2.1 Linear Systems and Matrix Exponentials

General Linear Dynamical System

can we written in the form

\dot{x} (t) = A x (t), A \in R^{n \times n}

Where it is autonomous

Matrix Exponential

A solution to

\dot{x} (t) = A x (t)

where $x (t_{0}) = x_{0}$ is a function $x^{*} (t)$ defined on $R$ such that:
$\forall t \in R, x^{*} (t)$ satisfies the equation
$x^{*} (t)$ satisfies the initial condition

Since $A$ is a constant matrix it is globally Lipschitz so the flow (solution) exists and is unique for the time interval on all of $R$ .

Given a matrix $A \in R^{2 \times 2}$ , the time-dependent matrix

e^{A t} := I + t A + \frac{t^{2} A^{2}}{2!} + \dots + \frac{t^{k} A^{k}}{k!} = \sum_{k = 0}^{\infty} \frac{t^{k} A^{k}}{k!}, t \in R

is called the matrix exponential of the matrix $A$ .

Matrix Exponential Commute Theorem

If $A, B \in R^{n \times n}$ satisfy $A B = B A$ , then

e^{(A + B) t} = e^{A t} e^{B t} = e^{B t} e^{A t} .

Matrix Exponential Solution

Let $A \in R^{n \times n}$ and $x_{0} \in R^{n}$ .
Then the unique solution of the initial value problem

\dot{x} (t) = A x (t), x (0) = x_{0}

x (t) = e^{A t} x_{0} .

Proof:

Differentiate $x (t)$ directly:

\frac{d}{d t} x (t) = \frac{d}{d t} (e^{A t} x_{0}) = \frac{d}{d t} (\sum_{k = 0}^{\infty} \frac{t^{k} A^{k}}{k!}) x_{0} .

Differentiate term-by-term (justified by uniform convergence):

\frac{d}{d t} x (t) = (\sum_{k = 1}^{\infty} \frac{k t^{k - 1} A^{k}}{k!}) x_{0} = (\sum_{k = 1}^{\infty} \frac{t^{k - 1} A^{k}}{(k - 1)!}) x_{0} .

Factor out $A$ :

\frac{d}{d t} x (t) = A (\sum_{k = 1}^{\infty} \frac{t^{k - 1} A^{k - 1}}{(k - 1)!}) x_{0} = A (\sum_{m = 0}^{\infty} \frac{t^{m} A^{m}}{m!}) x_{0} = A e^{A t} x_{0} = A x (t) .

x (0) = e^{A \cdot 0} x_{0} = I x_{0} = x_{0} .

Thus, $x (t) = e^{A t} x_{0}$ satisfies both the differential equation and the initial condition, proving it is the unique solution.

Basis of the Solution Space

Define

F_{A} = {x (t), t \in R ∣ \dot{x} (t) = A x (t), \forall t \in R} .

Then $F_{A}$ is an $n$ -dimensional vector space.
Moreover, if ${x_{k}}_{k = 1}^{n}$ is a basis of $R^{n}$ , then

x_{k} (t) = e^{A t} e_{k}, 1 \leq k \leq n

form a basis of $F_{A}$ .

Similar Transformations

Let $A \in R^{n \times n}$ be a matrix and $P \in R^{n \times n}$ any invertible matrix.
Define

B = P^{- 1} A P .

Then:

$A$ and $B$ have the same eigenvalues.
If $λ$ is an eigenvalue of $A$ and $x_{λ}$ is the corresponding eigenvector
$(A x_{λ} = λ x_{λ})$ , then $P^{- 1} x_{λ}$ is an eigenvector of $B$ with the same eigenvalue $λ$ .
If $y (t)$ solves the IVP

\dot{y} (t) = B y (t), y (0) = y_{0} \in R^{n},

then $x (t) = P y (t)$ solves

\dot{x} (t) = A x (t), x (0) = P y_{0} .

Application: Change of Variables

Using the theorem above, we can solve linear systems via:

Apply a linear change of variables $x = P y$ so that $A$ is transformed into a simpler matrix $B$ .
Compute $e^{B t}$ explicitly to solve for $y (t)$ .
Transform back to get $x (t) = P y (t)$ .

