Exponential of a Matrix

The quantity e^{at} where a and t are scalar is an infinite series defined as:

e^{at} = 1 + \frac{at}{1!} + \frac{a^{2}t^{2}}{2!} + \frac{a^{3}t^{3}}{3!} + ... + \frac{a^{k}t^{k}}{k!} + ... (1)

When we have a matrix A instead of a scalar a, we can form an infinite series similar to the above equation and use it as a definition of the exponential of a matrix.

e^{At} = I + \frac{At}{1!} + \frac{A^{2}t^{2}}{2!} + \frac{A^{3}t^{3}}{3!} + ... + \frac{A^{k}t^{k}}{k!} + ... (2)

Properties of e^{At}

1. e^{At}]_{t=0} = I

2. \frac{d}{dt} (e^{At}) = A.e^{At} = e^{At}.A

3. A. (∫_{0}^{t} e^{Aj}dj) = (∫_{0}^{t} e^{Aj}dj) .A = e^{At} − I 4. e^{A(t+s)} = e^{At}e^{As}

5. e^{At}e^{−At} = e^{−At}e^{At} = I

Solution of Linear Matrix Differential Equation

Using the properties of e^{At}, we can find the solution of a linear matrix differential equation with constant coefficients.

\frac{dX}{dt} − AX = BU  (3)

Multiplying both sides by e^{−At} from the left, we get

e^{−At} \frac{dX}{dt} − e^{−At}AX = e^{−At}BU  (4)

Or

\frac{d}{dt} (e^{−At}.X) = e^{−At}BU  (5)

Integrating both sides from time 0 to t, we obtain

e^{−At}X(t) − IX(0) = ∫_{0}^{t} e^{−At}BU(t)dt  (6)

Solving for X(t), we get

X(t) = e^{At}.X(0) + e^{At}.∫_{0}^{t} e^{−At}BU(t)dt  (7)

The solution of the differential equation is given by:

X(t) = e^{At}.X(0) + e^{At}.∫_{0}^{t} e^{−At}BU(t)dt

Note that X(t) is a vector of dimension n e^{At} is a nxn matrix; thus e^{At}.X(0) is a vector of dimension n B is a nxm matrix and U is a vector of

dimension m; thus e^{At}.∫_{0}^{t} e^{−At}BU(t)dt is a vector of dimension n If we can calculate e^{At}, we can solve for X(t).

Analytical Calculation of e^{At}

If A is an nxn matrix and P_{A}(λ) = λ^{n} + a_{1}λ^{n−1} + ... + a_{n−1}λ + a_{n} is its characteristic polynomial, then

e^{At} = Ψ_{0}(t)I + Ψ_{1}(t)A + ... + Ψ_{n−1}A^{n−1}  (8)

where Ψ_{0}(t), Ψ_{1}(t), ..., Ψ_{n−1}(t) are scalar functions of time.

The functions Ψ_{i}(t) can be computed from the eigenvalues of the matrix A and for n = 2, 3, this computation can be done easily as shown in the next slide.

Analytical Calculation of e^{At} when n = 2

Suppose the eigenvalues of the matrix A are λ_{1} and λ_{2}.

1. When λ_{1} ≠ λ_{2}

\begin{aligned}Ψ_{0}(t) & = & −\frac{λ_{2}}{λ_{1} − λ_{2}} e^{λ_{1}t} − \frac{λ_{1}}{λ_{2} − λ_{1}} e^{λ_{2}t} \\ Ψ_{1}(t) & = & \frac{1}{λ_{1} − λ_{2}} e^{λ_{1}t} + \frac{1}{λ_{2} − λ_{1}} e^{λ_{2}t} \end{aligned} (9)

2. When λ_{1} = λ_{2}

\begin{aligned}Ψ_{0}(t) & = & e^{λ_{1}t} − λ_{1}te^{λ_{1}t} \\ Ψ_{1}(t) & = & te^{λ_{1}t} \end{aligned}  (11)

Example 1

Calculate e^{At} for

A = [ \begin{matrix}1 & 2 \\ 4 & 3 \end{matrix} ]

Eigenvalues of A:

det(λI − A) = det [ \begin{matrix}λ − 1 & −2 \\ −4 & λ − 3 \end{matrix} ]
= λ^{2} − 4λ − 5
= (λ − 5)(λ + 1)

Thus λ_{1} = 5, λ_{2} = −1.

