Arbitrary non-integer power of a matrix

Question

does there exist the notion of a non-integer power of a matrix? This seems to be accessible via semigroup-theory, yet I have not seen an actual definition so far.

I am not too firm at this right now, but I am curious. Can you give me a sketch of the definition and provide with some introductory information?

There is useful discussion in answers to this duplicate question. — Mars, Dec 16 '17 at 18:38

score 6 · Answer 1 · answered Jan 08 '11 at 02:21

6

If your matrix has positive eigenvalues, then one definition is to take non-integer powers of each eigenvalue (but keep the eigenvectors the same). This is a common definition used to take square roots, for example.

answered Jan 08 '11 at 02:21

Qiaochu Yuan

419,620

The more general condition is that the eigenvalues should not be negative numbers; see e.g. this. – J. M. ain't a mathematician Apr 18 '13 at 16:30

lhf · Accepted Answer · 2017-12-16T19:08:53.090

4

There are several techniques for extending scalar functions to matrices. Wikipedia mentions techniques based on power series, eigendecomposition, Jordan decomposition, Cauchy integral, and more.

edited Dec 16 '17 at 19:08

answered Jan 08 '11 at 03:08

lhf

216,483

1

(I'm surprised this is the accepted answer. I'm used to link-only answers being discouraged. In this case it's not immediate where one should look on the linked page.) – Mars Dec 16 '17 at 18:36
@Mars, this is an answer from the very early days. I've tried to improve it. Thanks for the nudge. – lhf Dec 16 '17 at 19:09

score 1 · Answer 3 · answered Jan 08 '11 at 02:10

1

You can use the binomial series to define powers for appropriate matrices.

answered Jan 08 '11 at 02:10

Mariano Suárez-Álvarez

135,076

Does the square root of a matrix defined with this approach coincide with the canoical square root $\sqrt(A^\ast A )$? – shuhalo Jan 14 '11 at 03:46

Arbitrary non-integer power of a matrix

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