There is a objective function that has the following form: $$ \alpha \|X^T AX\|_F^2-\text{tr}(B^T X) +\beta\|X-C\|_F^2 $$ where $\alpha,\beta$ are scalars, and $X,A,B,C$ are compatible matrices. $\|\cdot\|_F^2$ is the Frobenius norm, $\text{tr}(\cdot)$ is trace operator.
Is it possible to merge similar terms to get a perfect square form as follows: $$ \gamma \|X-D\|_F^2 $$ where $\gamma$ is related to $\alpha$ and $\beta$, $D$ is just related to $A,B,C$.
Could you help give some advice, please?
In general, the answer is no, because $\|X^TAX\|^2_F$ is a degree-4 polynomial in the entries of $X$, while $\|X-D\|_F^2$ is quadratic. For instance, consider the scalar case where $A=1, B=C=0, \alpha=1$ and $\beta=0$. Your objective function is then $X^4$, but your desired expression is a quadratic polynomial $\gamma(X^2 - 2DX + D^2)$. They obviously aren't equal in general.