# Exponential upper matrix

Consider the matrix

$M = \left[ \begin{matrix} A & v \\ 0 & 0\end{matrix} \right] \in \mathbb{R}^{(n+1)\times(n+1)},$

where $A \in \mathbb{R}^{n \times n}$ and $v \in \mathbb{R}^n$.

Compute the matrix exponential $e^{M}$ ( $= \sum_{k=0}^{\infty} M^k/k!$ ).

Comment: I think $e^{M}$ is of the form $\left[ \begin{matrix} e^A & * \\ 0 & I \end{matrix} \right]$, but I am not sure about the vector $*$ in position $(1,2)$.

• The star is equal to $e^{Av}$. If you know the matrix M, in oreder to compute that powers you should use eigenvalues and diagonalization. – SebiSebi Dec 2 '14 at 21:23
• @SebiSebi: I think you might want to write $e^A v$. – Robert Lewis Dec 2 '14 at 21:25
• So $* = e^A v$? – user693 Dec 2 '14 at 21:27
• Yep. Sorry! It's $e^{A}v$ – SebiSebi Dec 2 '14 at 21:31
• I thought there is some factor $1/2$ or so. – user693 Dec 2 '14 at 21:32

Note that $$M^{k}=\left[\begin{array}{cc} A^{k} & A^{k-1}v\\ 0 & 0 \end{array}\right]$$ for $k\geq1$. Therefore \begin{align*} e^{M} & =\sum_{k=0}^{\infty}M^{k}/k!\\ & =I+\left[\begin{array}{cc} A & v\\ 0 & 0 \end{array}\right]+\left[\begin{array}{cc} A^{2} & Av\\ 0 & 0 \end{array}\right]/2+\left[\begin{array}{cc} A^{3} & A^{2}v\\ 0 & 0 \end{array}\right]/6+\ldots\\ & =\left[\begin{array}{cc} e^{A} & \left(\sum_{k=1}^{\infty}A^{k-1}/k!\right)v\\ 0 & I \end{array}\right]\\ & =\left[\begin{array}{cc} e^{A} & \left(e^{A}-I\right)A^{-1}v\\ 0 & I \end{array}\right]\text{ (assuming }A\text{ is nonsingular)} \end{align*}