6. Reconstruct from redundant coordinates
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Three measurements can describe a two-dimensional vector. They may reconstruct it exactly even though their individual reconstruction components are not projectors.
Analyze and synthesize
Work over the real numbers and set
\[A=\begin{pmatrix}1&0\\0&1\\1&1\end{pmatrix},\qquad
S=\frac13\begin{pmatrix}2&-1&1\\-1&2&1\end{pmatrix}.
\]
\(A\) is \(3\) by \(2\) and sends a vector to three coordinates. \(S\) is \(2\) by \(3\) and sends those coordinates back. Multiplication gives
\[SA=\frac13\begin{pmatrix}2+1&-1+1\\-1+1&2+1\end{pmatrix}
=I_2.
\]
For \(x=(2,3)^{\mathsf T}\),
\[Ax=\begin{pmatrix}2\\3\\5\end{pmatrix},\qquad
S(Ax)=\frac13\begin{pmatrix}4-3+5\\-2+6+5\end{pmatrix}
=\begin{pmatrix}2\\3\end{pmatrix}.
\]
The third coordinate is redundant because it is the sum of the first two.
Reconstruct one coordinate at a time
Let \(E_i\) keep only coordinate \(i\) in the three-dimensional coordinate space. Define \(C_i=SE_iA\). Their contributions to \(x\) are
\[C_1x=\begin{pmatrix}4/3\\-2/3\end{pmatrix},\qquad
C_2x=\begin{pmatrix}-1\\2\end{pmatrix},\qquad
C_3x=\begin{pmatrix}5/3\\5/3\end{pmatrix}.
\]
Adding gives
\[\begin{pmatrix}4/3-1+5/3\\-2/3+2+5/3\end{pmatrix}
=\begin{pmatrix}2\\3\end{pmatrix}.
\]
The matrices are
\[C_1=\frac13\begin{pmatrix}2&0\\-1&0\end{pmatrix},\quad
C_2=\frac13\begin{pmatrix}0&-1\\0&2\end{pmatrix},\quad
C_3=\frac13\begin{pmatrix}1&1\\1&1\end{pmatrix}.
\]
They sum to \(I_2\). However,
\[C_1^2=\frac19\begin{pmatrix}4&0\\-2&0\end{pmatrix}
=\frac23C_1\ne C_1.
\]
So \(C_1\) is not a projector. Exact additive reconstruction alone does not establish idempotence.
A left inverse is not a two-sided inverse
In the other order,
\[AS=\frac13\begin{pmatrix}2&-1&1\\-1&2&1\\1&1&2\end{pmatrix}\ne I_3.
\]
For \(z=(0,0,1)^{\mathsf T}\), we get \(ASz=(1/3,1/3,2/3)^{\mathsf T}\), not \(z\). The original \(z\) cannot equal \(Ax\) for any \(x\), since valid analyzed coordinates satisfy \(z_3=z_1+z_2\).
Try it
Calculate the three contributions for \(x=(1,-1)^{\mathsf T}\).
Answer: \(Ax=(1,-1,0)^{\mathsf T}\). The contributions are \((2/3,-1/3)^{\mathsf T}\), \((1/3,-2/3)^{\mathsf T}\) and zero. They sum to the original vector.
Continue reading: Left inverse, redundant coordinates, and components versus projectors.
Papers, proofs, Lean, and code
Terms in this example: Analysis and synthesis · Left inverse · Redundant coordinates · Reconstruction
Compare the exact packet setting of Paper I with the broader analysis and synthesis construction in the library guide.
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: redundant_coordinates(). This checks the displayed rational, modular or finite-set calculations exactly.
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