1

I know that generalized binomial theorem for matrices, i.e. expansion of $$(A+B)^r\quad\quad (*)$$ when $A$ and $B$ are positive-definite matrices and $r$ is a real number, works only when $A$ and $B$ commute. For example, since the identity matrix commutes with all matrices, we have the expansion in this post: Does Newton's generalized binomial theorem work on a matrix?.

My question is that when $A$ and $B$ do not commute, is this possible to expand (*) in the special case that $r$ is very close to 1, i.e. $r=1+\delta$ and $\delta$ is a real number which is very close to zero?

Rodrigo de Azevedo
  • 1
Mah
  • 1,211
  • If one of your matrices is invertible, you can convert to $A(1+A^{-1}B)^r$ and apply the binomial theorem to this. – ziggurism Nov 03 '17 at 17:46
  • Both are invertible, but it is not possible because they don't commute and $r\neq 1$. – Mah Nov 03 '17 at 17:48
  • What is not possible? The identity $A+B = A(1+A^{-1}B)$ holds whenever $A$ is invertible. It's an application of the distributive law. In particular it holds when $A$ and $B$ do not commute. – ziggurism Nov 03 '17 at 17:49
  • $(A+B)^r\neq A^r(1+A^{-1}B)^r$ if $A$ and $B$ don't commute and $r\neq 1$. – Mah Nov 03 '17 at 17:50
  • You are right. I've made a horrible mistake. – ziggurism Nov 03 '17 at 17:53

1 Answers1

1

Well, purely formally, at least in physics, one routinely blunders into: $$ (A+B)^{1+\delta}=(A+B) e^{\delta \ln (A+B)}=(A+B) \left (I+\delta \ln (A+B) + \frac{\delta^2}{2} (\ln (A+B))^2+...\right ) , $$ until the alarm bells start ringing.

But, beyond this, you might need to know something about the norm of ln(A+B) to go any further, or else that of (A+B-I), so as to expand the log, formally, $$ \ln (A+B)= \ln (I +(A+B-I))= A+B-I-\frac{1}{2}(A+B-I)^2+... $$ using $\ln (1+x)= x-x^2/2+...$ for "small" x, cf. WP, or the eigenvalues of the sum, etc... Smallness of the matrix A+B-I corresponds to a small norm, as utilized in physics and engineering.

Cosmas Zachos
  • 6,028
  • Thank you for the answer. Do you have a proof for your answer? It is not very obvious to me why it is correct. – Mah Nov 06 '17 at 15:33
  • Thank you for clarifying. This is interesting and should be useful in some applications, but I need an expansion that separates $A$ and $B$. i.e. it is in terms of powers of $A$ and $B$ not $A+B$ ... which might not exist when $A$ and $B$ don't commute. – Mah Nov 06 '17 at 15:52
  • There is nothing general around... That was part of the point: practical progress is predicated on the specific properties of A and B or A+B-I from that point on... – Cosmas Zachos Nov 06 '17 at 15:56