Radius of convergence for matrix exponential

Question

In the context of systems of linear ODE with constant coefficients, my lecture notes on ODE mention that the matrix exponential $e^{tA}$ has an infinite radius of convergence. This shows up in a proof that $e^{tA}$ is a fundamental matrix of $y'(t)=Ay(t)$, where the derivative of a matrix-valued function can be defined component-wise.

The matrix exponential function is defined by the series

$exp(A):=e^A:=\sum \limits_{n=0}^{\infty} \frac{A^n}{n!}$.

Now I know that in the case of the power series in $\mathbb{C}$ we define the radius of convergence of a power series $\sum \limits_{n=0}^{\infty} c_n z^n$ as

$R:=\sup \{r \geq0:(c_n r^n)_{n \in \mathbb{N}} \text{ is a bounded sequence}\}$

where bounded means that $|c_n r^n| \leq M$ for some $M>0$.

How can we generalize this to matrix power series like the matrix exponential? In general, it does not make sense to take the supremum over a set of matrices. But if we have a power series where $z$ is still a complex number, and only the coefficients can be matrices, then we could use the same definition as before

$R:=\sup \{r \geq0:(C_n r^n)_{n \in \mathbb{N}} \text{ is a bounded sequence}\}$

where $C_n$ are matrices and bounded then means that $\|C_n r^n\|| \leq M$ for some $C$.

It follows that each component of the power series has the same radius of convergence as the matrix series.

So we can differentiate term-wise in each component to obtain the derivative of the matrix exponential.

Is this a common definition? I could not find any source that gives a general definition.

Thanks for any help and suggestions!

Answer 1

We can actually keep the same definition for matrix power series since it is in terms of real sequences $(c_n r^n)_{n \in \mathbb{N}}$ only, i.e $r \geq 0$ is a non-negative real number.

For complex series we then have:

Theorem: Let the power series $P(z)=\sum_{n=0}^{\infty} c_n z^n$ have radius of convergence $R$. Then

$(i)$ $P(z)$ diverges for all $z$ with $\lvert z \rvert >R$.

$(ii)$ For each $0<r<R$, the series $P(z)$ converges uniformly on $\bar{B}_r(0)=\{z \in \mathbb{C}:\lvert z \rvert \leq r\}$

$(iii)$ $P(z)$ converges absolutely for $z \in B_R(0)=\{z \in \mathbb{C}:\lvert z \rvert <R\}$

A similar statement follows for matrix power series by replacing the absolute value by the matrix norm (see here).

Answer 2

You are correct to observe that we need some suitable function from matrices to real numbers to define convergence. This is called a norm. Unfortunately for a vector space (and the set of matrices here can be seen as a vector space) there is more than one norm. Fortunately, there is a theorem that says on a finite dimensional vector space all norms are equivalent (see also the wikipedia article). Here this means that all norms will give us the same definition of convergence.

One possible norm for matrices is the sum of the absolute values of the eigenvalues of the matrix. Using this norm (maybe there is a more convenient choice of norm) one can show that the radius of convergence of the exponential series is indeed infinite.

Answer 3

Write $$A\le aU$$ where $a$ is the largest element of $A$ and $U$ is a $d\times d$ matrix of all ones. The comparison is made element-wise, ignoring the signs.

Then

$$A^n\le(aU)^n= d^{n-1}a^nU$$ and

$$\sum_{n=0}^\infty\frac{A^n}{n!}\le\frac{e^{da}}dU=:MU.$$