On the Wikipedia page for Jordan matrices, under the section on functions $f$ of matrices $A = PJP^{-1}$ (that being the Jordan Canonical Form), the following can be found (I've paraphrased it a tiny bit to make it concern my point):

Let $f(z)=\sum _{{n=0}}^{{\infty }}a_{n}(z-z_{0})^{n}$ be the power series expansion of $f$ around $z_{0}\in\mathbb{C}$, which will be hereinafter supposed to be $0$ for simplicity's sake. The matrix $f(A)$ is then defined via the following formal power series: $f(A)=\sum _{n=0}^{\infty }a_{n}A^{n}$.

Obviously, scalar functions of matrices can then be computed via $f(A) = P\Big(\sum^\infty_{n=0}a_nJ^n\Big)P^{-1}$, thanks to

$$A^n = PJP^{-1}PJP^{-1} ... PJP^{-1}PJP^{-1} = PJ^nP^{-1}.$$

However, that's assuming $z_0=0$. You might want to use a Taylor series (since, in well-behaved cases, such a series exists at any point on the domain of $f$) about an arbitrary point $z_0$. The Wikipedia page doesn't cover that case, however. That'd mean the summation would be summing matrix powers of the form:

$$(A-z_0)^n = (PJP^{-1}-z_0)(PJP^{-1}-z_0) ... (PJP^{-1}-z_0)(PJP^{-1}-z_0) = \;?$$

... where, I guess, $z_0\equiv z_0I$. My question is two-fold:

  1. Do those powers amount to anything as nicely separable as when $z_0=0$?
  2. Say I want to compute the value of $f(A)$ with $f(x) = (1-x)^{-1}$, where $A$ has only one eigenvalue, that is confined to $]-1,1[$. Would it make sense at all to use the Taylor series about that eigenvalue, instead of the Maclaurin series? Furthermore, say $A$ had many different eigenvalues, possibly on intervals certain Taylor series of $f$ don't converge on. Is there any guideline as to which point we'd choose the series about, in that case?

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