| Version 5 |
Version 4 |
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| \newcommand{\trace}{\mathop{\mathrm{tr}}} |
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| The \emph{exponential} of a real valued square matrix $A$, denoted |
The \emph{exponential} of a real valued square matrix $A$, denoted |
| by $e^A$, is defined as |
by $e^A$, is defined as |
| \begin{eqnarray*} |
\begin{eqnarray*} |
| e^A &=& \sum_{k=0}^\infty \frac{1}{k!}A^k \\ |
e^A &=& \sum_{k=0}^\infty \frac{1}{k!}A^k \\ |
| &=& I + A + \frac{1}{2} A^2 + \cdots |
&=& I + A + \frac{1}{2} A^2 + \cdots |
| \end{eqnarray*} |
\end{eqnarray*} |
| Let us check that $e^A$ is a real valued square matrix. |
Let us check that $e^A$ is a real valued square matrix. |
| Suppose $M$ is a real number such $|A_{ij}| < M$ for all |
Suppose $M$ is a real number such $|A_{ij}| < M$ for all |
| entries $A_{ij}$ of $A$. |
entries $A_{ij}$ of $A$. |
| Then $|(A^2)_{ij}| < nM^2$ for all entries in $A^2$, |
Then $|(A^2)_{ij}| < nM^2$ for all entries in $A^2$, |
| where $n$ is the dimension of $A$. Thus, |
where $n$ is the dimension of $A$. Thus, |
| in general, we have $|(A^k)_{i,j}| < n^k M^{k+1}$. Since |
in general, we have $|(A^k)_{i,j}| < n^k M^{k+1}$. Since |
| $\sum_{k=0}^\infty \frac{n^k}{k!} M^{k+1}$ converges, we see that |
$\sum_{k=0}^\infty \frac{n^k}{k!} M^{k+1}$ converges, we see that |
| $e^A$ converges to real valued $n\times n$ matrix. |
$e^A$ converges to real valued $n\times n$ matrix. |
|
|
| {\bf Example 1.} Suppose $A$ is nilpotent, i.e., $A^r = 0$ for some natural |
{\bf Example 1.} Suppose $A$ is nilpotent, i.e., $A^r = 0$ for some natural |
| number $r$. Then |
number $r$. Then |
| \begin{eqnarray*} |
\begin{eqnarray*} |
| e^A &=& I + A + \frac{1}{2!} A^2 + \cdots + \frac{1}{(r-1)!} A^{r-1}. |
e^A &=& I + A + \frac{1}{2!} A^2 + \cdots + \frac{1}{(r-1)!} A^{r-1}. |
| \end{eqnarray*} |
\end{eqnarray*} |
|
|
| {\bf Example 2.} If $A$ is diagonalizable, i.e., of the form |
{\bf Example 2.} If $A$ is diagonalizable, i.e., of the form |
| $A=L D L^{-1}$, where $D$ is a diagonal matrix, then |
$A=L D L^{-1}$, where $D$ is a diagonal matrix, then |
| \begin{eqnarray*} |
\begin{eqnarray*} |
| e^A &=& \sum_{k=0}^\infty \frac{1}{k!}(LDL^{-1})^k \\ |
e^A &=& \sum_{k=0}^\infty \frac{1}{k!}(LDL^{-1})^k \\ |
| &=& \sum_{k=0}^\infty \frac{1}{k!}LD^kL^{-1} \\ |
&=& \sum_{k=0}^\infty \frac{1}{k!}LD^kL^{-1} \\ |
| &=& L e^D L^{-1}. |
&=& L e^D L^{-1}. |
| \end{eqnarray*} |
\end{eqnarray*} |
| Further, if |
Further, if |
| $D=\diag\{a_1,\cdots, a_n\}$, then |
$D=\diag\{a_1,\cdots, a_n\}$, then |
| $D^k = \diag\{a_1^k, \cdots, a_n^k\}$ whence |
$D^k = \diag\{a_1^k, \cdots, a_n^k\}$ whence |
| \begin{eqnarray*} |
\begin{eqnarray*} |
| e^A &=& L \diag\{e^{a_1}, \cdots, e^{a_n}\} L^{-1}. |
e^A &=& L \diag\{e^{a_1}, \cdots, e^{a_n}\} L^{-1}. |
| \end{eqnarray*} |
\end{eqnarray*} |
| For diagonalizable matrix $A$, it follows that |
For diagonalizable matrix $A$, it follows that |
| $\det e^A = e^{\trace A}$. |
$\det e^A = e^{\trace A}$. |
| However, this formula is, in fact, valid for all $A$. |
However, this formula is, in fact, valid for all $A$. |
|
|
| {\bf \PMlinkescapetext{Properties}} \\ |
{\bf \PMlinkescapetext{Properties}} \\ |
| Let $A$ be a square $n\times n$ real valued matrix. |
Let $A$ be a square $n\times n$ real valued matrix. |
| Then the matrix exponential satisfies the following properties |
Then the matrix exponential satisfies the following properties |
| \begin{enumerate} |
\begin{enumerate} |
| \item For the $n\times n$ zero matrix $O$, $e^O=I$, where $I$ is the |
\item For the $n\times n$ zero matrix $O$, $e^O=I$, where $I$ is the |
| $n\times n$ identity matrix. |
$n\times n$ identity matrix. |
| \item If $A=L\diag\{a_1,\cdots, a_n\} L^{-1}$ for an invertible $n\times n$ |
\item If $A=L\diag\{a_1,\cdots, a_n\} L^{-1}$ for an invertible $n\times n$ |
| matrix $L$, then |
matrix $L$, then |
| $$ e^A = L \diag\{e^{a_1},\cdots, e^{a_n}\} L^{-1}.$$ |
$$ e^A = L \diag\{e^{a_1},\cdots, e^{a_n}\} L^{-1}.$$ |
| \item If $A$ and $B$ commute, |
\item If $A$ and $B$ commute, |
| then $e^{A+B} = e^{A} e^B$. |
then $e^{A+B} = e^{A} e^B$. |
| \item The trace of $A$ and the determinant of $e^A$ are related by the formula |
\item The trace of $A$ and the determinant of $e^A$ are related by the |
|
by the formula |
| $$ \det e^A = e^{\trace A}.$$ |
$$ \det e^A = e^{\trace A}.$$ |
| In effect, $e^A$ is always invertible. The inverse is given by |
In effect, $e^A$ is always invertible. The inverse is given by |
| $$ (e^A)^{-1} = e^{-A}.$$ |
$$ (e^A)^{-1} = e^{-A}.$$ |
| \item If $e^A$ is a rotational matrix, then $\trace A=0$. |
\item If $e^A$ is a rotational matrix, then $\trace A=0$. |
| %\item Let $F(t) = e^{At}$ for $t\in \mathbb{R}$. Then |
%\item Let $F(t) = e^{At}$ for $t\in \mathbb{R}$. Then |
| %$F(s+t)=F(s)F(t)$ and $\frac{d}{dt} F(t) = F e^{Ft}$. |
%$F(s+t)=F(s)F(t)$ and $\frac{d}{dt} F(t) = F e^{Ft}$. |
| \end{enumerate} |
\end{enumerate} |
