Given two square real matrices $A$ and $B$ (with no specific properties), find $r$ and $p$ (integers but reals could be good as well) and $x$ such that: $$(A^rB^p-I)x=0$$ has non-trivial solutions (I is the identity matrix). At this time, every combination of couples $(r,p)$ is numerically explored but this is long. There might be some sophisticated matrix theory that could help. I've also thought of first diagonalising $A$ and $B$ but they do not share the same eigenvectors in general and it does not seem to help.

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    $\begingroup$ Defining non-integer powers of an arbitrary matrix is problematic, though you can do it if you assume $A$ and $B$ are positive definite (or positive definite if the powers are positive). $\endgroup$ – Robert Israel Jun 3 '16 at 22:40
  • $\begingroup$ ... positive semidefinite if the powers are positive, I meant to say. $\endgroup$ – Robert Israel Jun 3 '16 at 23:45

Let's assume you can define $A^r$ and $B^p$ for arbitrary real $r$ and $p$ in such a way that the function $F(r,p) = \det(A^r B^p - I)$ is real-valued and continuous on $\mathbb R^2$. If solutions exist (other than the trivial $r=p=0$), you should be able to find them using numerical methods. In particular, $F(r,p)=0$ should typically define a curve in $\mathbb R^2$, and given some $r_i,p_i$ with $F(r_1,p_1) < 0 < F(r_2, p_2)$, there will be some $(r,p)$ on the line segment from $(r_1,p_1)$ to $(r_2, p_2)$ where $F(r,p) = 0$.

For example, if $A$ has no eigenvalues in $(-\infty, 0]$, you can define $A^r = \exp(r \log(A))$, where $\log(A)$ is defined using the holomorphic functional calculus with the principal branch of the logarithm, and similarly for $B^p$.

EDIT: If the dimension $n$ is odd, $F(r,p)$ and $F(-r,-p)$ (as defined above) have opposite sign, so there will be nontrivial solutions. On the other hand, consider the case $n=2$, $A = eI$, $B = \pmatrix{\cos(\theta) & \sin(\theta)\cr -\sin(\theta) & \cos(\theta)\cr}$, so $A^r = e^r I$ and $B^p = \pmatrix{\cos(p\theta) & \sin(p\theta)\cr -\sin(p\theta) & \cos(p\theta)\cr}$. Then $\det(A^r B^p - I) = e^{2r} - 2 e^r \cos(p\theta) + 1$, which is $0$ only for $r=0$, $p=2\pi m$ with $m$ an integer.

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  • $\begingroup$ Thanks. Just a piece of information: the dimension of $A$ and $B$ is about $n=500$. $\endgroup$ – pluton Jun 4 '16 at 13:36
  • $\begingroup$ Then the numerical calculation of $F(r,p)$ might be rather challenging; high precision (perhaps hundreds of digits) might be needed. $\endgroup$ – Robert Israel Jun 10 '16 at 18:03
  • $\begingroup$ It might be better in practice to avoid determinants and track a real eigenvalue of $A^r B^p$ as it crosses the value $1$. $\endgroup$ – Robert Israel Jun 10 '16 at 18:13

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