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Matrix Series#

Definition: Partial Sum

Let \((\boldsymbol{A}_k)_{k \in \mathcal{D}}\) be a sequence of \(m \times n\)-matrices over the same topological field \(F\).

The \(p\)-th partial sum of \((\boldsymbol{A}_k)_{k \in \mathcal{D}}\) is the sum of the first \(p\) matrices in the sequence:

\[\sum_{\substack{j \in \mathcal{D} \\ j \le p}} \boldsymbol{A}_j\]

Definition: Matrix Series

A matrix series is the sequence of the partial sums of some matrix sequence.

Example

Consider the following matrix sequence \((\boldsymbol{A}_k)_{k \in \mathbb{N}_0}\):

\[\boldsymbol{A}_k = \begin{bmatrix}\frac{1}{k!} & 0 \\ \frac{1}{2^k} & \frac{1}{5^k}\end{bmatrix}\]

The corresponding matrix series is the following:

\[\begin{aligned}\sum_{k = 0}^{\infty} \boldsymbol{A}_k & = \sum_{k = 0}^{\infty}\begin{bmatrix}\frac{1}{k!} & 0 \\ \frac{1}{2^k} & \frac{1}{5^k}\end{bmatrix} \\ & = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & \frac{1}{5} \end{bmatrix} + \begin{bmatrix} \frac{1}{2} & 0 \\ \frac{1}{4} & \frac{1}{25} \end{bmatrix} + \begin{bmatrix} \frac{1}{6} & 0 \\ \frac{1}{8} & \frac{1}{125} \end{bmatrix} + \cdots\end{aligned}\]
Example

Consider the following matrix sequence \((\boldsymbol{A}_k)_{k \in \mathbb{N}}\):

\[\boldsymbol{A}_k = \begin{bmatrix}\frac{2^k}{k!} & \frac{1}{k} \\ \frac{1}{2^k} & \frac{1}{5^k}\end{bmatrix}\]

The corresponding matrix series is the following:

\[\begin{aligned}\sum_{k = 1}^{\infty} \boldsymbol{A}_k & = \sum_{k = 1}^{\infty}\begin{bmatrix}\frac{2^k}{k!} & \frac{1}{k} \\ \frac{1}{2^k} & \frac{1}{5^k}\end{bmatrix} \\ & = \begin{bmatrix} 2 & 1 \\ \frac{1}{2} & \frac{1}{5} \end{bmatrix} + \begin{bmatrix} 2 & \frac{1}{2} \\ \frac{1}{4} & \frac{1}{25} \end{bmatrix} + \begin{bmatrix} \frac{4}{3} & \frac{1}{3} \\ \frac{1}{8} & \frac{1}{125} \end{bmatrix} + \cdots\end{aligned}\]

Convergence#

Since a matrix series is just a matrix sequence, the notion of convergence carries over directly.

Example

Consider the following matrix series of real matrices:

\[\begin{aligned}\sum_{k = 0}^{\infty}\begin{bmatrix}\frac{1}{k!} & 0 \\ \frac{1}{2^k} & \frac{1}{5^k}\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & \frac{1}{5} \end{bmatrix} + \begin{bmatrix} \frac{1}{2} & 0 \\ \frac{1}{4} & \frac{1}{25} \end{bmatrix} + \begin{bmatrix} \frac{1}{6} & 0 \\ \frac{1}{8} & \frac{1}{125} \end{bmatrix} + \cdots\end{aligned}\]

It is convergent:

\[\sum_{k = 0}^{\infty}\begin{bmatrix}\frac{1}{k!} & 0 \\ \frac{1}{2^k} & \frac{1}{5^k}\end{bmatrix} = \begin{bmatrix} \sum_{k = 0}^{\infty} \frac{1}{k!} & \sum_{k = 0}^{\infty} 0 \\ \sum_{k = 0}^{\infty} \frac{1}{2^k} & \sum_{k = 0}^{\infty} \frac{1}{5^k}\end{bmatrix} = \begin{bmatrix}\mathrm{e} & 0 \\ 2 & \frac{5}{4}\end{bmatrix}\]
Example

Consider the following matrix series of real matrices:

\[\sum_{k = 1}^{\infty}\begin{bmatrix}\frac{2^k}{k!} & \frac{1}{k} \\ \frac{1}{2^k} & \frac{1}{5^k}\end{bmatrix} = \begin{bmatrix} 2 & 1 \\ \frac{1}{2} & \frac{1}{5} \end{bmatrix} + \begin{bmatrix} 2 & \frac{1}{2} \\ \frac{1}{4} & \frac{1}{25} \end{bmatrix} + \begin{bmatrix} \frac{4}{3} & \frac{1}{3} \\ \frac{1}{8} & \frac{1}{125} \end{bmatrix} + \cdots\]

It is divergent because \(\sum_{k = 1}^{\infty} \frac{1}{k}\) is divergent.

Definition: Absolute Convergence

A matrix series \(\sum_{k \in \mathcal{D}} \boldsymbol{A}_k\) is absolutely convergent if the real series \(\sum_{k \in \mathcal{D}} ||\boldsymbol{A}_k||\) of any of its norms is convergent.

Theorem: Absolute Convergence \(\implies\) Convergence

If a matrix series is absolutely convergent, then it is also convergent.

Proof

TODO