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I am solving the following problem:

For given position vectors $\vec{x}_1,...,\vec{x}_n$ and $\vec{y}_1,...,\vec{y}_n$ we have following equations

$Q\vec{x}_i+\vec{b}=\vec{y}_i$, $i=1,...,n$

$Q^TQ=I$

with unknowns

$ Q= \begin{bmatrix} q_{11} & q_{12} & ... &q_{1k}\\ q_{21} & q_{22} & ... &q_{2k}\\ ... & ... & ... & ...\\ q_{k1} & q_{k2} & ... &q_{kk} \end{bmatrix} $ and $b=\begin{bmatrix} b_{1}\\ b_{2}\\ ...\\ b_{k} \end{bmatrix}$

(This is a system of $kn+k^2$ equations for $k^2+k$ unknowns, where $k=\dim\mathbb{R}^k$.) The problem is that $k^2$ equations from $Q^TQ=I$ are not linear. The naive solution: We ignore equality $Q^TQ=I$ and are left with $kn$ equations for $k^2+k$ unknowns:

$Q\vec{x}_i+\vec{b}=\vec{y}_i$, $i=1,...,n$

We assume that $n\geq 2k$ so that we get an overdetermined system of linear equations.

Task: Write down the matrix of this system and find the solution of the system, matrix $Q'$ and vector $b$ by least squares method. Because $Q'$ is not necessarily orthogonal, we correct it by taking $Q$ from QR decomposition $Q'=QR$. The solution to the problem are now matrix $Q$ and vector $b$

My solving:

$Q\vec{x}_i+\vec{b}=\vec{y}_i$ is equal to $\vec{x}_{i}^TQ^T + \vec{b}^T=\vec{y}_i^T$

For $n=3$ the system of equations looks like

$ \begin{bmatrix} \vec{x}_{1}^TQ^T + \vec{b}^T\\ \vec{x}_{2}^TQ^T + \vec{b}^T\\ \vec{x}_{3}^TQ^T + \vec{b}^T\\ \end{bmatrix} = \begin{bmatrix} \vec{y}_{1}^T\\ \vec{y}_{2}^T\\ \vec{y}_{3}^T\\ \end{bmatrix} $

to put in in the form $Ax=b$

$ \begin{bmatrix} \vec{x}_{1}^T& 1\\ \vec{x}_{2}^T& 1\\ \vec{x}_{3}^T& 1 \end{bmatrix} \cdot \begin{bmatrix} Q^T\\ \vec{b}^T \end{bmatrix} = \begin{bmatrix} \vec{y}_{1}^T\\ \vec{y}_{2}^T\\ \vec{y}_{3}^T \end{bmatrix} $

And then got $x=\begin{bmatrix} Q^T\\ \vec{b}^T \end{bmatrix} $ by solving $A^TAx=A^Tb$

and from there got $b$ from the last row and $Q'$ from the other rows.

For $Q'\vec{x}_i+b=\vec{y}_i$ i get corrent solutions. But after getting corrected $Q$ from QR decomposition, i get wrong solutions from this equation when swapping $Q'$ with $Q$.

How am i supposed to represent the system of equations differently

  • 1
    https://math.stackexchange.com/questions/2215359/showing-that-matrix-q-uvt-is-the-nearest-orthogonal-matrix-to-a might help? – Eric Mar 29 '24 at 13:47
  • I believe this is equivalent to the Orthogonal Procrustes problem https://en.wikipedia.org/wiki/Orthogonal_Procrustes_problem – whpowell96 Mar 29 '24 at 16:53

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