Substituting in the formulae for Ψ_{0}(t) and Ψ_{1}(t)

\begin{aligned}Ψ_{0} & = & \frac{1}{6} (e^{5t} + 5e^{−t}) \\ Ψ_{1} & = & \frac{1}{6} (e^{5t} − e^{−t}) \end{aligned}

Thus,

\begin{aligned}e^{At} & = & Ψ_{0}(t)I + Ψ_{1}(t)A \\ & = & [ \begin{matrix}\frac{1}{3}e^{5t} + \frac{2}{3}e^{−t} & \frac{1}{3}e^{5t} − \frac{1}{3}e^{−t} \\ \frac{2}{3}e^{5t} − \frac{2}{3}e^{−t} & \frac{2}{3}e^{5t} + \frac{1}{3}e^{−t} \end{matrix} ] \end{aligned}

Example 2

Calculate e^{At} for

A = [ \begin{matrix}1 & 2 \\ −8 & −5 \end{matrix} ]

Eigenvalues of A:

\begin{aligned}det(λI − A) & = & det [ \begin{matrix}λ − 1 & −2 \\ 8 & λ + 5 \end{matrix} ] \\ & = & λ^{2} + 4λ + 11 \end{aligned}

Thus λ_{1} = −2 + i\sqrt{7} λ_{2} = −2 − i\sqrt{7} Substituting in the formulae for Ψ_{0}(t) and Ψ_{1}(t)

\begin{aligned}Ψ_{0} & = & \frac{(2 + i\sqrt{7})e^{(−2+i\sqrt{7})t} − (2 − i\sqrt{7})e^{(−2−i\sqrt{7})t}}{2i\sqrt{7}} \\ Ψ_{1} & = & \frac{e^{(−2+i\sqrt{7})t} − e^{(−2−i\sqrt{7})t}}{2i\sqrt{7}} \end{aligned}

Using the formula

\begin{aligned}\frac{e^{iθ} + e^{−iθ}}{2} & = & Cos(θ) \\ \frac{e^{iθ} − e^{−iθ}}{2i} & = & Sin(θ) \end{aligned}  (13)

we get

\begin{aligned}Ψ_{0}(t) & = & e^{−2t} [cos(\sqrt{7}t) + \frac{2}{\sqrt{7}}sin(\sqrt{7}t)] \\ Ψ_{1}(t) & = & \frac{e^{−2t}}{\sqrt{7}} sin(\sqrt{7}t) \end{aligned}

Thus

\begin{matrix}e^{At} = Ψ_{0}(t)I + Ψ_{1}(t)A \\ = e^{−2t} [ \begin{matrix}cos(\sqrt{7}t) + \frac{3}{\sqrt{7}}sin(\sqrt{7}t) & \frac{2}{\sqrt{7}}sin(\sqrt{7}t) \\ −\frac{8}{\sqrt{7}}sin(\sqrt{7}t) & cos(\sqrt{7}t) − \frac{3}{\sqrt{7}}sin(\sqrt{7}t) \end{matrix} ] \end{matrix}

Analytical Calculation of e^{At} when n = 2

Suppose the eigenvalues of the matrix A are complex.

\begin{aligned}λ_{1} & = & R + iΩ \\ λ_{2} & = & R − iΩ \end{aligned}  (14)

Then,

\begin{aligned}Ψ_{0}(t) & = & (cos(Ωt) − \frac{R}{Ω}Sin(Ωt)) e^{Rt} \\ Ψ_{1}(t) & = & (\frac{1}{Ω}Sin(Ωt)) e^{Rt} \end{aligned}  (15)