matrix exponentialMatrixExponential2003-04-06 07:15:042006-08-10 13:25:29Definition% this is the default PlanetMath preamble. as your knowledge
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The \emph{exponential} of a real valued square matrix $A$, denoted
by $e^A$, is defined as
\begin{eqnarray*}
e^A &=& \sum_{k=0}^\infty \frac{1}{k!}A^k \\
&=& I + A + \frac{1}{2} A^2 + \cdots
\end{eqnarray*}
Let us check that $e^A$ is a real valued square matrix.
Suppose $M$ is a real number such $|A_{ij}| < M$ for all
entries $A_{ij}$ of $A$.
Then $|(A^2)_{ij}| < nM^2$ for all entries in $A^2$,
where $n$ is the order of $A$. (Alternatively, one could argue using matrix norms: We have $||e^A||\leq e^{||A||}$ for the 2-norm, and hence the entries of $e^A$ are bounded by $M=||e^A||$.) Thus,
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
$e^A$ converges to real valued $n\times n$ matrix.
{\bf Example 1.} Suppose $A$ is nilpotent, i.e., $A^r = 0$ for some natural
number $r$. Then
\begin{eqnarray*}
e^A &=& I + A + \frac{1}{2!} A^2 + \cdots + \frac{1}{(r-1)!} A^{r-1}.
\end{eqnarray*}
{\bf Example 2.} If $A$ is diagonalizable, i.e., of the form
$A=L D L^{-1}$, where $D$ is a diagonal matrix, then
\begin{eqnarray*}
e^A &=& \sum_{k=0}^\infty \frac{1}{k!}(LDL^{-1})^k \\
&=& \sum_{k=0}^\infty \frac{1}{k!}LD^kL^{-1} \\
&=& L e^D L^{-1}.
\end{eqnarray*}
Further, if
$D=\diag\{a_1,\cdots, a_n\}$, then
$D^k = \diag\{a_1^k, \cdots, a_n^k\}$ whence
\begin{eqnarray*}
e^A &=& L \diag\{e^{a_1}, \cdots, e^{a_n}\} L^{-1}.
\end{eqnarray*}
For diagonalizable matrix $A$, it follows that
$\det e^A = e^{\trace A}$.
However, this formula is, in fact, valid for all $A$.
{\bf \PMlinkescapetext{Properties}} \\
Let $A$ be a square $n\times n$ real valued matrix.
Then the matrix exponential satisfies the following properties
\begin{enumerate}
\item For the $n\times n$ zero matrix $O$, $e^O=I$, where $I$ is the
$n\times n$ identity matrix.
\item If $A=L\diag\{a_1,\cdots, a_n\} L^{-1}$ for an invertible $n\times n$
matrix $L$, then
$$ e^A = L \diag\{e^{a_1},\cdots, e^{a_n}\} L^{-1}.$$
\item If $A$ and $B$ commute,
then $e^{A+B} = e^{A} e^B$.
\item The trace of $A$ and the determinant of $e^A$ are related by the formula
$$ \det e^A = e^{\trace A}.$$
In effect, $e^A$ is always invertible. The inverse is given by
$$ (e^A)^{-1} = e^{-A}.$$
\item If $e^A$ is a rotational matrix, then $\trace A=0$.
%\item Let $F(t) = e^{At}$ for $t\in \mathbb{R}$. Then
%$F(s+t)=F(s)F(t)$ and $\frac{d}{dt} F(t) = F e^{Ft}$.
\end{enumerate}