Let $H$ be the real separable Hilbert space with orthonormal basis $\{e_n\}$ and consider the operator $T:H \times H \to H \times H$ given by
$T(\sum a_ne_n, \sum b_ne_n) = \sum A_n(a_ne_n, b_ne_n)$
where $A_n$ is a $2 \times 2$ matrix with real entries for all $n$. Here we interpret $A_n(a_ne_n, b_ne_n)$ as
$\begin{pmatrix}a_1 && a_2 \\ a_2 && a_4\\ \end{pmatrix}\begin{pmatrix}a_ne_n \\ b_ne_n \\ \end{pmatrix} = \begin{pmatrix}(a_1a_n + a_2b_n)e_n\\ (a_3a_n + a_4b_n)e_n\\ \end{pmatrix}$
Assume the $A_n$ are such that $T$ is a bounded operator.
Is it true that $\text{dim } N(T) = \sum_n \text{dim } N(A_n)$ and $\text{dim }R(T) = \sum_n \text{dim }R(A_n)$ ? And if so, how can we prove this?