4. Assemble an operator from transfer blocks
All examples · Glossary: transfer block · Papers, proofs, Lean, and code
Use the coordinate projectors \(P_1,P_2\) from Example 1, and let
\[T=\begin{pmatrix}1&2\\3&4\end{pmatrix},\qquad
x=\begin{pmatrix}2\\3\end{pmatrix}.
\]
The block \(T_{ji}=P_jTP_i\) selects source component \(i\), applies \(T\) and selects target component \(j\). The source selection acts first.
Calculate the four blocks
Right multiplication keeps one column. Left multiplication keeps one row:
\[T_{11}=\begin{pmatrix}1&0\\0&0\end{pmatrix},\quad
T_{12}=\begin{pmatrix}0&2\\0&0\end{pmatrix},\quad
T_{21}=\begin{pmatrix}0&0\\3&0\end{pmatrix},\quad
T_{22}=\begin{pmatrix}0&0\\0&4\end{pmatrix}.
\]
Adding these four matrices gives \(T\).
Track the contribution of each block
| Route | Contribution to the output |
|---|---|
| Source 1 to target 1 | \(T_{11}x=(2,0)^{\mathsf T}\) |
| Source 2 to target 1 | \(T_{12}x=(6,0)^{\mathsf T}\) |
| Source 1 to target 2 | \(T_{21}x=(0,6)^{\mathsf T}\) |
| Source 2 to target 2 | \(T_{22}x=(0,12)^{\mathsf T}\) |
The sum is \((8,18)^{\mathsf T}\). Direct multiplication gives the same answer:
\[Tx=\begin{pmatrix}1\cdot2+2\cdot3\\3\cdot2+4\cdot3\end{pmatrix}
=\begin{pmatrix}8\\18\end{pmatrix}.
\]
The reconstruction law comes from inserting the complete family on both sides:
\[T=(P_1+P_2)T(P_1+P_2)=\sum_{j=1}^{2}\sum_{i=1}^{2}P_jTP_i.
\]
Compose another operation
Let \(U\) exchange the two coordinates:
\[U=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad
UT=\begin{pmatrix}3&4\\1&2\end{pmatrix}.
\]
The two-step output is \(U(Tx)=(18,8)^{\mathsf T}\), which also equals \((UT)x\). More generally, inserting the identity between the operations gives
\[P_kUTP_i=\sum_{j=1}^{2}(P_kUP_j)(P_jTP_i).
\]
Here \(P_j^2=P_j\) makes the repeated middle selection harmless. The sum runs over the intermediate packets.
Try it
Compute the four contributions for \(x=(1,-1)^{\mathsf T}\).
Answer: They are \((1,0)^{\mathsf T}\), \((-2,0)^{\mathsf T}\), \((0,3)^{\mathsf T}\) and \((0,-4)^{\mathsf T}\). The output is \((-1,-1)^{\mathsf T}\).
Continue reading: Composition and the transfer operations guide. The example checks finite matrix identities. It makes no claim about the cost of a particular hardware implementation.
Papers, proofs, Lean, and code
Terms in this example: Transfer block · Composition · Block matrix
Paper II supplies transfer background. Paper VIII develops operator algorithms and compilation.
Paper II: Refinement Towers, Weighted Transfer, and Integral Separation of Operator Packets
Read · Written proofs · Lean coverage · Code and setup · Evidence
Paper VIII: Packet Operator Algorithms, Exact Compilation, and GPU Realization
Read · Written proofs · Lean coverage · Code and setup · Evidence
Related library guide and code examples. The guide covers related constructions and their software interfaces.
Python arithmetic check for this example: transfer(). This checks the displayed rational, modular or finite-set calculations exactly.
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