For scalar variables $x$, we have a simple solution for the following problem. \begin{eqnarray} \min_x&&\alpha(x-a)^2+\beta(x-b)^2 \\\ \mathrm{s.t. }&&x\leq a\\\ &&x\leq b \end{eqnarray} where $\alpha,\beta>0$.

The optimal solution $x=\min(a,b)$ is straightforward and independent of $\alpha,\beta$.

However, in the case of real symmetrical matrix variables, the problem seems to be much more complex because the relationship $\leq$ in constraints has to be replaced with $\preceq$ which is a more complex relationship defined as $$ A\preceq B\Leftrightarrow A-B\preceq0 $$ where $A\preceq0$ means $A$ is negative semi-definite.

Then the problem above is reformulated as following with matrix variables. Assume that all matrices in this problem are real symmetrical. \begin{eqnarray} \min_X&&\alpha\|X-A\|_F^2+\beta\|X-B\|_F^2\\\ \mathrm{s.t.}&&X\preceq A\\\ &&X\preceq B \end{eqnarray} where $\alpha,\beta\geq0$.

If $A$ and $B$ can be diagonalized by the same orthogonal matrix $U$, the problem reduces to a the problems of eigenvalues since $$ \alpha\|X-A\|_F^2+\beta\|X-B\|_F^2=\alpha\|\tilde{X}-\Lambda\|_F^2+\beta\|\tilde{X}-\Theta\|_F^2 $$ and \begin{eqnarray} X\preceq A\Leftrightarrow\tilde{X}\preceq\Lambda\\\ Y\preceq B\Leftrightarrow\tilde{X}\preceq\Theta \end{eqnarray} where $\tilde{X}=U^\top XU$, $\Lambda=U^\top AU$, $\Theta=U^\top BU$. Since $\Lambda$ and $\Theta$ are diagonal matrices, the problem can decompose to sum of some scalar variables and solved independently.

However, how to solve it if $A$ and $B$ have different eigenvectors? Is the solution independent of $\alpha$ and $\beta$ yet?

Could any one be so kind to help me with the question or give some suggestions? Thank you very much!