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Let $X$ be an $n\times n$ complex matrix. The exponential of $X$ is defined by $$\exp (X)=\sum_{k=0}^\infty \frac{1}{k!}X^k$$ where $X^0$ is the $n\times n$ identity matrix. Importantly, the matrix infinite series always converges.

This led me to the following question: Let $X$ be defined as above. If $$f(z)=\sum_{k=0}^\infty a_k z^k$$ represents an entire function $f$, is it true that the matrix infinite series $$\sum_{k=0}^\infty a_k X^k$$ always converges?

The powers series expansion at the origin of an entire function converges for all complex numbers, but I have no idea how would I prove/disprove convergence for complex matrices.

Nomas2
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1 Answers1

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First of all, before you define such a series you should mention in which norm do you want it to converge. In the space of matrices it doesn't really matter, because over a finite dimensional space all norms are equivalent, but still.

A very good norm on the space of matrices is the so called operator norm:

$||A||=\sup\limits_{0\ne x\in\mathbb{C^n}}\frac{||Ax||}{||x||},$

where $||x||$ is the usual Euclidean norm on $\mathbb{C^n}$. This norm has the very nice property $||AB||\leq ||A||\cdot ||B||$ for all matrices $A,B\in M_n(\mathbb{C})$. Hence, with this norm we get:

$$\sum\limits_{k=0}^{\infty}||a_kX^k||\leq\sum\limits_{k=0}^{\infty}|a_k|\cdot||X||^k.$$

And this series converges by our assumption, because $||X||$ is a real number. So the series $\sum\limits_{k=0}^{\infty}a_kX^k$ converges absolutely, and since finite dimensional normed spaces are complete, this implies the series converges itself.

Mark
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