Logarithm of Matrix exponential

Question

Can we say $\log(e^X e^Y)=X+Y \tag 1$ ?

where $X$,$Y$ are general skew symmetric matrices of order $3 \times 3$ (Just mentioned skew symmetric matrices to indicate that these are rotational matrices)

Answer 1

EDIT 1. Remark 1. For general complex (or real ) matrices $e^Xe^Y=e^{X+Y}$ does not imply $XY=YX$. For example, let $A=60i\pi\begin{pmatrix}1&0\\0&-1\end{pmatrix},B=\pi\begin{pmatrix}-150i&-91\\391&150i\end{pmatrix}$. Then $e^{tA+B}=e^{tA}e^B$ for $t=1,2,3,4,5$ and $e^{t(A+B)}=e^{tA}e^{tB}$ for every positive integer ; yet $AB\not=BA$.

Remark 2. Let $\log$ be the principal logarithm ; it is a matrix function (cf. Higham, matrix functions). Even if we assume that the eigenvalues of $X,Y$ have imaginary parts in $(-\pi,\pi)$ and $XY=YX$, the formula $\log(e^Xe^Y)=X+Y$ is false. Choose, for instance, these skew symmetric matrices: $X=Y=\begin{pmatrix}0&2\pi/3\\-2\pi/3&0\end{pmatrix}$.

Remark 3. One might wonder if the implication in Remark 1 would be true for real skew symmetric matrices $X,Y$. In dimension $2$, it is clear because $2$ skew symmetric matrices commute. In dimension $3$, it is not, in particular because the couples $(X,Y)$ in $M_3(\mathbb{C})$ or $M_3(\mathbb{R})$ s.t. $e^Xe^Y=e^{X+Y}$ are unknown. To a skew symmetric matrix $A=\begin{pmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{pmatrix}$, we associate the vector $a=[a_1,a_2,a_3]^T$ and $e^A$ is the rotation $Rot(a,||a||)$. Then the question is: does the relation $Rot(a,||a||)\circ Rot(b,||b||)=Rot(a+b,||a+b||)$ imply that $AB=BA$, that is, $a,b$ are parallel vectors ? I do not know the answer.

EDIT 2. The above assertion (in Remark 3.) is almost true. We identify a vector with a point.

Case 1. Assume that $||a||$ is not in $2\pi\mathbb{N}^*$. Let $c=Rot(a,||a||)(b)=Rot(a+b,||a+b||)(b)$. Then $c\in\Gamma_1$, the circle, with axis $a$, that goes through $b$ ; moreover $c\in\Gamma_2$, the circle, with axis $a+b$, that goes through $b$. Both circles have same tangent in $b$. Then $c=b$ (that is impossible) except if $\Gamma_1=\Gamma_2$. Finally $a,a+b$ are parallel vectors and we are done.

Case 2. If $||a||$ and $||b||$ are in $2\pi\mathbb{N}^*$, then there are counter-examples as the following one: $A=\begin{pmatrix}0&0&0\\0&0&-6\pi\\0&6\pi&0\end{pmatrix},B=\begin{pmatrix}0&0&8\pi\\0&0&0\\-8\pi&0&0\end{pmatrix}$.

EDIT 3. Answer to Nick. Your wikipedia reference gives a false BCH formula! Indeed, if $\log$ is the principal logarithm, then it is false even over $\mathbb{C}$: take $X=Y=2i\pi/3$ as in my Remark 2. The correct formula is (for example) here: http://en.wikipedia.org/wiki/Lie_group $e^ue^v=e^{u+v+\cdots}$. Thus $XY=YX$ does not imply that $\log(e^Xe^Y)=X+Y$.

The converse is also false (even in dimension $2$ over $\mathbb{C}$): assume that $\log(e^Xe^Y)=X+Y$, that is equivalent to $e^Xe^Y=e^{X+Y},spectrum(X+Y)\subset \mathbb{R}\times (-i\pi,i\pi]$. A counter-example is given by $X=\begin{pmatrix}u&0\\0&0\end{pmatrix},Y=\begin{pmatrix}-u&1\\0&0\end{pmatrix}$, where $e^u=1+u,u\not= 0$ (for example $u\approx 2.0888+7.4615i$).

Answer 2

Assuming that $\left\|e^{X}e^{Y}-I\right\| < 1$ (which assures convergence of the series defining the matrix logarithm) and that $\left\|X\right\|$ and $\left\|Y\right\|$ are small, then the Baker-Campbell-Hausdorff formula says \begin{align} \log(\exp(X)\exp(Y)) = X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}\left([X,[X,Y]] +[Y,[Y,X]] \right) + \cdots \end{align} and hence $\log(\exp(X)\exp(Y)) = X + Y$ iff $[X,Y] = 0$. From the previous answer we have seen that skew-symmetry does not imply commutativity, so in general we cannot say $\log(\exp(X)\exp(Y)) = X + Y$ for skew-symmetric matrices.