Matrix Exponent

July 14, 2023

Given an uncontrolled system,

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

the solution could be stated as

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

where $e^{A t}$ is known as the matrix exponent.

The following rules hold for a matrix exponent

$e^{A t} e^{A s} = e^{A (t + s)}$ (semi-group property)
$A e^{A t} = e^{A t} A$
${(e^{A t})}^{- 1} = e^{- A t}$
If $P$ is a nonsingular square matrix, then $e^{P^{- 1} A P} = P^{- 1} e^{A} P$ (similarity transformation=change of the basis)
If $A = d i a g (a_{1}, \dots, a_{n})$ , then $e^{A} = d i a g (e^{a_{1}}, \dots, e^{a_{n}})$
If $A$ and $B$ commute (i.e., $A B = B A$ ), then $e^{A + B} = e^{A} e^{B}$

\begin{matrix} (1) & e^{A t} = b_{0} (t) I + b_{1} (t) A + \dots + b_{n - 1} (t) A^{n - 1} = \sum_{i = 0}^{n - 1} b_{i} (t) A^{i} \end{matrix}

The values of $b_{i} (t)$ can be found using the following expressions

\begin{array}{r} (2) & \begin{aligned} e^{λ_{1} t} & = b_{0} (t) + b_{1} (t) λ_{1} + \dots + b_{n - 1} (t) λ_{1}^{n - 1} \\ e^{λ_{2} t} & = b_{0} (t) + b_{1} (t) λ_{2} + \dots + b_{n - 1} (t) λ_{2}^{n - 1} \\ ⋮ ⋮ ⋮ \\ e^{λ_{n} t} & = b_{0} (t) + b_{1} (t) λ_{n} + \dots + b_{n - 1} (t) λ_{n}^{n - 1} \end{aligned} \end{array}

The last expression (2) could be expressed more compactly as follows

e^{λ_{j} t} = \sum_{i = 0}^{n - 1} b_{i} (t) λ_{j}^{i}, j = 1, 2, \dots, n

Given a $2 \times 2$ matrix,

A = [\begin{matrix} 1 & 2 \\ 0 & - 1 \end{matrix}]

The matrix exponential can be found using the following equation

\begin{matrix} (3) & e^{A t} = \sum_{i = 0}^{1} b_{i} (t) A^{i} = b_{0} (t) I + b_{1} (t) A \end{matrix}

So, we need to find the values of $b_{0} (t)$ and $b_{1} (t)$ using the following two equations.

\begin{array}{r} (4) & \begin{aligned} e^{λ_{1} t} & = b_{0} (t) + λ_{1} b_{1} (t) \\ e^{λ_{2} t} & = b_{0} (t) + λ_{2} b_{1} (t) \end{aligned} \end{array}

Since the eigenvalues of $A$ have the values $1$ and $- 1$ , or

λ_{1} = 1 and λ_{2} = - 1

The expression (4) could be written as follows

\begin{array}{r} (5) & \begin{aligned} e^{t} & = b_{0} (t) + b_{1} (t) \\ e^{- t} & = b_{0} t) - b_{1} (t) \end{aligned} \end{array}

And expression (5) could be expressed in matrix form as follows

[\begin{matrix} e^{t} \\ e^{- 1} \end{matrix}] = [\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix}] [\begin{matrix} b_{0} (t) \\ b_{1} (t) \end{matrix}]

Then,

\begin{aligned} [\begin{array}{c} b_{0} (t) \\ b_{1} (t) \end{array}] & = {[\begin{array}{c} 1 & 1 \\ 1 & - 1 \end{array}]}^{- 1} [\begin{array}{c} e^{t} \\ e^{- t} \end{array}] \\ = \frac{1}{2} [\begin{array}{c} 1 & 1 \\ 1 & - 1 \end{array}] [\begin{array}{c} e^{t} \\ e^{- t} \end{array}] \\ = \frac{1}{2} [\begin{array}{c} e^{t} + e^{- t} \\ e^{t} - e^{- t} \end{array}] \end{aligned}

Therefore,

\begin{array}{r} (6) & \begin{aligned} b_{0} (t) & = \frac{1}{2} (e^{t} + e^{- t}) \\ b_{1} (t) & = \frac{1}{2} (e^{t} - e^{- t}) \end{aligned} \end{array}

Then by substituting by (6) into (3) leads to

\begin{aligned} e^{A t} & = \frac{1}{2} (e^{t} + e^{- t}) [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] + \frac{1}{2} (e^{t} - e^{- t}) [\begin{array}{c} 1 & 2 \\ 0 & - 1 \end{array}] \\ = \frac{1}{2} ([\begin{array}{c} e^{t} + e^{- t} & 0 \\ 0 & e^{t} + e^{- t} \end{array}] + [\begin{array}{c} e^{t} - e^{- t} & 2 (e^{t} - e^{- t}) \\ 0 & - (e^{t} - e^{- t}) \end{array}]) \\ = \frac{1}{2} [\begin{array}{c} 2 e^{t} & 2 (e^{t} - e^{- t}) \\ 0 & 2 e^{- t} \end{array}] \\ = [\begin{array}{c} e^{t} & e^{t} - e^{- t} \\ 0 & e^{- t} \end{array}] \end{aligned}

Then,

e^{A t} = [\begin{matrix} e^{t} & e^{t} - e^{- t} \\ 0 & e^{- t} \end{matrix}]

The matrix exponential could be expressed as follows

e^{A t} = P e^{Λ t} P^{- 1}

where $P$ is the matrix encompassing the eigenvectors, or

P = [\begin{matrix} v_{1} & v_{2} & \dots & v_{n} \end{matrix}]

and $Λ$ is a diagonal matrix of all eigenvalues, or

Λ = P^{- 1} A P = d i a g (λ_{1}, λ_{2}, \dots, λ_{n})

Given a $2 \times 2$ matrix,

A = [\begin{matrix} 1 & 2 \\ 0 & - 1 \end{matrix}]

The matrix exponential can be found using the following equation

e^{A t} = P e^{Λ t} P^{- 1}

The eigenvectors of $A$ are as follows

v_{1} = [\begin{matrix} 1 \\ 0 \end{matrix}] and v_{2} = [\begin{matrix} - 1 \\ 1 \end{matrix}]

Then

P = [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}]

The value of $Λ$ is

\begin{aligned} Λ & = P^{- 1} A P \\ = [\begin{array}{c} 1 & 1 \\ 0 & 1 \end{array}] [\begin{array}{c} 1 & 2 \\ 0 & - 1 \end{array}] [\begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array}] \\ = [\begin{array}{c} 1 & 0 \\ 0 & - 1 \end{array}] \end{aligned}

Therefore, the matrix exponential of $Λ$ is

e^{Λ t} = [\begin{matrix} e^{t} & 0 \\ 0 & e^{- t} \end{matrix}]

Then

\begin{aligned} e^{A t} & = P e^{Λ t} P^{- 1} \\ = [\begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array}] [\begin{array}{c} e^{t} & 0 \\ 0 & e^{- t} \end{array}] [\begin{array}{c} 1 & 1 \\ 0 & 1 \end{array}] \\ = [\begin{array}{c} e^{t} & - e^{- t} \\ 0 & e^{- t} \end{array}] [\begin{array}{c} 1 & 1 \\ 0 & 1 \end{array}] \\ = [\begin{array}{c} e^{t} & e^{t} - e^{- t} \\ 0 & e^{- t} \end{array}] \end{aligned}

Therefore,

e^{A t} = [\begin{matrix} e^{t} & e^{t} - e^{- t} \\ 0 & e^{- t} \end{matrix}]