2
$\begingroup$

I have a skew-symmetric matrix $A \in \mathbb{R}^{4 \times 4} $

$$ A = \begin{bmatrix} 0 & a & b & 0 \\ -a & 0 & c & 0 \\-b & -c & 0 & 0 \\ 0& 0 & 0 & 0\end{bmatrix} $$

another matrix $B\in \mathbb{R}^{4 \times 4} $

$$ B = \begin{bmatrix} 0 & 0 & 0 & t_1 \\ 0 & 0 & 0 & t_2 \\0 & 0 & 0 & t_3 \\ 0 & 0 & 0 & 0\end{bmatrix} $$

Also $C = A+ B$ and the question is what is

$$ D = e^{A + B} $$

I have checked that

$$ D = \begin{bmatrix} e^A & x \\ 0 & 1\end{bmatrix} \in \mathbb{R}^{4 \times 4}$$

But I cannot figure out what $x$ is. Has anyone thoughts on this? Thanks

$\endgroup$
1
$\begingroup$

Write $$ (A+B)^n = \pmatrix{A^{(n)} & x^{(n)}\\0&0} $$ We have $$ (A+B)(A+B)^{n} = \pmatrix{AA^{(n)} & Ax^{(n)}\\0&0} $$ So, we may show inductively that $A^{(n)} = A^n$, and $x^{(n)} = A^n x$ for $n \geq 1$, where $x = (t_1,t_2,t_3)^T$. That is, $(A + B)^0 = I$, and $$ (A + B)^n = \pmatrix{A^n & A^n x\\0&0} \quad n \geq 1 $$ As such, we have $$ \exp(A + B) = \sum_{n=0}^\infty \frac 1{n!}(A+B)^n = \pmatrix{\sum_{n=0}^\infty \frac 1{n!} A^n & \sum_{\color{red}{n=1}}^\infty \frac 1{n!}A^n x\\0&1} = \\ =\pmatrix{e^A & (e^A - I) x\\0&1} $$

$\endgroup$
  • $\begingroup$ It seems good but I cannot verify that. A numerical example returns not the correct result. $\endgroup$ – Armen Avetisyan Jan 14 '17 at 19:16
  • $\begingroup$ I miscalculated! I'll change it $\endgroup$ – Omnomnomnom Jan 14 '17 at 19:43
  • $\begingroup$ See my latest edit. $\endgroup$ – Omnomnomnom Jan 14 '17 at 19:46
  • $\begingroup$ that does not seem to work neither... $\endgroup$ – Armen Avetisyan Jan 14 '17 at 20:09
  • $\begingroup$ @ArmenAvetisyan strange... I wonder what's going wrong here $\endgroup$ – Omnomnomnom Jan 14 '17 at 20:14

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.