# Optimizing for an unknown conjugation matrix

I am trying to find the invertible matrix $$M$$ satisfying: $$v_t'= MA_tM^{-1}v_t$$

For a dataset of observed transformations $$(v_t',A_t,v_t)_t$$. Note, there is a different linear transformation $$A_t$$ for each sample tuple.

Basically I have two isomorphic vector spaces $$U$$ and $$V$$, where transformations on $$U$$ are known but the space itself is not observed. I am trying to describe transformations on $$V$$ from observing the transformations $$A_t$$ on the space $$U$$ and their effects on $$V$$ through observed vectors $$(v_t', v_t)_t$$.

For now I am trying to solve it through Least squares: $$M = \underset{X}{argmin} \sum_t||v_t' - XA_tX^{-1}v_t||^2_2$$

But I was wondering if this is a known problem or if a closed solution exists.

I would also be interested in any factorizations of $$M$$ that could make it simpler, maybe orthogonal.

• @Reinderien Thank you for the pointer but I actually meant the sample index $t$, as for every $t$ there is a linear transformation $A_t$. Commented Jan 28 at 21:12
• Ah; good to know - it would help for you to add that to the question. Commented Jan 28 at 21:31

An approach to (approximately) solving your problem is to first reformulate it as: $$\min_{\mathbf{X} \in \mathbb{R}^d} \sum_{t} \|\mathbf{X} \mathbf{v}_t' - \mathbf{A}_t \mathbf{X} \mathbf{v}_t\|_2^2,$$ assuming that $$\mathbf{v}_t,\mathbf{v}_t' \in \mathbb{R}^d$$ and $$\mathbf{A}_t \in \mathbb{R}^{d \times d}$$ for all $$t$$. Thus, inverting the solution to the above solves OP's problem. The benefit is that it is easier to differentiate the objective now wrt. $$\mathbf{X}$$ (see The Matrix Cookbook for details): $$2\sum_t \mathbf{X} \mathbf{v}_t' \mathbf{v}_t'^\top + \mathbf{A}_t^\top \mathbf{A}_t \mathbf{X}\mathbf{v}_t \mathbf{v}_t^\top -(\mathbf{A}_t^\top\mathbf{X} \mathbf{v}_t' \mathbf{v}_t^\top+\mathbf{A}_t\mathbf{X} \mathbf{v}_t \mathbf{v}_t'^\top).$$ Then using gradient descent with the derivative above should solve the problem. Also, looking at the expression above, it seems to me that finding a matrix with zero derivative is as hard as the original matrix problem, so I doubt that a closed-form can be obtained this way.
• You need some additional constraint, otherwise $X = 0$ is a trivial solution. I think you'd want to make sure it's invertible as well. Under linear equality constraints, there is a closed-form solution by setting the gradient to be orthogonal to the equalities (see Wikipedia). Commented Feb 21 at 13:06
• A unique solution does not exist in general and is not expected, basically you get solutions up to multiplication by matrices which commute with the $A_t$'s. Commented Feb 26 at 9:18