1
$\begingroup$

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 $(1,2)$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – SebiSebi Dec 2 '14 at 21:23
  • $\begingroup$ @SebiSebi: I think you might want to write $e^A v$. $\endgroup$ – Robert Lewis Dec 2 '14 at 21:25
  • $\begingroup$ So $* = e^A v$? $\endgroup$ – user693 Dec 2 '14 at 21:27
  • $\begingroup$ Yep. Sorry! It's $e^{A}v$ $\endgroup$ – SebiSebi Dec 2 '14 at 21:31
  • $\begingroup$ I thought there is some factor $1/2$ or so. $\endgroup$ – user693 Dec 2 '14 at 21:32
1
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.