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Posted here too: https://math.stackexchange.com/questions/1711026/two-minimization-problems-using-singular-value-decomposition

Let $q_0, q_1:[0,1]\to \mathbb{R}^n$ be two maps whose components are $L^2[0,1]$, i.e. $q_0, q_1 \in L^2([0,1],\mathbb{R}^n)$. Denote by $||.||$ the Euclidean distance in $\mathbb{R}^n$. Consider the following two minimization problems:

1) minimize $\int_{0}^{1} || q_0(t) - A q_1(t) ||^2 dt $ over $A \in SO(n)$.

2) Consider $U \subset L^2([0,1],\mathbb{R}^n)$ to be the unit sphere in $L^2([0,1],\mathbb{R}^n)$. Let $q_0, q_1 \in U$ now. Then $Aq_1 \in U$ as well for any $A \in SO(n)$. Consider the geodesic $c(A)$ in $U$ joining $q_0, Aq_1$. Minimize the length of $c(A)$.

How can we solve these two problems with/without, preferably with, using singular value decomposition? It was mentioned in the paper http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=5601739

But I'd appreciate a detailed explanation for solving these two problems.

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Problem 1 is essentially finite dimensional and quite geometric in nature.

For a real square matrix $M$ of order $n$ and $1\le p\le\infty$ denote $\big| M\big|_p$ its $p$-trace norm.

Consider the $n\times n$ matrix $Q$ with entries $Q_{ij}:=(q_0^i, q_1^j)_{ 2}:=\int_0^1q_0^i(x)q_1^j(x)dx$. Then, for any $A\in SO(n)$

$$\|q_0 - Aq_1\|_2^2:=\int_0^1\big|q_0(x) - Aq_1(x)\big|^2dx=\|q_0\|_2^2 +\|q_1\|_2^2- 2\int_0^1\sum_{ij} A_{ij}q_0^i(x) q_1^j(x)dx $$

$$=\|q_0\|_2^2 +\|q_1\|_2^2- 2 \sum_{ij} A_{ij}Q_{ij}=\|q_0\|_2^2 +\|q_1\|_2^2- 2 \operatorname{tr}(A^TQ). $$

Therefore, the initial minimization reduces (up to an additive constant) to maximizing the trace of a matrix in the $SO(n)$-orbit of $Q$ by (left or right) multiplication. Consider first the case $\operatorname{det}(Q)\ge0$. For any $A\in SO(n)$ we have, by duality,

$$ \operatorname{tr}(A^TQ)\le |A|_\infty |Q|_1= |Q|_1 .$$

The nuclear norm of $Q$ on the RHS is indeed attained by a suitable $A\in SO(n)$ on the LHS, and here the SVD of $Q$ may help. Write $Q=UDV^T$ with $U, V$ in $SO(n)$ and $D$ non-negative and diagonal, and choose $A:=UV^T$; then $A^TQ=VDV^T$ and by invariance of the trace $\operatorname{tr}(A^TQ)=\operatorname{tr}(D)= |Q|_1 .$ $$*$$

Note that the latter optimization problem is also equivalent to a point-set distance minimization, that is, finding the closest rotation to $Q$ in the Frobenius distance. Indeed, for any rotation $A$, since $\big|A\big|_2^2=n$ we have

$$ \big| A-Q\big|_2^2 = n + \big|Q\big|_2^2 -2\operatorname{tr}(A^TQ)$$ so that maximizing $\operatorname{tr}(A^TQ)$ is again equivalent to minimizing a distance.

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    $\begingroup$ Note that the matrix $Q$ produced by $q_0$ and $q_1$ can be any square matrix. If det(Q)≤0, then, I think $\max_{A\in SO(n)} tr(A^TQ)$ is equal to $\sigma_1+\sigma_2+⋯+\sigma_{n−1}−\sigma_n$, where $\sigma_1\ge\sigma_2\ge,\dots,\sigma_{n−1}\ge\sigma_n$ are the singular values of $Q$ in decreasing order. $\endgroup$ Commented Mar 25, 2016 at 17:10

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