A linear operator on the space of all sequences that cannot be thought of as matrix This not homework but a general question about how to think about linear operators on not so nice spaces.
Consider the space of all real-valued sequences $\mathbb{R}^{\mathbb{N}}$ and a linear operator $T \colon \mathbb{R}^{\mathbb{N}} \to \mathbb{R}^{\mathbb{N}}$ defined on it.
Can it happen that
$$
T(\sum_{j \in \mathbb{N}} \alpha_j e_j) \neq \sum_{j \in \mathbb{N}} \alpha_j T(e_j)
$$ holds (where $e_j$ is the sequence which is 1 at index $j$ and 0 elsewhere)?
What about the case where we assume that $T$ is symmetric, i.e. $(T e_j)_i = (T e_i)_j$?
 A: The space $ V = \mathbb{R}^\mathbb{N} $ is indeed a vector space, but without topology, the sequence of vectors $ e_i $ is not in any sense a basis for $ V $.  For example the vector $ \xi = \langle 1, 1, 1 \ldots \rangle $ is not a (finite) linear combination of the $ e_i $.  So even if you have already specified coefficients $ \alpha_{ij} $ with $ T e_i = \sum \alpha_{ij} e_j $, you are free to define $ T \xi $ however you want, without contradiction.
You can, however, find an uncountable Hamel basis for $ V $ (look up Hamel basis on Wikipedia).  Call this basis $\langle \xi_i\rangle_{i \in I}$ for some uncountable indexing set $ I $.  You can then write down coefficients $ \alpha_{ij} $ for $ (i,j) \in I \times I $, with $ \alpha_{ij} = 0 $ for all but at most finitely many $ j $ for each fixed $ i $, and then you can, in a sense, have a "matrix" representation of $ T $ where
$$
T \left( \sum_{i \in I} c_i \xi_i \right) = \sum_{i \in I} {
  \sum_{j \in I} { c_i \alpha_{ij} \xi_j}
}
$$
for when all but finitely many $ c_i $ are zero.  Notice, all sums on both sides are finite by assumption.  I should warn you, this is not of any practical utility known to me.  The construction of the uncountable Hamel basis uses the axiom of choice, which should divorce it from any practical application.
A: As comments pointed out, first thing would be to define properly. But i think what you said can happen, Lets say $T$ is linear operator with domain on the set of all convergent sequences in $\mathbb{R}$. Then define $$T(\{s_n\}) = \lim_{n \rightarrow \infty} s_n$$. Let $s_n = 1-1/n$. Then $T(\{s_n\}) = 1$ and $T(s_i e_i) = 0$ where $e_i$ is the sequence with $1$ at $i^{th}$ position and $0$ at other places. Hence we have $\lim_{\ell \rightarrow \infty} \sum_{i=1}^{\ell} T(s_ie_i) = 0$ but $T(\{s_n\}) = T(\lim_{\ell \rightarrow \infty} \sum_{i=1}^{\ell} s_i e_i) = 1$. See if this is useful.
