# Deriving the Optimal Solution of the Orthogonal Procrustes Problem

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?

• The traces of $\rm A^\top A$ and $\rm B^\top B$ are constants. Minimizing $-f$ is equivalent to maximizing $f$. Hence, the switch from $\arg\min$ to $\arg\max$. Jan 8, 2018 at 18:08
• Thank you @RodrigodeAzevedo Jan 8, 2018 at 18:09

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 $$\arg \max$$.