The Exponential of a Matrix

# The Exponential of a Matrix

Lemma: Let $A$ be an $n \times n$ matrix. For each $n \in \mathbb{N}$ let $\displaystyle{s_N(t) = I + \sum_{k=1}^{N} \frac{t^k}{k!} A^k}$. Then the sequence of matrices $(s_N(t))_{N=1}^{\infty}$ converges absolutely and uniformly on any interval $(-a, a)$ ($a \in \mathbb{R}$, $a > 0$). |

**Proof:**We take a matrix norm of $s_n(t)$:

\begin{align} \quad | s_N(t) | & = \biggr \lvert I + \sum_{k=1}^{N} \frac{t^k}{k!} A^k \biggr \rvert \\ & \leq | I | + \sum_{k=1}^{N} \frac{|t|^k}{k!} |A|^k \\ & \leq | I | + \sum_{k=1}^{N} \frac{|a|^k}{k!} |A|^k \\ & \leq | I | - 1 + \sum_{k=0}^{N} \frac{|a|^k}{k!} |A|^k \\ & \leq (n - 1) + e^{a|A|} \end{align}

- So $| s_N(t) | \leq (n - 1) + e^{a|A|}$ for each $N \in \mathbb{N}$ on the open interval $(-a, a)$ where $a \in \mathbb{R}$ and $a > 0$. By the Weierstrass M-test, $(s_N(t))_{N=1}^{\infty}$ converges absolutely and uniformly on $(-a, a)$ to some matrix. $\blacksquare$

We now give this matrix an important name.

Definition: If $A$ is an $n \times n$ matrix we defined the Exponential Matrix of $A$ as $\displaystyle{e^{A} = I + \sum_{k=1}^{\infty} \frac{A^k}{k!}}$. |