10.3: The Matrix Exponential as a Sum of Powers
You may recall from Calculus that for any numbers aa and tt one may achieve eateat via
\[e^{at} = \sum_{k=0}^{\infty} \frac{(at)^k}{k!} \nonumber\]
The natural matrix definition is therefore
\[e^{At} = \sum_{k=0}^{\infty} \frac{(At)^k}{k!} \nonumber\]
where \(A^{0} = I\) is the n-by-n identity matrix.
The easiest case is the diagonal case, e.g.,
\[A = \begin{pmatrix} {1}&{0}\\ {0}&{2} \end{pmatrix} \nonumber\]
for then
\[(At)^k = \begin{pmatrix} {t^k}&{0}\\ {0}&{(2t)^k} \end{pmatrix} \nonumber\]
and so
\[e^{At} = \begin{pmatrix} {e^t}&{0}\\ {0}&{e^{2t}} \end{pmatrix} \nonumber\]
Note that this is NOT the exponential of each element of \(A\).
As a second example let us suppose
\[A = \begin{pmatrix} {0}&{1}\\ {-1}&{0} \end{pmatrix} \nonumber\]
We recognize that its powers cycle, i.e.,
\[A^2 = \begin{pmatrix} {-1}&{0}\\ {0}&{-1} \end{pmatrix} \nonumber\]
\[A^3 = \begin{pmatrix} {0}&{-1}\\ {1}&{0} \end{pmatrix} \nonumber\]
\[A^4 = \begin{pmatrix} {1}&{0}\\ {0}&{1} \end{pmatrix} \nonumber\]
\[A^5 = \begin{pmatrix} {0}&{1}\\ {-1}&{0} \end{pmatrix} = A \nonumber\]
and so
\[e^{At} = \begin{pmatrix} {1-\frac{t^2}{2}+\frac{t^4}{4}+\cdots}&{t-\frac{t^3}{3!}+\frac{t^5}{5!}-\cdots}\\ {-t+\frac{t^3}{3!}-\frac{t^5}{5!}+\cdots}&{1-\frac{t^2}{2}+\frac{t^4}{4}+\cdots} \end{pmatrix} = \begin{pmatrix} {\cos(t)}&{\sin(t)}\\ {-\sin(t)}&{\cos(t)} \end{pmatrix} \nonumber\]
If
\[A = \begin{pmatrix} {0}&{1}\\ {0}&{0} \end{pmatrix} \nonumber\]
then
\[A^2 = A^3 = A^k = \begin{pmatrix} {0}&{1}\\ {0}&{0} \end{pmatrix} \nonumber\]
and so
\[e^{At} = (I+tA) \begin{pmatrix} {1}&{t}\\ {0}&{1} \end{pmatrix} \nonumber\]