Matrix exponential of $4 \times 4$ (almost) skew-symmetric matrix

Question

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

Answer 1

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 $$ \exp(A + B) = \sum_{n=0}^\infty \frac 1{n!}(A+B)^n = \pmatrix{\sum_{n=0}^\infty \frac 1{n!} A^n & \sum_{\color{red}{n=1}}^\infty \frac 1{n!}A^n x\\0&1} = \\ =\pmatrix{e^A & (e^A - I) x\\0&1} $$