# General formula for the exponential of a block matrix

Let $$M = \begin{pmatrix}A& B\\C& D\end{pmatrix}.$$ Is there a general formula for $e^M$ in terms of the sub-matrices $A$, $B$, $C$ and $D$? If not, can anything useful be said about the properties of $e^M$, given properties of its component blocks?

I've tried a few different approaches but haven't got anywhere, mostly because the formula $e^{A + B}=e^Ae^B$ only works if $A$ and $B$ commute.

If it helps, I'm most interested in the case where $M$ is a symmetryic real matrix, and an 'infinitesimal stochastic' matrix in particular. (An infinitesimal stochastic matrix is a symmetric real matrix whose column sums are all zero, and which has no negative off-diagonal elements.) If there are other special sub-classes of matrices for which something useful can be said here, I would be interested to know that.

• the case where $M$ is a real symmetric matrix, or where the sub-matrices are? – Thoth Oct 23 '15 at 5:33
• Since any matrix $M$ can be represented in block form as indicated, it is unclear why any special expression for the exponential of $M$ ought to be expected. – hardmath Oct 23 '15 at 5:39
• @Thoth the case where $M$ is - I've clarified in the question – Nathaniel Oct 23 '15 at 7:01
• @hardmath that's a good point. (I was coming from the point of view of "it would be really useful to me if there was a formula for this," not "I have some reason to expect there would be a formula for this.") I've edited the question to ask whether there's anything else that can be said about $e^M$, given properties of its component blocks. – Nathaniel Oct 23 '15 at 7:02
• I doubt you can say much about $A,B,C,D$, but if $M$ is symmetric then it has an eigen-decomp as $M=P^TDP$ and so I think you have $e^M=P^Te^DP$. I don't know if that helps any. $D$ contains the (possibly absolute value or squared can't remember) eigenvalues of $M$, but unless the top right or bottom left block is the zero block, there is no relationship known between the eigenvalues of the blocks and the eigenvalues of $M$. – Thoth Oct 23 '15 at 8:35

If $$M$$ is a squared symmetric real matrix then $$\begin{equation} M=\begin{pmatrix} A & B \\ B^\top & C\end{pmatrix}. \end{equation}$$ If $$B=0$$, then $$\begin{equation} \exp(M)=\begin{pmatrix} \exp(A) & 0 \\ 0 & \exp(C)\end{pmatrix}. \end{equation}$$
Moreover, since $$M$$ is symmetric positive semidefinite then it is diagonalizable, i.e. $$M=V D V^\top$$ with $$D$$ diagonal matrix and $$V$$ orthogonal matrix where each column is an eigenvector of $$M$$. If you work a bit with the relation $$\begin{equation}\begin{pmatrix} D_1 & 0 \\ 0 & D_2\end{pmatrix}= D=V^\top M V = \begin{pmatrix} V_1^\top & V_3^\top \\ V_2^\top & V_4^\top\end{pmatrix}\begin{pmatrix} A & B \\ B^\top & C\end{pmatrix}\begin{pmatrix} V_1 & V_2 \\ V_3 & V_4\end{pmatrix}.\end{equation}$$ you might find the block diagonal matrices $$D_1$$ and $$D_2$$ and block matrices $$V_1$$, $$V_2$$, $$V_3$$ and $$V_4$$. Hence you can apply $$\exp(M)=V\exp(D)V^\top$$.