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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?

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up vote 1 down vote accepted

For the $N$, the formula doesn't work. Take $A_n=\frac{(-1)^n}n\pmatrix{1&0\\\ 0&1}$. Then $\dim N(A_n)=0$ and taking for example $a_{2k}=b_{2k}=2k$, $a_{2k+1}=b_{2k+1}=2k+1$ and $a_j=b_j=0$ for $j\neq 2k$ and $j\neq 2k+1$ we can see that $\dim N(T)$ is infinite.

For the ranges, consider two cases. If the series $\sum_{n\geq 0}\dim R(A_n)$ is convergent then $\dim R(A_n)=0$ for $n$ large enough so $A_n=0$ for $n$ large enough and the discussion is in fact only finite dimensional.

If the series $\sum_{n\geq 0}\dim R(A_n)$ is divergent then there is $J\subset \mathbb N$ infinite such that $R(A_j)\geq 1$ for all $j\in J$, and you can construct the sequences $\{a_n\}$ and $\{b_n\}$ in order to see that the dimension of $R(T)$ is infinite. For example, take $v_j\in\mathbb R^2, v_j\neq 0$ and $u_j$ such that $A_j(u_j)=v_j$ and writing $J=\{j_1,\ldots,j_n\ldots,\}$, $(w_1^{(n)},w_2^{(n)}) =u_{j_n}e_{j_n}$ we get that the range of $T$ is infinite so in this case the formula is true.

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