# solving equation also involving unknown matrix in trace

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

• your expression is of the form $tr(\cdot)Q^T C$, which obviously is symmetric iff $Q^TC$ is symmetric, therefore, you already know the solution. Commented Aug 3, 2014 at 21:06
• I think we do not know if $Q^TC$ is symmetric. We only know that the sum of the two parts is symmetric, which is IMO not the same. Commented Aug 4, 2014 at 13:33
• My apologies, your expression is not of the form I wrote in my first comment and the solution is, of course, not the same (unfortunately, I do not know the solution!) Commented Aug 4, 2014 at 14:36
• I am impressed that a psychologist knows so well linear algebra. Commented Aug 6, 2014 at 18:40
• Firstly, you can redefine $\tilde A_1 = A_1/\sqrt{tr(A_1^T A_1)}$. Then your problem will look simpler $$tr(Q^TC_1) (Q^T C_1 -C_1^T Q) +Q^T C_2 - C_2^T Q=0$$ where $C_1=\tilde A_1^TB_1$, $C_2=A_2^TB_2$ Commented Aug 6, 2014 at 19:07