IVP (2.3) has a unique solution for all $t \in R$ which is given by

$$x (t) = e^{At} x_{0} \, .$$

Proof: (Omitted)

Corollary 2.2.1     By Theorem 2.2.1, we have

(1)

$e^{\tiny\left[ \begin{array}{ll} \lambda_{1} & 0 \\ 0 & \lambda_{2} \end{array}... ...{array}{ll} e^{\lambda_{1}t} & 0 \\ 0 & e^{\lambda_{2}t} \end{array}\right]\, ;$

(2)

$e^{\tiny\left[ \begin{array}{ll} \lambda & 1 \\ 0 & \lambda \end{array} \right]t} = e^{\lambda t} \left[ \begin{array}{ll} 1 & t \\ 0 & 1 \end{array} \right]\, ;$

(3)

$e^{\tiny\left[ \begin{array}{rl} \alpha & \beta \\ - \beta & \alpha \end{array}... ...cos \beta t & \sin \beta t \\ - \sin \beta t & \cos \beta t \end{array} \right]$.

Corollary 2.2.2     Let P be an $n \times n$ nonsingular matrix. If P^-1 AP = B, then x(t) = Pe^Bt P^-1 x₀ is a solution of the IVP (2.3).

Exercise Perko, P. 19-20; 4, 6, 7, 8.