Claim: Given a Schauder basis $\{ \boldsymbol{e}_j \} \subset X$ and a linear operator $T: \mathrm{D}(T) \subset X \to X$, the vectors $T(\boldsymbol{e}_j)$ aren't always sufficient to describe the action of the operator on a general vector of $X$.
Or in other words: There exists a non-zero linear operator, whose matrix is zero.
Proof: Let $X = L^2([0, 2\pi])$ be the space of all square-integrable functions equipped with the Fourier basis $\{ \boldsymbol{e}_j \} = (1,\, \sin x,\, \cos x,\, \sin 2x,\, ...)$. We now use the distribution theory to construct an unbounded linear operator $T$.
Let us take $\mathrm{D}(T)$ to be the space of bounded piece-wise continuous functions. Since $\mathrm{D}(T) \subset L^1_{\mathrm{loc.}}(\mathbb{R})$, all vectors $f \in X$ can be considered regular distributions on $\mathcal{D}(\mathbb{R})$ and assigned a distributional derivative $Df$. The result will be a sum of a regular function and (singular) delta distributions in discontinuities: $Df = (Df)_{\mathrm{reg.}} + (Df)_{\mathrm{sing.}}$. Finally we chose a test function $\varphi \in \mathcal{D}(\mathbb{R})$ that is nonzero on $[0, 2\pi]$ (ie. $\varphi(x)=1$ on $[0, 2\pi]$). We define:
$$
T(f) = (Df)_{\mathrm{sing.}}[\varphi] \; \boldsymbol{e}_1
$$
This operator is zero for all continuous functions, including the entire basis $\{ \boldsymbol{e}_j \}$, but it is generally nonzero for discontinuous functions.
Corollary: Since, given a Schauder basis, there's a natural isomorphism $\iota: L^2 \to \ell_2$:
$$
\underbrace{f}_{L^2} = \sum_j \underbrace{c_j}_{\ell_2} \boldsymbol{e}_j
\: , \quad
\iota(f) = \{ c_j \}
$$
we can construct a non-zero operator on $\ell_2$ whose matrix is only zeros:
$$
S(a_j) = \iota \, T \, \iota^{-1} \; a_j
\quad \neq \quad
\sum_{j,k} S_{jk} \, a_j \, \boldsymbol{e}_k = 0
$$
Claim: Let $X$ be a separable Banach space and $T: \mathrm{D}(T) \subset X \to X$ a densely defined closed operator. Given a Schauder basis $\{ \boldsymbol{e}_j \} \subset \mathrm{D}(T)$ (such a basis always exists), we construct the coordinates $T_{jk}$ of the operator $T$:
$$
T(\boldsymbol{e}_j) =: \sum_k T_{jk} \, \boldsymbol{e}_k
$$
Now we define an operator $T_0: \mathrm{D}(T_0) \to X$ using those coordinates:
$$
T_0(v) = T_0( \, \sum_j v^j \boldsymbol{e}_j \, ) = \sum_{j,k} T_{jk} \, v^j \, \boldsymbol{e}_k
\: , \qquad
\mathrm{D}(T_0) = \{ v \in X \,|\, \text{the sum converges}\}
$$
Then $T_0$ is also closed and $T_0 \subset T$.
Proof: The Schauder basis in question exists, because $\mathrm{D}(T)$ is dense in $X$ – therefore it is a generating set in the sense of countable linear combinations. We can throw away its linearly dependent vectors until we're left with a basis.
The defining feature of closed operators is:
$$
T \text{ is closed} \;\wedge\;
x_n \to x \;\wedge\;
T(x_n) \to y
\quad \implies \quad
x \in \mathrm{D}(A) \;\wedge\;
T(x) = y
$$
To show that $T_0 \subset T$, we chose an arbitrary $v \in \mathrm{D}(T)$ and define:
$$
x_n = \sum_{j=1}^n v^j \, \boldsymbol{e}_j
\: , \quad
x_n \to v
$$
Then:
$$
T(x_n)
= \sum_{j=1}^n v^j \, T(\boldsymbol{e}_j)
= \sum_{j=1}^n \sum_k T_{jk} \, v^j \, \boldsymbol{e}_k
$$
If that sum converges for $n\to\infty$, by the definition of $T_0$, it holds that $v \in \mathrm{D}(T_0)$ and the result is equal to $T_0(v)$. However, from the closedness criterion, we also know that $v\in\mathrm{D}(T)$ and the result is equal to $T(v)$. If the sum doesn't converge, by definition $v\notin\mathrm{D}(T_0)$, but it doesn't tell us anything about $T$. Thus, $T_0 \subset T$.
It remains to prove that $T_0$ is also closed. We assume $① \Leftrightarrow x_n \to x$ and $② \Leftrightarrow T_0(x_n) \to y$ and want to prove $T_0(x) = y$. If we expand the expressions, we get:
$$
① \; \implies \;
\sum_{j=1}^m (x_n)^j \, \boldsymbol{e}_j \xrightarrow{n}
\sum_{j=1}^m x^j \, \boldsymbol{e}_j
\\[8pt]
② \; \Longleftrightarrow \;
\sum_{j,k} T_{jk} \, (x_n)^j \, \boldsymbol{e}_k \xrightarrow{n} y
$$
To simplify the matters, we introduce a double-sequence $\Phi_{mn}$:
$$
\Phi_{mn} = \sum_{j=1}^m \sum_k T_{jk} \, (x_n)^j \, \boldsymbol{e}_k
$$
The assumption $②$ is equivalent to $\lim_{n\to\infty} \lim_{m\to\infty} \Phi_{mn} = y$. The assumption $①$ implies that $\lim_{n \to \infty} \Phi_{mn}$ exists – as it turns out, the convergence is even uniform wrt. $m$ (is it really tho?). Then it follows from the Moore-Osgood Theorem, that $\lim_{m\to\infty} \lim_{n\to\infty} \Phi_{mn}$ exists and is equal to $y$, too.