As-Rigid-As-Possible Surface Modeling

paper source

2.1

if M is a psd(positive semi definite) matrix, then for any orthogonal R, $Tr(M) \ge Tr(RM)$

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note: In this monograph positive(semi) define matrices are necessarily symmetric. In the literature a matrix X is sometimes called positive (semi)definite if its symmetric part is positive(semi) definite.

Using spectral decomposition, M can be denoted as: $M = Q\Lambda Q^T$, where matrix Q is an orthogonal matrix.
It is well known that $Tr(Q\Lambda Q^T)=Tr(Q^TQ\Lambda )=Tr(\Lambda )$, therefore, for any orthogonal matrix R,$Tr(RM) = Tr(RQ\Lambda Q^T )=Tr(Q^TRQ\Lambda )$, expressing matrix $Q^TRQ$ as H, we know H is also an orthogonal matrix, i.e. each column vector of H is a unit vector and any two of its column vectors are perpendicular. That is to say the absolute value of every diagonal entry is less than or equal 1. It is simply obtained that $Tr(H\Lambda )\le Tr(\Lambda )$.
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the result pictures after running look like below:

please refer to my github for source codes:
https://github.com/seamanj/ARAP