Matrix Exponent
Overview
Given an uncontrolled system,
$$ \dot{x}(t) = Ax(t) $$the solution could be stated as
$$ x(t) = e^{At}x_0 $$where $e^{At}$ is known as the matrix exponent.
Properties
The following rules hold for a matrix exponent
- $e^{At}e^{As} = e^{A(t+s)}$ (semi-group property)
- $Ae^{At}=e^{At}A$
- $\left(e^{At}\right)^{-1}=e^{-At}$
- If $P$ is a nonsingular square matrix, then $e^{P^{-1}AP}=P^{-1}e^AP$ (similarity transformation=change of the basis)
- If $A={\rm diag}(a_1,\dots,a_n)$, then $e^A={\rm diag}(e^{a_1},\dots,e^{a_n})$
- If $A$ and $B$ commute (i.e., $AB=BA$), then $e^{A+B}=e^Ae^B$
Computing the Matrix Exponent
Method 1
$$ \begin{equation} e^{At} = b_0(t)I+b_1(t)A+ \dots + b_{n-1}(t)A^{n-1} = \sum_{i=0}^{n-1}b_i(t)A^i \end{equation} $$The values of $b_i(t)$ can be found using the following expressions
$$ \begin{align} \begin{split} e^{\lambda_1t} &= b_0(t) + b_1(t)\lambda_1 + \dots+b_{n-1}(t)\lambda_1^{n-1}\\ e^{\lambda_2t} &= b_0(t) + b_1(t)\lambda_2 + \dots+b_{n-1}(t)\lambda_2^{n-1}\\ &~~\vdots\hspace{2cm}\vdots\hspace{2cm}\vdots\\ e^{\lambda_nt} &= b_0(t) + b_1(t)\lambda_n + \dots+b_{n-1}(t)\lambda_n^{n-1} \end{split} \end{align} $$The last expression (2) could be expressed more compactly as follows
$$ e^{\lambda_jt}=\sum_{i=0}^{n-1}b_i(t)\lambda_j^{i},~~~~j=1,2,\dots,n $$Example
Given a $2\times 2$ matrix,
$$ A = \begin{bmatrix}1 & 2\\0&-1 \end{bmatrix} $$The matrix exponential can be found using the following equation
$$ \begin{equation} e^{At} = \sum_{i=0}^{1}b_i(t)A^{i}=b_0(t)I+b_1(t)A \end{equation} $$So, we need to find the values of $b_0(t)$ and $b_1(t)$ using the following two equations.
$$ \begin{align} \begin{split} e^{\lambda_1t}&=b_0(t)+\lambda_1b_1(t)\\ e^{\lambda_2t}&=b_0(t)+\lambda_2b_1(t) \end{split} \end{align} $$Since the eigenvalues of $A$ have the values $1$ and $-1$, or
$$ \lambda_1 = 1 ~~~\text{and}~~~ \lambda_2=-1 $$The expression (4) could be written as follows
$$ \begin{align} \begin{split} e^{t} &= b_0(t)+b_1(t)\\ e^{-t} &= b_0t)-b_1(t) \end{split} \end{align} $$And expression (5) could be expressed in matrix form as follows
$$ \begin{bmatrix} e^t \\ e^{-1}\end{bmatrix}=\begin{bmatrix}1 & 1\\1 & -1\end{bmatrix}\begin{bmatrix}b_0(t)\\b_1(t) \end{bmatrix} $$Then,
$$ \begin{align*} \begin{bmatrix}b_0(t) \\ b_1(t)\end{bmatrix}&=\begin{bmatrix}1 & 1\\ 1 & -1\end{bmatrix}^{-1}\begin{bmatrix}e^t\\e^{-t}\end{bmatrix}\\&=\frac{1}{2}\begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}\begin{bmatrix}e^t\\e^{-t}\end{bmatrix}\\&= \frac{1}{2}\begin{bmatrix}e^t + e^{-t}\\e^{t}-e^{-t}\end{bmatrix} \end{align*} $$Therefore,
