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Examples of derivations of matrix exponentials#
2-by-2 diagonal matrix (with complex diagonal entries)#
\[\begin{split}
A = \begin{bmatrix}
\alpha_1 & 0\\
0 & \alpha_2
\end{bmatrix}, \qquad \alpha_1,\alpha_2 \in \mathbb{C}.
\end{split}\]
\[
e^{At} = \sum_{k=0}^{\infty} \frac{(At)^k}{k!}
= \sum_{k=0}^{\infty} \frac{t^k A^k}{k!}.
\]
\[\begin{split}
A^k = \begin{bmatrix}
\alpha_1^k & 0\\
0 & \alpha_2^k
\end{bmatrix}
\quad\Rightarrow\quad
t^k A^k =
\begin{bmatrix}
(\alpha_1 t)^k & 0\\
0 & (\alpha_2 t)^k
\end{bmatrix}.
\end{split}\]
\[\begin{split}
e^{At}
= \sum_{k=0}^{\infty} \frac{1}{k!}
\begin{bmatrix}
(\alpha_1 t)^k & 0\\
0 & (\alpha_2 t)^k
\end{bmatrix}
= \begin{bmatrix}
\displaystyle \sum_{k=0}^{\infty} \frac{(\alpha_1 t)^k}{k!} & 0\\[10pt]
0 & \displaystyle \sum_{k=0}^{\infty} \frac{(\alpha_2 t)^k}{k!}
\end{bmatrix}.
\end{split}\]
\[\begin{split}
e^{At}
= \begin{bmatrix}
e^{\alpha_1 t} & 0\\
0 & e^{\alpha_2 t}
\end{bmatrix}.
\end{split}\]
2-by-2 skew-symmetric matrix (with zero diagonal entries and real off-diagonal entries)#
\[\begin{split}
A = \begin{bmatrix}
0 & \beta \\
-\beta & 0
\end{bmatrix}, \qquad \beta \in \mathbb{R}.
\end{split}\]
\[
e^{At} = \sum_{k=0}^{\infty} \frac{(At)^k}{k!}.
\]
\[\begin{split}
A^2
= \begin{bmatrix}
0 & \beta \\
-\beta & 0
\end{bmatrix}^2
= \begin{bmatrix}
-\beta^2 & 0\\
0 & -\beta^2
\end{bmatrix}
= -\beta^2 I.
\end{split}\]
\[
A^3 = A A^2 = -\beta^2 A,
\qquad
A^4 = A^2 A^2 = \beta^4 I,
\]
\[
A^{2k} = (-1)^k \beta^{2k} I,
\qquad
A^{2k+1} = (-1)^k \beta^{2k} A.
\]
\[
(At)^{2k} = t^{2k} A^{2k}
= (-1)^k (\beta t)^{2k} I,
\]
\[
(At)^{2k+1} = t^{2k+1} A^{2k+1}
= (-1)^k (\beta t)^{2k} t A.
\]
\[
e^{At}
= \sum_{k=0}^{\infty} \frac{(At)^{2k}}{(2k)!}
+ \sum_{k=0}^{\infty} \frac{(At)^{2k+1}}{(2k+1)!}.
\]
\[
e^{At}
= \sum_{k=0}^{\infty} \frac{(-1)^k (\beta t)^{2k}}{(2k)!} I
\;+\;
\sum_{k=0}^{\infty} \frac{(-1)^k (\beta t)^{2k}}{(2k+1)!} t A.
\]
\[
\sum_{k=0}^{\infty} \frac{(-1)^k (\beta t)^{2k}}{(2k)!} = \cos(\beta t),
\]
\[
\sum_{k=0}^{\infty} \frac{(-1)^k (\beta t)^{2k} t}{(2k+1)!}
= \frac{\sin(\beta t)}{\beta}.
\]
\[
e^{At}
= \cos(\beta t)\, I
+ \frac{\sin(\beta t)}{\beta} A.
\]
\[\begin{split}
e^{At}
= \begin{bmatrix}
\cos(\beta t) & \sin(\beta t) \\
-\sin(\beta t) & \cos(\beta t)
\end{bmatrix}.
\end{split}\]
2-by-2 real, skew-symmetric matrix with diagonal entries equal to each other#
\[\begin{split}
A =
\begin{bmatrix}
\alpha & \beta \\
-\beta & \alpha
\end{bmatrix}, \qquad \alpha,\beta \in \mathbb{R}.
\end{split}\]
\[\begin{split}
I = \begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix},
\qquad
J = \begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}.
\end{split}\]
\[
A = \alpha I + \beta J.
\]
\[
e^{At}
= e^{(\alpha I + \beta J)t}
= e^{\alpha t I + \beta t J}.
\]
\[
IJ = JI \quad\Rightarrow\quad
e^{\alpha t I + \beta t J}
= e^{\alpha t I} e^{\beta t J}.
\]
\[
e^{\alpha t I} = e^{\alpha t} I
\quad\text{(case 1 with } \alpha_1=\alpha_2=\alpha\text{)}.
\]
\[\begin{split}
e^{\beta t J}
= \begin{bmatrix}
\cos(\beta t) & \sin(\beta t) \\
-\sin(\beta t) & \cos(\beta t)
\end{bmatrix}
\quad\text{(case 2 with } A=J\text{)}.
\end{split}\]
\[\begin{split}
e^{At}
= e^{\alpha t} I \,
e^{\beta t J}
= e^{\alpha t}
\begin{bmatrix}
\cos(\beta t) & \sin(\beta t) \\
-\sin(\beta t) & \cos(\beta t)
\end{bmatrix}.
