The 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.

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*} 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 $A$ .

Properties
Let $A$ be a square $n\times n$ real valued matrix. Then the matrix exponential satisfies the following properties

1. For the $n\times n$ zero matrix $O$ , $e^O=I$ , where $I$ is the $n\times n$ identity matrix.
2. If $A=L\diag\{a_1,\cdots, a_n\} L^{-1}$ for an invertible $n\times n$ matrix $L$ , then $$ e^A = L \diag\{e^{a_1},\cdots, e^{a_n}\} L^{-1}.$$
3. If $A$ and $B$ commute, then $e^{A+B} = e^{A} e^B$ .
4. 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 $$ (e^A)^{-1} = e^{-A}.$$
5. If $e^A$ is a rotational matrix, then $\trace A=0$ .