$$ \begin{align} \begin{split} b_0(t) &= \frac{1}{2}\left(e^t + e^{-t}\right)\\ b_1(t) &= \frac{1}{2}\left(e^t - e^{-t}\right) \end{split} \end{align}$$Then by substituting by (6) into (3) leads to
$$ \begin{align*} e^{At} &= \frac{1}{2}(e^{t}+e^{-t})\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}+ \frac{1}{2}(e^{t}-e^{-t})\begin{bmatrix}1 & 2\\0 &-1\end{bmatrix}\\&= \frac{1}{2}\left(\begin{bmatrix}e^t + e^{-t} & 0\\0 & e^t + e^{-t}\end{bmatrix}+\begin{bmatrix}e^t - e^{-t} & 2(e^t - e^{-t})\\0 & -(e^t - e^{-t})\end{bmatrix}\right)\\&= \frac{1}{2}\begin{bmatrix}2e^t & 2(e^t - e^{-t})\\0 & 2e^{-t}\end{bmatrix}\\&= \begin{bmatrix}e^t & e^{t}-e^{-t}\\0 & e^{-t}\end{bmatrix} \end{align*} $$Then,
$$ \boxed{e^{At}=\begin{bmatrix}e^t & e^{t}-e^{-t}\\0 & e^{-t}\end{bmatrix}}$$Method 2
The matrix exponential could be expressed as follows
$$ e^{At} = P e^{\Lambda t}P^{-1} $$where $P$ is the matrix encompassing the eigenvectors, or
$$ P = \begin{bmatrix} \mathbf{v_1} & \mathbf{v_2} & \cdots & \mathbf{v_n} \end{bmatrix} $$and $\Lambda$ is a diagonal matrix of all eigenvalues, or
$$ \Lambda = P^{-1}AP={\rm diag}(\lambda_1,\lambda_2,\dots,\lambda_n) $$Example
Given a $2\times 2$ matrix,
$$ A = \begin{bmatrix} 1 & 2\\0&-1 \end{bmatrix} $$The matrix exponential can be found using the following equation
$$ e^{At} = P e^{\Lambda t} P^{-1} $$The eigenvectors of $A$ are as follows
$$ \mathbf{v_1} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} ~~~\text{and} ~~~ \mathbf{v_2} = \begin{bmatrix} -1 \\ 1 \end{bmatrix} $$Then
$$ P = \begin{bmatrix} 1 & -1\\ 0 & 1 \end{bmatrix} $$The value of $\Lambda$ is
$$ \begin{align*} \Lambda &= P^{-1}AP\\ &= \begin{bmatrix} 1 & 1\\0 &1 \end{bmatrix} \begin{bmatrix} 1 & 2\\0 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1\\0 & 1 \end{bmatrix}\\ &= \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \end{align*} $$Therefore, the matrix exponential of $\Lambda$ is
$$ e^{\Lambda t} = \begin{bmatrix} e^t & 0\\0 & e^{-t} \end{bmatrix} $$Then
$$ \begin{align*} e^{At} &= Pe^{\Lambda t}P^{-1}\\ &= \begin{bmatrix} 1 & -1\\0 & 1 \end{bmatrix} \begin{bmatrix} e^t & 0\\ 0 & e^{-t} \end{bmatrix} \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}\\ &= \begin{bmatrix} e^{t} & -e^{-t}\\ 0 & e^{-t} \end{bmatrix} \begin{bmatrix} 1 & 1\\0 & 1 \end{bmatrix}\\ &= \begin{bmatrix} e^t & e^t-e^{-t}\\ 0 & e^{-t} \end{bmatrix} \end{align*} $$Therefore,
$$ \boxed{e^{At} = \begin{bmatrix} e^t & e^t-e^{-t}\\ 0 & e^{-t} \end{bmatrix}} $$