Dynamical Systems:

Matrix Exponential

Table of Contents





1. Matrix Exponential

The matrix exponential provides a powerful tool for solving systems of linear differential equations, generalizing the familiar scalar exponential function.


Scalar Case: A Starting Point

Consider the first-order linear scalar differential equation:


$$ \dot{u}(t) = a u(t). $$


Its solution is:


$$ u(t) = e^{at} u(0). $$



Extension to Vector Case (Matrix)

For a linear system of ODEs:


$$ \dot{\vec{u}}(t) = A \vec{u}(t), $$


the solution is:


$$ \vec{u}(t) = e^{A t} \vec{u}(0), $$


where $e^{A t}$ is the matrix exponential.



1.1. Matrix Exponential

Matrix Exponential of a Diagonal Matrix (Intuition)

If $A = \Lambda$ is diagonal:


$$ \Lambda = \begin{bmatrix} \lambda_1 & & \\ & \ddots & \\ & & \lambda_n \end{bmatrix}, $$


then:


$$ e^{\Lambda t} = \begin{bmatrix} e^{\lambda_1 t} & & 0 \\ & \ddots & \\ 0 & & e^{\lambda_n t} \end{bmatrix}. $$



Diagonalization and Similarity

If $A$ is diagonalizable, we have:


$$ A = S \Lambda S^{-1}, \quad \text{where } \Lambda \text{ is diagonal}, S \text{ contains eigenvectors}. $$


Thus:


$$ e^{A t} = S e^{\Lambda t} S^{-1}. $$


The Exponential Function

Recall the scalar exponential series:


$$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots. $$


The derivative satisfies:


$$ \frac{d}{dx} e^x = e^x, $$


which is why $e^x$ solves $\dot{y} = y$.


The Matrix Exponential Series

Similarly, for matrices:


$$ e^A = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots, $$


and:


$$ e^{A t} = I + A t + \frac{A^2 t^2}{2!} + \frac{A^3 t^3}{3!} + \cdots. $$


Differentiating the matrix exponential with respect to $t$:


$$ \frac{d}{dt} e^{A t} = A e^{A t}. $$


Similarity Transformation of Exponentials

If $A = S \Lambda S^{-1}$, then:


$$ \begin{align*} e^{A} &= e^{S \Lambda S^{-1}} \\ &= I + S \Lambda S^{-1} + \frac{S \Lambda^2 S^{-1}}{2!} + \cdots \\ &= S \left( I + \Lambda + \frac{\Lambda^2}{2!} + \cdots \right) S^{-1} \\ &= S e^{\Lambda} S^{-1}. \end{align*} $$


Similarly:


$$ e^{A t} = S e^{\Lambda t} S^{-1}. $$


This framework connects solutions of linear systems, eigenvalues, and eigenvectors, and leads naturally to applications such as system stability analysis, modal decomposition, and control design.




1.2. Examples of Matrix Exponential

We now illustrate matrix exponentials with explicit examples.


Example 1: Diagonal Matrix

Let:


$$ A = \begin{bmatrix} 2 & 0 \\ 0 & -1 \end{bmatrix}. $$


Since $A$ is diagonal, its exponential is:


$$ e^{A t} = \begin{bmatrix} e^{2t} & 0 \\ 0 & e^{-t} \end{bmatrix}. $$


Thus, the solution of:


$$ \dot{\vec{u}} = A \vec{u}, \quad \vec{u}(0) = \begin{bmatrix} u_1(0) \\ u_2(0) \end{bmatrix}, $$


is:


$$ \vec{u}(t) = \begin{bmatrix} e^{2t} u_1(0) \\ e^{-t} u_2(0) \end{bmatrix}. $$



Example 2: Non-Diagonalizable Matrix

Let:


$$ A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}. $$


Compute:


$$ A^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}. $$


Thus:


$$ e^{A t} = I + A t = \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}. $$


The system:


$$ \dot{\vec{u}} = A \vec{u}, \quad \vec{u}(0) = \begin{bmatrix} u_1(0) \\ u_2(0) \end{bmatrix}, $$


has solution:


$$ \vec{u}(t) = \begin{bmatrix} u_1(0) + t u_2(0) \\ u_2(0) \end{bmatrix}. $$



Example 3: Diagonalizable Matrix

Let:


$$ A = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix}. $$


Find eigenvalues: $\lambda_1=3$, $\lambda_2=2$.


Eigenvectors:


$$ \vec{x}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad \vec{x}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}. $$


Thus:


$$ S = \begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}, \quad \Lambda = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}. $$


We can compute:


$$ S^{-1} = \begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}^{-1} = \frac{1}{-1} \begin{bmatrix} -1 & -1 \\ 0 & 1 \end{bmatrix}. $$


Then:


$$ e^{A t} = S e^{\Lambda t} S^{-1} = \begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} e^{3t} & 0 \\ 0 & e^{2t} \end{bmatrix} S^{-1}. $$





2. MATLAB Implementation

Consider the state-space system:


$$ \begin{align*} \left[ {\begin{matrix} \dot{x}_1 \\ \dot{x}_2 \\ \end{matrix} } \right] &= \left[ {\begin{matrix} 0 & 1 \\ -\omega_n^2 & -2\zeta \omega_n \\ \end{matrix} } \right] \left[ {\begin{array}{cc} x_1 \\ x_2 \\ \end{array} } \right] \\ \\y & = \left[ {\begin{matrix} 1 & 0 \end{matrix} } \right] \left[ {\begin{array}{cc} x_1 \\ x_2 \\ \end{array} } \right] \end{align*} $$


With parameters: $\omega_n = 2$, and $\zeta = 0.1$


MATLAB Code

In [ ]:
% method 1: use expm (matrix exponential)

wn = 2; zeta = 0.1;
A = [0 1;-wn^2 -2*zeta*wn];

x0 = [1; 0];
t = linspace(0,20,200);
x = [];

for i = 1:length(t)
    x = [x, expm(A*t(i))*x0];
end

plot(t,x(1,:),t,zeros(size(t)),'--k')












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