MIN tr(ABCD)
s.t BC=I
An optimization problem with trace in the objective function, but matrix equation in the constraint. So, you see the problem right away when constructing the Lagrangian (you end up adding scalar with matrices).
I'm reading the solution and it goes like this when building the Lagrangian:
L=tr(ABCD) - tr((BC-I)*Z)
Can you explain this solution? Where did the Z come from? how does it help solving the problem mentioned above. How was tr() added to the constraint.
Best Answer
I think, you need the trace in this setting, since the constraint is not scalar, but matrix-valued. The $Z$ is your matrix-valued Lagrange multiplier, which cannot be scalar in this setting.
I have written down a the Lagrange approach for such very general equality constraints: \begin{align} \max_{x \in X} f(x) \\ \text{s.t. g(x) = 0}, \end{align} where $f:X \rightarrow \mathbb{R}$, and $g: X \rightarrow Y$, where $X$ and $Y$ are Hilbert spaces (like $\mathbb{R}^k$, $\mathcal{L}^2$,...).
The Lagrange function is given by $$L(x, \Lambda) = f(x) + \langle \Lambda, g(x) \rangle_Y,$$
where $\Lambda \in Y$ is the Lagrange multiplier, and $\langle \cdot, \cdot \rangle_Y$ is the inner product of $Y$.
The question is: Does the set of matrices that is some superset of the image of $BC-I$ in your notation, form a Hilbert space $Y$ with the trace of the product of two matrices being their inner product; i.e. $\langle A, B \rangle_X := \mathrm{trace}(AB)$? That is at least not obvious for me.