Given two real $m$ x $k$ matrices $A_1$ and $B_1$ and two $k$ x $k$ real matrices $A_2$ and $B_2$ I want to solve the following equation for $Q$. $Q$ is an orthogonal matrix, i.e. $Q^TQ=I$.
$$\frac{tr(Q^TA_1^TB_1)}{tr(A_1^TA_1)} Q^TA_1^TB_1 + Q^TA_2^TB_2 = symmetric$$
The solution to a similar problem (without the trace expression) has e.g. been described by Schönemann (1966, p. 2) and goes like this.
$$ Q^TA_1^TB_1 + Q^TA_2^TB_2 = symmetric \\ Q^T(\underbrace{A_1^TB_1 + A_2^TB_2}_{C}) = symmetric \\ Q^TC = C^TQ \\ C = QC^TQ \\ CC^T = QC^TQQ^TCQ^T = QC^TCQ^T $$
With $CC^T$ and $C^TC$ being diagonizable and having the same latent roots, let
$$ CC^T = WDW^T \text{and} C^TC = VDV^T \\ \text{with} \\ I = W^TW = WW^T = V^TV = VV^T$$
We get $$ WDW^T = QVDV^TQ^T$$ and thus $$W=QV \\ \text{and} \\ Q=WV^T$$
I tried work out an argument along the same lines but do not know how to do that with the trace expression which also contains $Q$. Any ideas?
PS. I am a psychologist, no mathematician, so please bear with me ;)
Schönemann, P. H. (1966). A generalized solution of the orthogonal procrustes problem. Psychometrika, 31(1), 1–10. doi:10.1007/BF02289451
The problem I see here is that it was derived from another model. This equation is from a minimization using least squares. If you can provide the original model, maybe we could derive the second expression needed to solve the constant.
Here is a nice document describing this:
http://www2.isikun.edu.tr/personel/akca/devrim/2003CH_Praktikum_Procrustes.pdf
More information regarding where did you got this problem could be useful.