# Matrix equation involving commutator

Given two matrices $N\times N$ $A,B$, is there some method to solve the matrix equation: $$e^{[A,B]}=A$$ where the symbol $[A,B]$ means the commutator of the two matrices: $[A,B]=AB-BA$?

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If $A, B$ are given the equation is either true or not. What is the unknown? – user20266 Mar 9 '12 at 13:51
Presumably the intention is to aks whether one can find $B$, given a choice of $A$ (perhaps). Note that $A$ must have determinant $1$ to have any chance of doing this because ${ \rm det}( \exp([A,B]) ) = \exp( {\rm trace}([A,B]))$ and $[A,B]$ has trace zero. – Geoff Robinson Mar 9 '12 at 13:58
@Thomas:sorry. I forgot to say $B$ is known and $A$ is the matrix to find – Riccardo.Alestra Mar 9 '12 at 14:02
(Trivial) observation: $A = \mathbb{I}_{N \times N}$ and arbitrary $B$ solve the equation. – Sasha Mar 9 '12 at 14:07
(Minor) observations: The equation is invariant under conjugation, so we may assume that $B$ is in Jordan canonical form. If $B = \mathrm{diag}(\lambda_1, \ldots, \lambda_n)$ and we do not have $\lambda_i-\lambda_j=1$ for any $(i,j)$, then the solution at $A = \mathrm{Id}$ is isolated. – David Speyer Mar 9 '12 at 15:12
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In the $2 \times 2$ case, the only solutions are when $A = \mathrm{Id}$, and concievably some other solution when $B$ is of the form $\left( \begin{smallmatrix} \lambda & 1 \\ 0 & \lambda \end{smallmatrix} \right)$. (I doubt it, but I didn't check this case.)
Since the problem is unvariant under conjugation, we may assume that $B$ is in Jordan canonical form. Let's do the case where $B$ is diagonalizable: $B = \left( \begin{smallmatrix} \lambda & 0 \\ 0 & \mu \end{smallmatrix} \right)$ and $A = \left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right)$ then $$[A,B] =(\lambda - \mu) \begin{pmatrix} 0 & - x \\ y & 0 \end{pmatrix} \ \mathrm{and} \ [A,B]^2 = - (\lambda-\mu)^2 xy \ \mathrm{Id}.$$
So $$e^{[A,B]} = \cosh \left( (\lambda-\mu)^2 xy \right) \mathrm{Id} - (\lambda-\mu) \sinh \left( (\lambda-\mu)^2 xy \right) \begin{pmatrix} 0 & -x \\ y & 0 \end{pmatrix}. \quad (\dagger)$$
Comparing off diagonal entries: $$\begin{array}{rl} x =& \phantom{-} (\lambda-\mu) \sinh \left( (\lambda-\mu)^2 xy \right) x \\ y =& - (\lambda-\mu) \sinh \left( (\lambda-\mu)^2 xy \right) y \end{array}$$
These equations are only consistent if at least one of $x$ and $y$ is $0$. In this case, $(\lambda - \mu)xy=0$ so $(\dagger)$ collapses to $e^{[A,B]} = \mathrm{Id}$ and $A=\mathrm{Id}$.