\end{split}\]
Special cases#
\(e^{A t}\) for diagonal \(A\)#
Let
\[\begin{split} A = \begin{bmatrix}
\beta_1 & 0 & \cdots & 0 \\
0 & \beta_2 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \beta_n
\end{bmatrix}, \end{split}\]
where \(\beta_1, \dots, \beta_n \in \mathbb{R}\).
Then
\[\begin{split} A^k = \begin{bmatrix}
\beta_1^k & 0 & \cdots & 0 \\
0 & \beta_2^k & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \beta_n^k
\end{bmatrix}. \end{split}\]
Therefore, for \(i = 1,\dots,n\),
\[
(e^{A t})_{ii}
= 1 + t \beta_i + { t^2 \beta_i^2 }/{ 2! } + { t^3 \beta_i^3 }/{ 3! } + \cdots
= e^{ \beta_i t }.
\]
\[(e^{A t})_{i j} = 0 \text{ for } i \neq j.\]
So
\[\begin{split} e^{A t} = \begin{bmatrix}
e^{\beta_1 t} & 0 & \cdots & 0 \\
0 & e^{\beta_2 t} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & e^{\beta_n t}
\end{bmatrix}. \end{split}\]
\(e^{A t}\) for block-diagonal \(A\)#
Consider a block-diagonal matrix
\[\begin{split} A = \begin{bmatrix}
A_1 & 0 & \cdots & 0 \\
0 & A_2 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & A_N
\end{bmatrix}, \end{split}\]
where each \(A_i \in \mathbb{R}^{n_i \times n_i}\).
Then
\[\begin{split} A^k = \begin{bmatrix}
A_1^k & 0 & \cdots & 0 \\
0 & A_2^k & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & A_N^k
\end{bmatrix}. \end{split}\]
From the series definition,
\[\begin{split} e^{A t} = \begin{bmatrix}
e^{A_1 t} & 0 & \cdots & 0 \\
0 & e^{A_2 t} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & e^{A_N t}
\end{bmatrix}. \end{split}\]
\(e^{A t}\) for a \(2\times 2\) Jordan block#
Let
\[\begin{split}
A =
\begin{bmatrix}
\lambda & 1 \\
0 & \lambda
\end{bmatrix},
\qquad \lambda \in \mathbb{R}.
\end{split}\]
Compute powers of \(A\)#
First,
\[\begin{split}
A^2 = A A =
\begin{bmatrix}
\lambda & 1 \\
0 & \lambda
\end{bmatrix}
\begin{bmatrix}
\lambda & 1 \\
0 & \lambda
\end{bmatrix}
=
\begin{bmatrix}
\lambda^2 & 2\lambda \\
0 & \lambda^2
\end{bmatrix}.
\end{split}\]
Next,
\[\begin{split}
A^3 = A^2 A =
\begin{bmatrix}
\lambda^2 & 2\lambda \\
0 & \lambda^2
\end{bmatrix}
\begin{bmatrix}
\lambda & 1 \\
0 & \lambda
\end{bmatrix}
=
\begin{bmatrix}
\lambda^3 & 3\lambda^2 \\
0 & \lambda^3
\end{bmatrix}.
\end{split}\]
We see the pattern:
\[\begin{split}
A^k =
\begin{bmatrix}
\lambda^k & k \lambda^{k-1} \\
0 & \lambda^k
\end{bmatrix}.
\end{split}\]
Compute \(e^{A t}\) from the power series#
By definition:
\[
e^{A t}
= I + tA + {t^2 A^2}/{2!} + {t^3 A^3}/{3!} + \cdots.
\]
Write this entrywise.
\[\begin{split}
e^{A t}
=
\begin{bmatrix}
1 + \lambda t + {\lambda^2 t^2}/{2!} + {\lambda^3 t^3}/{3!} + \cdots
&
t
+ 2\,{\lambda t^2}/{2!}
+ 3\,{\lambda^2 t^3}/{3!}
+ 4\,{\lambda^3 t^4}/{4!}
+ \cdots
\\[10pt]
0
&
1 + \lambda t + {\lambda^2 t^2}/{2!} + {\lambda^3 t^3}/{3!} + \cdots
\end{bmatrix}.
\end{split}\]
Notice the pattern and factorizing:
\[\begin{split}
e^{A t}
=
\begin{bmatrix}
e^{\lambda t} & t e^{\lambda t} \\
0 & e^{\lambda t}
\end{bmatrix}.
\end{split}\]
\(e^{At}\) for an \(n\times n\) Jordan block#
Consider now the general \(n \times n\) Jordan block
\[\begin{split}
A =
\begin{bmatrix}
\lambda & 1 & 0 & \cdots & 0\\
0 & \lambda & 1 & \ddots & \vdots\\
\vdots & \ddots & \ddots & \ddots & 0\\
\vdots & & \ddots & \lambda & 1\\
0 & \cdots & \cdots & 0 & \lambda
\end{bmatrix}.
\end{split}\]
It follows that the previous result generalizes to
\[\begin{split}
e^{At} =
\begin{bmatrix}
e^{\lambda t} & t e^{\lambda t} & t^2 e^{\lambda t}/2! & \cdots & t^{n-1} e^{\lambda t}/(n-1)! \\
0 & e^{\lambda t} & t e^{\lambda t} & \cdots & t^{n-2} e^{\lambda t}/(n-2)! \\
\vdots & \vdots & \ddots & \ddots & \vdots \\
0 & 0 & \cdots & e^{\lambda t} & t e^{\lambda t} \\
0 & 0 & \cdots & 0 & e^{\lambda t}
\end{bmatrix}.
\end{split}\]