1. Matrix Exponential¶
- Starting from single variable (scalar case)
$$\dot{u}=au \quad \implies \quad u(t)=e^{at}u(0)$$
- Extending to multivariate case (matrix)
$$\dot{\vec{u}}=A\vec{u} \quad \implies \quad \vec{u}(t) = \underbrace{e^{At}}_{\text{matrix exponential}}\vec{u}(0)$$
- Matrix Exponential with Diagonal Matrix (intuitive)
$$\begin{align*}
e^{\Lambda t} &= \exp{\left( \,
\begin{bmatrix}
\lambda_1 & &\\
& \ddots & \\
& & \lambda_n
\end{bmatrix} t \,\right)} \\ \\
& = \begin{bmatrix}
e^{\lambda_1 t} & & 0 \\
& \ddots & \\
0 & & e^{\lambda_n t}
\end{bmatrix}
\end{align*}
$$
$$\begin{align*}
S &= \begin{bmatrix}
\vec{x}_1 & \vec{x}_2
\end{bmatrix} \quad \text{ where } \vec{x}_i \text{ is eigenvectors}\\ \\
AS &= S\Lambda\\
\implies
\Lambda &= S^{-1}AS \\
\implies
A &= S\Lambda S^{-1}
\end{align*}
$$
$$
\begin{align*}
e^x &= 1+x+{x^2\over2!}+{x^3\over3!}+\cdots+{x^k\over k!} + \cdots \\
e^0 &= 1 \\ \\
{d \over dx}e^x &= 0+1+x+{x^2\over2!}+\cdots+{x^{k-1}\over ({k-1})!} + \cdots \\
& = e^x \\ \\
\therefore \; e^x &\text{ is a solution of } \dot{y}(x) = y(x)
\end{align*}
$$
$$\begin{align*}
e^A &= I+A+{A^2\over2!}+{A^3\over3!}+\cdots \\ \\
e^{At} &= I+At+{A^2t^2\over2!}+{A^3t^3\over3!}+\cdots
\end{align*}
$$
$$\frac{d}{dt}e^{At} = \frac{d}{dt} \sum_{k=0}^{\infty} \frac{(At)^k}{k!} = 0 + \sum_{k=1}^{\infty} \frac{kA^kt^{k-1}}{k!} = A\sum_{k=1}^{\infty} \frac{(At)^{k-1}}{(k-1)!} = A \sum_{k=0}^{\infty} \frac{(At)^k}{k!} = A e^{At}$$
- Similarity transformation
$$\begin{align*}
e^A = e^{S\Lambda S^{-1}} &= I + S\Lambda S^{-1} + {S\Lambda^2 S^{-1} \over 2!} + {S\Lambda^3 S^{-1} \over 3!} + \cdots \\
& = S
\begin{bmatrix}
I + \Lambda + {\Lambda^2 \over 2!} + {\Lambda^3 \over 3!} + \cdots
\end{bmatrix}
S^{-1}\\
& = Se^{\Lambda}S^{-1}\\ \\
e^{At} = e^{S \Lambda S^{-1} t} & = Se^{\Lambda t}S^{-1}
\end{align*}
$$