Let $\mathcal{H}$ be a complex finite-dimensional Hilbert space with basis $\mathcal{B} = \{e_{1},...,e_{n}\}$. Suppose that, for each $t \in \mathbb{R}$, $M(t)$ is a linear operator on $\mathcal{H}$. There are two ways of understanding the following equation: $$\frac{dM(t)}{dt} = A(t) \tag{1}\label{1}$$ with $A(t)$ being another linear operator on $\mathcal{H}$. The first one, call it (A1), is in terms of the operator norm: if $\varepsilon > 0$ is given, then there exists $\delta > 0$ such that: $$\|\frac{M(t+h)-M(t)}{h}-A(t)\| < \varepsilon \tag{2}\label{2}$$ whenever $0 < |h| < \delta$. Here, the norm $\|\cdot \|$ is the usual operator norm: $$\|T\| := \sup_{\|\psi\| = 1}{\|T\psi\|}.$$
The other way of understanding (\ref{1}), say (A2), is by thinking of each $M$ and $A$ as matrices, so (\ref{1}) would actually mean: $$\frac{dM_{ij}(t)}{dt} = A_{ij}(t) \tag{3}\label{3}$$ for every entry $ij$. I am trying to prove that (A1) and (A2) are equivalent.
My proof: For simplicity, I will denote: $$T_{h}M := \frac{M(t+h)-M(t)}{h}.$$
(A1) $\Rightarrow$ (A2): Using the Cauchy-Schwarz inequality: $$|\langle e_{i}, (T_{h}M-A)e_{j}\rangle| \le \|e_{i}\|\|(T_{h}M-A)e_{j}\| =\|(T_{h}M-A)e_{j}\| \le \sup_{\|\psi\|=1}\|(T_{h}M-A)\psi\| = \|T_{h}M-A\| < \epsilon \tag{4}\label{4}$$ whenever $0 < |h| < \delta$.
(A2) $\Rightarrow$ (A1) Let us take $\psi = \sum_{i=1}^{n}\alpha_{i}e_{i}\in \mathcal{H}$ arbitrary. Then, the $i$-th entry of the vector $M(t)\psi$ is given by: $$(M(t)\psi)_{i} = \sum_{j=1}^{n}M_{ij}(t)\alpha_{j}$$ so that: $$M(t)\psi = \sum_{i,j=1}^{n}M_{ij}(t)\alpha_{j}e_{i} \tag{5}\label{5}$$ By (\ref{3}), given $\varepsilon > 0$ there exists $\delta_{ij}>0$ such that: $$\bigg{|}\frac{M_{ij}(t+h)-M(t)}{h}-A_{ij}(t)\bigg{|} < \varepsilon_{ij}$$ whenever $0 < |h| < \delta_{ij}$. Take $\delta := \min_{ij}\delta_{ij}$. We get: $$\|(T_{h}M-A(t))\psi\| = \|\sum_{i,j=1}^{n}[(T_{h}M)_{ij}-A_{ij}(t))]\alpha_{j}e_{i}\| \le \|\sum_{i,j=1}^{n}|(T_{h}M)_{ij}-A_{ij}(t)||\alpha_{j}| \le \epsilon \sum_{i,j=1}^{n}|\alpha_{j}| \le \epsilon \sqrt{n}\|\psi\| \tag{6}\label{6}$$ whenever $0 < |h| < \delta$. (In the last inequality, I used that $\|\psi\|_{1} = \sum_{i=1}^{n}|\alpha_{i}| \le \sqrt{n}\|\psi\|$).
Is my proof correct? Are there other ways (maybe more straightforward) to prove the equivalence?