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Let $T\in B(H_1\oplus H_2)$ have matrix representation $\begin{pmatrix} T_{11} & T_{12}\\ T_{21} & T_{22} \end{pmatrix}$, where $H_1$ and $H_2$ are separable infinite dimensional Hilbert spaces and $T_{ab}\in B(H_i,H_j)$. Are there exist bounded operators $U_a$ and $V_b$ such that $V_aTU_b = T_{ab}$, for $a=1,2$ and $b=1,2$.

I have defined the bounded linear operators $U_1:H_1\longrightarrow H_1\oplus H_2$ by $U_1(h_1) = (h_1,0)$, $U_2:H_2\longrightarrow H_1\oplus H_2$ by $U_2(h_2) = (0,h_2)$, $V_1:H_1\oplus H_2\longrightarrow H_1$ by $V_1(h_1,h_2) = h_1$ and $V_2:H_1\oplus H_2\longrightarrow H_2$ by $V_2(h_1,h_2) = h_2$.

It is easy to check $V_aTU_b = T_{ab}$.

Now, i am thinking can i always define like this because i don't know how operator $T$ is partitioned with respect to any basis that we consider on $H_1$ and $H_2$. On the other hand, i am thinking if i know basis then i can suitably define the operators $U_b$ and $V_a$.

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I’ll assume that by $B(H)$ you denote the set of all bounded linear operators on a linesr space $H$. A matrix representation of $T$ depends on a choice of basises in $H_1$ and $H_2$. But the description of the operator $T$ by operators $T_{ab}$ is basis independent. Operators $U_a$ and $V_b$ are embeddings and projections, whose description are given in terms of vectors of the spaces $H_i$ and are not tied to their basis decomposition. This also concerns the operators $T_{ab}$ which you have defined.

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