## Properties of the derivative of a matrix power series

Let $G(x)$ be a formal power series on $\mathbb{C}$ having an infinite radius of convergence and fix an $n \times n$ matrix $A$ over $\mathbb{C}$. The mapping $t \mapsto G(At)$ carries a complex number $t$ to a complex matrix $G(At)$; we can think of this as the ‘direct sum’ of $n \times n$ different functions on $\mathbb{C}$, one for each entry of $G(At)$.

We now define the derivative of $G(At)$ with respect to $t$ to be a mapping $\mathbb{C} \rightarrow \mathsf{Mat}_n(\mathbb{C})$ as follows: $\left[\frac{d}{dt} G(At)\right]_{i,j} = \frac{d}{dt} \left[ G(At)_{i,j}\right]$. In other words, thinking of $G(At)$ as a matrix of functions, $\frac{d}{dt} G(At)$ is the matrix whose entries are the derivatives of the corresponding entries of $G(At)$.

We will use the limit definition of derivative (that is, $\frac{d}{dt} f(t) = \mathsf{lim}_{h \rightarrow 0} \dfrac{f(t+h) - f(t)}{h}$, where it doesnt matter how $h$ approaches 0 in $\mathbb{C}$) and will assume that all derivatives exist everywhere.

Prove the following properties of derivatives:

1. If $G(x) = \sum_{k \in \mathbb{N}} \alpha_kx^k$, then $\frac{d}{dt} G(At) = A \sum_{k \in \mathbb{N}} (k+1)\alpha_{k+1}(At)^k$.
2. If $V$ is an $n \times 1$ matrix with constant entries (i.e. not dependent on $t$) then $\frac{d}{dt} (G(At)V) = \left( \frac{d}{dt} G(At) \right) V$.

[My usual disclaimer about analysis applies here: as soon as I see words like ‘limit’ and ‘continuous’ I become even more confused than usual. Read the following with a healthy dose of skepticism, and please point out any errors.]

Note the following.

 $\left[ \dfrac{d}{dt} G(At) \right]_{i,j}$ = $\dfrac{d}{dt} G(At)_{i,j}$ = $\mathsf{lim}_{h \rightarrow 0} \dfrac{G(A(t+h)_{i,j}) - G(At)_{i,j}}{h}$ = $\mathsf{lim}_{h \rightarrow 0} \left[ \dfrac{G(A(t+h)) - G(At)}{h} \right]_{i,j}$ = $\mathsf{lim}_{h \rightarrow 0} \left[ \dfrac{\sum_k \alpha_k(A(t+h))^k - \sum_k \alpha_k(At)^k}{h} \right]_{i,j}$ = $\mathsf{lim}_{h \rightarrow 0} \left[ \dfrac{\sum_k \alpha_k A^k ((t+h)^k - t^k)}{h} \right]_{i,j}$ = $\mathsf{lim}_{h \rightarrow 0} \left[ \dfrac{\sum_{k > 0} \alpha_k A^k \left( \sum_{m=0}^k {k \choose m} t^m h^{k-m} - t^k \right)}{h} \right]_{i,j}$ = $\mathsf{lim}_{h \rightarrow 0} \left[ \dfrac{\sum_{k > 0} \alpha_k A^k \left( \sum_{m=0}^{k-1} {k \choose m} t^m h^{k-m} \right)}{h} \right]_{i,j}$ = $\mathsf{lim}_{h \rightarrow 0} \displaystyle\sum_{k > 0} \alpha_k A^k \sum_{m=0}^{k-1} {k \choose m} t^m h^{k-1-m}$ (Now we can substitute $h = 0$.) = $\displaystyle\sum_{k > 0} \alpha_k A^k {k \choose {k-1}} t^{k-1}$ (All terms but $m=k-1$ vanish.) = $\displaystyle\sum_{k > 0} k \alpha_k A^k t^{k-1}$ = $\displaystyle\sum_k (k+1) \alpha_{k+1}A^{k+1}t^k$ = $A \sum_k (k+1)\alpha_{k+1}(At)^k$

As desired.

Now say $V = [v_i]$ and $G(At) = [c_{i,j}(t)]$; we then have the following.

 $\dfrac{d}{dt} \left( G(At)V \right)$ = $\dfrac{d}{dt} \left( [c_{i,j}(t)][v_{i,j}] \right)$ = $\dfrac{d}{dt} [\sum_k c_{i,k}(t)v_k]$ = $[\frac{d}{dt} \sum_k c_{i,k}(t)v_k]$ = $[\sum_k (\frac{d}{dt} c_{i,k}(t)) v_k]$ = $[\frac{d}{dt} c_{i,j}(t)][v_i]$ = $\left(\frac{d}{dt} G(At) \right)V$

As desired.