Matrix exponential in terms of submatrices

Question

I want to compute the exponential of a matrix: $$ A = \begin{bmatrix} A_{11} & I_{3\times3} & \boldsymbol{0}_{3\times3}\\ \boldsymbol{0}_{3\times3} & \boldsymbol{0}_{3\times3} & I_{3\times3} \\ \boldsymbol{0}_{3\times3} & \boldsymbol{0}_{3\times3} & b \ I_{3\times3} \end{bmatrix} $$ where $I_{3\times3}$ is a 3-by-3 identity matrix and $b \ (>0)$ is a scalar. The submatrix $A_{11}$ is a skew-symmetric matrix, given by: $$ A_{11} = \begin{bmatrix} 0 & -a_{3} & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{bmatrix} $$ Is there any straightforward way to compute $\exp{(A)}$? After a quick search on Wiki, the most common approach seems to be a diagonalization of a given matrix, but I'm wondering if there is a more straightforward way in this case. Is there any useful theorem to express $\exp{(A)}$ in terms of $\exp{(A_{11})}$?

Answer 1

Unfortunately, you do have to do something akin to diagonalization. However, we can manage to do so symbolically. Here, I attempt transform the matrix into its Jordan form.

First, we deal with the generalized eigenspace associated with $0$. As the only generalized eigenvector for $A$, we can take $$ \vec v_1 = \frac 1{a_1a_3}\pmatrix{a_1 a_2\\a_2^{2} + a_3^{2}\\0\\a_1^{2} a_3 + a_2^{2} a_3 + a_3^{3}\\0\\0\\0\\0\\0}. $$ Note that $A\vec v_1 = (a_1,a_2,a_3,0,\dots,0)$. For the Jordan basis of the zero-eigenspace, we can use the columns of the following matrix: $$ \pmatrix{a_{1} & \frac{a_{2}}{a_{3}} & - a_{1} & 0\\a_{2} & \frac{a_{2}^{2} + a_{3}^{2}}{a_{1} a_{3}} & - a_{2} & -1\\a_{3} & 0 & 0 & 0\\0 & \frac{a_{1}^{2} a_{3} + a_{2}^{2} a_{3} + a_{3}^{3}}{a_{1} a_{3}} & - a_{2} a_{3} & - a_{3}\\0 & 0 & a_{1} a_{3} & 0\\0 & 0 & 0 & a_{1}\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0} $$ Next, to handle the eigenspaces associated with $\pm i\|\vec a\|$, use the columns of the matrix $$ \pmatrix{- a_{2} \|\vec a\| & - a_{1} a_{3}\\a_{1} \|\vec a\| & - a_{2} a_{3}\\0 & a_{1}^{2} + a_{2}^{2}\\0 & 0\\0 & 0\\0 & 0\\0 & 0\\0 & 0\\0 & 0}. $$ Finally, the eigenspace associated with $b$ has a basis given by the columns of the following: $$ \pmatrix{a_{1}^{2} + b^{2} & a_{1} a_{2} - a_{3} b & a_{1} a_{3} + a_{2} b\\a_{1} a_{2} + a_{3} b & a_{2}^{2} + b^{2} & - a_{1} b + a_{2} a_{3}\\a_{1} a_{3} - a_{2} b & a_{1} b + a_{2} a_{3} & a_{3}^{2} + b^{2}\\b \left(\|\vec a\|^2 + b^{2}\right) & 0 & 0\\0 & b \left(\|\vec a\|^2 + b^{2}\right) & 0\\0 & 0 & b \left(\|\vec a\|^2 + b^{2}\right)\\b^{2} \left(\|\vec a\|^2 + b^{2}\right) & 0 & 0\\0 & b^{2} \left(\|\vec a\|^2 + b^{2}\right) & 0\\0 & 0 & b^{2} \left(\|\vec a\|^2 + b^{2}\right)}. $$ If $P$ denotes the matrix whose columns are the columns of the matrices listed above (in order and consecutively), then we can write $A = PJP^{-1}$, where $$ J = \pmatrix{ 0&1\\ &0\\ &&0\\ &&&0\\ &&&&0&-\|\vec a\|\\ &&&&\|\vec a\| & 0\\ &&&&&&b\\ &&&&&&&b\\ &&&&&&&&b }, $$ where the "blank" entries in the above are zeroes.

The exponential $\exp(A)$ can be computed as $P\exp(J)P^{-1}$, where $$ \exp(J) = \pmatrix{ 1&1\\ &1\\ &&1\\ &&&1\\ &&&&\cos(\|\vec a\|) & -\sin(\|\vec a\|)\\ &&&&\sin(\|\vec a\|) & \cos(\|\vec a\|)\\ &&&&&&e^b\\ &&&&&&&e^b\\ &&&&&&&&e^b }. $$