Deriving the Optimal Solution of the Orthogonal Procrustes Problem
Sophia Terry 
I am trying to work through the Orthogonal Procrustes Problem but I do not understand a particular step. I would appreciate any help in understanding the steps the author goes from the first line to the second line.
Given the closest linear mapping $\Omega$ between one set of vectors $A$ and another $B$ such that $\Omega$ is an orthogonal matrix. In layman’s terms, we seek the best Euclidean rotation (possibly including mirroring) that maps points $A$ to points $B$.
$$\hat{Ω} = \arg \min( [|ΩA−B|])$$
where $|\cdot|$ denotes the Frobenius norm of a matrix – the sum of the square of all of the elements. To make progress, we recall that the trace of a matrix is the sum of its diagonal entries and so $|X| = \mbox{tr} [X^{T}X]$ which gives the new criterion
$$\begin{array}{rl} \hat{Ω} &= \arg \min [\mbox{tr}[A^{T}A] + \mbox{tr} [B^{T}B] − 2 \mbox{tr}[A^{T}Ω^{T}B]]\\ &= \arg \max [\mbox{tr}[A^{T}Ω^{T}B]]\\ &= \arg \max[\mbox{tr}[Ω^{T}BA^{T}]]\end{array}$$
How can one go from the first line of this derivation to the second?
$\endgroup$ 21 Answer
$\begingroup$The traces of $ {A}^{⊤} A $ and $ {B}^{⊤} B $ are constants.
For any function Minimizing $ −f \left( \cdot \right) $ is equivalent to maximizing $ f \left( \cdot \right) $.
Hence, the switch from $ \arg \min $ to $ \arg \max $.
Remark
Wiki solution based on Rodrigo de Azevedo's comment.
