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My question is that does Newton's generalized binomial theorem work on a matrix? In other words, for a Hermitian matrix $X$ and a real number $r$, is the following statement correct? Is there any condition on $X$ that makes the series valid/invalid? $$(I+X)^r=\sum_{k=0}^{\infty} \binom{r}{k} X^k$$

Mah
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    Banach-space structure on Hermitian matrices allows you to translate the proof for complex case mutatis mutandis to the proof for matrix case provided $|X| < 1$. – Sangchul Lee Oct 13 '16 at 14:36
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    @SangchulLee it's not clear how one should define the $r$th power of an arbitrary Hermitian matrix. For example, what exactly should $(-I)^{1/2}$ be? edit: never mind, that case doesn't arise with your constraint. – Ben Grossmann Oct 13 '16 at 14:38
  • @Azoodish it is certainly possible to do so if $X$ is Hermitian with eigenvalues at least $-1$. In any other case, one needs to be careful about what precisely the $r$th power of a matrix means. – Ben Grossmann Oct 13 '16 at 14:39
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    This will indeed work when $|X| < 1$ ($|\cdot|$ meaning the spectral norm). In any other case, there is no guarantee. – Ben Grossmann Oct 13 '16 at 14:42
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    @Omnomnomnom, I agree that the statement does depend on how the $r$-th power is defined. As you may know, however, for $(1+X)^r$ with $|X|<1$, many definitions (definition using spectral theorem, definition using matrix logarithm and matrix exponential, taking the series in question as definition) coincide, though proving the equivalence is another story. – Sangchul Lee Oct 13 '16 at 14:42
  • @SangchulLee also, "mutatis mutandis" is a nice turn of phrase, I'll be hanging on to that one – Ben Grossmann Oct 13 '16 at 14:44
  • @Omnomnomnom, I found that expression in a textbook (complex-analysis textbook if I remember correctly) and thought that it was really cool, but I was unsure how it reads like for native speakers... until now. :) – Sangchul Lee Oct 13 '16 at 14:51
  • Thanks for your comments. So it is correct if $||X||<1$. – Mah Oct 13 '16 at 14:58
  • @Omnomnomnom : Can you introduce a reference for this? I have difficulty in translating the proof for complex case to the proof for matrix case. Thanks – Mah Oct 14 '16 at 22:10
  • Possible duplicate of Binomial theorem for matrices – Rodrigo de Azevedo Jan 17 '18 at 22:04
  • @SangchulLee, well, the only speakers for whom mutatis mutandis was native have long been dead... – Mariano Suárez-Álvarez Jan 18 '18 at 18:51
  • I suggest you decide on a specific dofinition of of what $X^r$ means, and then maybe someoner can offer a proof of what you want. – Mariano Suárez-Álvarez Jan 18 '18 at 18:52

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