Example 1 — Diagonal matrix

Suppose

B = [\begin{matrix} λ_{1} & 0 \\ 0 & λ_{2} \end{matrix}] .

Use the series

e^{B t} = \sum_{k = 0}^{\infty} \frac{t^{k}}{k!} B^{k} .

Because $B^{k} = diag (λ_{1}^{k}, λ_{2}^{k})$ ,

\begin{aligned} e^{B t} & = \sum_{k = 0}^{\infty} \frac{t^{k}}{k!} B^{k} \\ = \sum_{k = 0}^{\infty} \frac{t^{k}}{k!} diag (λ_{1}^{k}, λ_{2}^{k}) \\ = diag (\sum_{k = 0}^{\infty} \frac{(λ_{1} t)^{k}}{k!}, \sum_{k = 0}^{\infty} \frac{(λ_{2} t)^{k}}{k!}) \\ = diag (e^{λ_{1} t}, e^{λ_{2} t}) . \end{aligned}

Example 2 — Scaling–Rotation Matrix

Suppose

B = [\begin{matrix} a & b \\ - b & a \end{matrix}] = a I + b J, J := [\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}] .

Note that $J^{2} = - I$ , $J^{3} = - J$ , $J^{4} = I$ , and $I$ commutes with $J$ .

Then

e^{B t} = e^{(a I + b J) t} = e^{a t I} e^{b t J} = e^{a t} e^{b t J}

Expand $e^{b t J}$ and split even and odd powers:

\begin{aligned} e^{b t J} & = \sum_{k = 0}^{\infty} \frac{(b t)^{k}}{k!} J^{k} \\ = \sum_{m = 0}^{\infty} \frac{(b t)^{2 m}}{(2 m)!} J^{2 m} + \sum_{m = 0}^{\infty} \frac{(b t)^{2 m + 1}}{(2 m + 1)!} J^{2 m + 1} \\ = \sum_{m = 0}^{\infty} \frac{(b t)^{2 m}}{(2 m)!} (- 1)^{m} I + \sum_{m = 0}^{\infty} \frac{(b t)^{2 m + 1}}{(2 m + 1)!} (- 1)^{m} J \\ = \cos (b t) I + \sin (b t) J . \end{aligned}

Therefore,

e^{B t} = e^{a t} (\cos (b t) I + \sin (b t) J) = e^{a t} [\begin{matrix} \cos (b t) & \sin (b t) \\ - \sin (b t) & \cos (b t) \end{matrix}] .

Example 3 — Degenerate (Jordan) Matrix

Suppose

B = [\begin{matrix} λ & 1 \\ 0 & λ \end{matrix}] = λ I + N, N := [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}], N^{2} = 0.

Since $N$ is nilpotent of index $2$ ,

e^{B t} = e^{(λ I + N) t} = e^{λ t I} e^{t N} = e^{λ t} e^{t N} .

Expand $e^{t N}$ :

e^{t N} = \sum_{k = 0}^{\infty} \frac{t^{k}}{k!} N^{k} = I + t N (because N^{2} = 0) .

Hence

e^{B t} = e^{λ t} (I + t N) = e^{λ t} [\begin{matrix} 1 & t \\ 0 & 1 \end{matrix}] = [\begin{matrix} e^{λ t} & t e^{λ t} \\ 0 & e^{λ t} \end{matrix}] .

Similar Matrices

Use the series $e^{t M} = \sum_{k = 0}^{\infty} \frac{t^{k}}{k!} M^{k}$ and $B^{k} = (P^{- 1} A P)^{k} = P^{- 1} A^{k} P$ .

\begin{aligned} e^{t B} & = \sum_{k = 0}^{\infty} \frac{t^{k}}{k!} B^{k} \\ = \sum_{k = 0}^{\infty} \frac{t^{k}}{k!} P^{- 1} A^{k} P \\ P e^{t B} P^{- 1} & = \sum_{k = 0}^{\infty} \frac{t^{k}}{k!} P (P^{- 1} A^{k} P) P^{- 1} \\ = \sum_{k = 0}^{\infty} \frac{t^{k}}{k!} A^{k} = e^{t A} . \end{aligned}

So $e^{t A} = P e^{t B} P^{- 1}$