10.2: The Matrix Exponential as a Limit of Powers
You may recall from Calculus that for any numbers aa and tt one may achieve \(e^{at}\) via
\[e^{at} = \lim_{k \rightarrow \infty} (1+\frac{at}{k})^k \nonumber\]
The natural matrix definition is therefore
\[e^{At} = \lim_{k \rightarrow \infty} (I+\frac{At}{k})^k \nonumber\]
where \(I\) 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
\[(I+\frac{At}{k})^k = \begin{pmatrix} {(1+\frac{t}{k})^k}&{0}\\ {0}&{(1+\frac{2t}{k})^k} \end{pmatrix} \nonumber\]
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 concrete example let us suppose
\[A = \begin{pmatrix} {0}&{1}\\ {-1}&{0} \end{pmatrix} \nonumber\]
From
\[I+At = \begin{pmatrix} {1}&{t}\\ {-t}&{1} \end{pmatrix} \nonumber\]
\[(I+\frac{At}{2})^2 = \begin{pmatrix} {1}&{\frac{t}{2}}\\ {\frac{-t}{2}}&{1} \end{pmatrix} \begin{pmatrix} {1}&{\frac{t}{2}}\\ {\frac{-t}{2}}&{1} \end{pmatrix} = \begin{pmatrix} {1-\frac{t^2}{4}}&{t}\\ {-t}&{1-\frac{t^2}{4}} \end{pmatrix} \nonumber\]
\[(I+\frac{At}{2})^3 = \begin{pmatrix} {1-\frac{t^2}{3}}&{t-\frac{t^3}{27}}\\ {-t+\frac{t^3}{27}}&{1-\frac{t^2}{3}} \end{pmatrix} \nonumber\]
\[(I+\frac{At}{2})^4 = \begin{pmatrix} {-\frac{3t^2}{8}+\frac{t^4}{256}+1}&{t-\frac{t^3}{16}}\\ {-t+\frac{t^3}{16}}&{-\frac{3t^2}{8}+\frac{t^4}{256}+1} \end{pmatrix} \nonumber\]
\[(I+\frac{At}{2})^5 = \begin{pmatrix} {-\frac{2t^2}{5}+\frac{t^4}{125}+1}&{t-\frac{2t^3}{25}+\frac{t^5}{3125}}\\ {-t+\frac{2t^3}{25}-\frac{t^5}{3125}}&{-\frac{2t^2}{5}+\frac{t^4}{125}+1} \end{pmatrix} \nonumber\]
We discern a pattern: the diagonal elements are equal even polynomials while the off diagonal elements are equal but opposite odd polynomials. The degree of the polynomial will grow with kk and in the limit we 'recognize'
\[e^{At} = \begin{pmatrix} {\cos(t)}&{-\sin(t)}\\ {\sin(t)}&{\cos(t)} \end{pmatrix} \nonumber\]
If
\[A = \begin{pmatrix} {0}&{1}\\ {0}&{0} \end{pmatrix} \nonumber\]
then
\[(I+\frac{At}{k})^k = \begin{pmatrix} {1}&{t}\\ {0}&{1} \end{pmatrix} \nonumber\]
for each value of \(k\) and so
\[e^{At} = \begin{pmatrix} {1}&{t}\\ {0}&{1} \end{pmatrix} \nonumber\]