Does a matrix logarithm expansion work when the identity matrix is rank deficient?

83 Views Asked by At

Apologies if this seems trivial to some, but I've been unable to find any decisive literature on this (physicist asking here).

Consider the following Taylor expansion of the matrix logarithm:

$log(1_{P} + M ) \approx M + ...$

where $1_{P}$ is a $N \times N$ matrix with $1$ on a subset $P < N$ of its components, and $M$ is an $N \times N$ full rank matrix, which we can assume to be Schur stable such that its eigenvalues lie within the unit circle. It seems like it could, since the matrix norm of $1_{P}$ is also $1$ as in the case if $P=N$.

By analogy (and correct me if I'm wrong on this), the expansion of $(1_{P} - M)^{-1}$ wouldn't work, since $1_{P}$ and $M$ do not commute in general, and hence the regular binomial expansion can't be used. Is such a condition for commutativity absent from the matrix logarithm? It seems like the case by simply looking up the definition, but I'm really ignorant on this and it would be great if someone could provide insight.