How to get the perfect square for the following equation

Question

The problem is defined as follows: $$ \min_X tr(X^T A X)-\alpha tr(X^T B) $$ I want to get the equal perfect square equation as that above, that is $$ \min_X \| X-C\|_F^2 $$ where $C$ is related to $A$ and $B$. Is it possible? How to derive the perfect square equation?

UPDATED: This problem is from this paper. In Eqn. (15), the authors have derived a Lagrange function, but I'm not sure whether the following term in Eqn. (15) is right when optimizing the $J_n$. From the Eqn. (15), we have $$ \min_{J_n} \sum_{n=1}^{N}\|J_n\|_* + \lambda \|J_n^T(W_b-\alpha B_n)J_n\|_F^2 -tr(V_2^TJ_n) +\frac{\mu}{2}\|U_n-J_n\|_F^2 $$ The authors have wrote an equal optimization problem in the first equation of Eqn. (16) of the paper. But I was confused with it for a long time. And I also made a post in this link to ask for a favor, but @user1551 also thought the term $\lambda \|J_n^T(W_b-\alpha B_n)J_n\|_F^2$ in the equation above couldn't be changed into the perfect square form in Eqn. (16).

Thus, I've tried to change the term $\lambda \|J_n^T(W_b-\alpha B_n)J_n\|_F^2$ into $\lambda tr(J_n^T(W_b-\alpha B_n)J_n)$, and make a derivation as follows:

But I still couldn't get the perfect square form. Thus I made this post for your kind help. If convenient, could you help derive it, please?

Answer 1

As noted by copper.hat, this is not always possible.

In particular, in your second problem, there is no coupling between the different entries in the array $X$. In particular, each entry $X_{ij}$ solves the one-dimensional problem $$\min_{X_{ij}} (X_{ij} - C_{ij})^2.$$

In the general case, this decoupling is not possible for the first problem.

Answer 2

The other commenters are correct, in that you can't get it precisely into the form you've shown. But you can get a perfect square, if that's all you're after, as long as $A$ is positive definite. Let $A^{1/2}$ be the symmetric square root of $A$. Then $$\begin{aligned} &\mathop{\textrm{Tr}}(X^TAX)-\alpha\mathop{\textrm{Tr}}(X^TB) = \langle A^{1/2}X,A^{1/2}X\rangle - \alpha\langle A^{1/2}X,A^{-1/2}B\rangle\\ &\qquad = \langle A^{1/2}X-\tfrac{\alpha}{2}A^{-1/2}B,A^{1/2}X-\tfrac{\alpha}{2}A^{-1/2}B\rangle -\tfrac{\alpha^2}{4}\langle A^{-1/2}B,A^{-1/2}B\rangle \\ &\qquad \|A^{1/2}X-\tfrac{\alpha}{2}A^{-1/2}B\|_F^2-\tfrac{\alpha^2}{4}\|A^{-1/2}B\|_F^2 \end{aligned} $$ You can drop the constant term in the optimization, and you have $$\min_X \|A^{1/2} X-\tfrac{\alpha}{2}A^{-1/2}B\|_F^2$$ If you're wiling to change variables to $Y=A^{1/2}X$, you now have $$\min_Y \|Y-\tfrac{\alpha}{2}A^{-1/2}B\|_F^2$$