1. Split and reconstruct a vector
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We will split a vector into two components and put it back together. Work over the real numbers and take
\[x=\begin{pmatrix}2\\3\end{pmatrix},\qquad
P_1=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad
P_2=\begin{pmatrix}0&0\\0&1\end{pmatrix}.
\]
Select the components
Multiply each row by the column vector:
\[P_1x=\begin{pmatrix}1\cdot2+0\cdot3\\0\cdot2+0\cdot3\end{pmatrix}
=\begin{pmatrix}2\\0\end{pmatrix},\qquad
P_2x=\begin{pmatrix}0\\3\end{pmatrix}.
\]
The first matrix keeps the first coordinate. The second keeps the second.
Check the packet identities
Multiplying the matrices gives
\[P_1^2=P_1,\qquad P_2^2=P_2,\qquad
P_1P_2=P_2P_1=0,\qquad
P_1+P_2=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I_2.
\]
Idempotence says selecting a component twice has the same effect as selecting it once. The zero products say the two selections annihilate each other. The sum-to-identity equation gives reconstruction:
\[P_1x+P_2x=\begin{pmatrix}2\\0\end{pmatrix}
+\begin{pmatrix}0\\3\end{pmatrix}
=\begin{pmatrix}2\\3\end{pmatrix}=x.
\]
This last identity holds for every vector in this two-dimensional space because \(P_1+P_2=I_2\), not just because it worked for the chosen input.
An oblique decomposition
A projector need not be orthogonal. Replace the matrices by
\[Q_1=\begin{pmatrix}1&-1\\0&0\end{pmatrix},\qquad
Q_2=\begin{pmatrix}0&1\\0&1\end{pmatrix}.
\]
For example,
\[Q_1Q_2=\begin{pmatrix}0&1-1\\0&0\end{pmatrix}=0.
\]
Likewise \(Q_2Q_1=0\), \(Q_1^2=Q_1\), \(Q_2^2=Q_2\) and \(Q_1+Q_2=I_2\). The components of the same input are now
\[Q_1x=\begin{pmatrix}-1\\0\end{pmatrix},\qquad
Q_2x=\begin{pmatrix}3\\3\end{pmatrix}.
\]
They still add to \(x\). Their ordinary dot product is \(-3\), so they are not perpendicular. Algebraic reconstruction and orthogonality are different requirements.
Try it
Use \(x=(4,-1)^{\mathsf T}\) with both families.
Answer: The \(P\) components are \((4,0)^{\mathsf T}\) and \((0,-1)^{\mathsf T}\). The \(Q\) components are \((5,0)^{\mathsf T}\) and \((-1,-1)^{\mathsf T}\). Both pairs sum to \((4,-1)^{\mathsf T}\).
Continue reading: Packet family, orthogonal projector, and the library's first packet program.
Papers, proofs, Lean, and code
Terms in this example: Projector · Packet · Complete packet family · Orthogonal projector · Direct sum
Exact packet coordinates provide the source setting for the elementary decomposition.
Paper I: Typed Operator Systems, Seed Rank, and Exact Packet Coordinates
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: packets(). This checks the displayed rational, modular or finite-set calculations exactly.
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