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$\max_{~\mathbf{X}\in \mathbb{C}^{m\times n}}~ ~~Tr(\mathbf{A}^H \mathbf{X})$

s.t. $\mathbf{X}^H \mathbf{X} = \mathbf{I}_n$, where $\mathbf{A} \in \mathbb{C}^{m\times n}$ and has full column rank.

Can we relax the = in constraint to be $\le$, and use KKT condition to solve it? On the other hand, the optimal solution seems to be related to the SVD of $\mathbf{A}$.

Update: assume the SVD of $\mathbf{A}$ to be $\mathbf{A} = \mathbf{V}\mathbf{\Sigma}\mathbf{U}^H$, then we just take $\mathbf{X}$ to be $\mathbf{X}^* = \mathbf{V}\mathbf{\Lambda}\mathbf{U}^H$, where the diagonal entries of $\mathbf{\Lambda}$ (diagonal matrix) is $\lambda_i = sign(\sigma_i)$. It can be verified that $\mathbf{X}^*$ is the optimal solution satisfying the constriant and with the optimal value to be $\sum_{i=1}^n |\sigma_i|$ (w.l.o.g., assume $n <= m$).

bitvt
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  • Note that askers are expected to provide context for their questions, as is explained here;, it would be helpful if you could [edit] your question to provide some more. For example, where did you encounter this problem? You mention that the optimal solution is related to the SVD of $A$; what do you know about the solution in this case and how do you know it? – Ben Grossmann Jun 14 '21 at 14:42

1 Answers1

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Because $X^HX = I$, all singular values of $X$ are equal to $1$. With this inequality, we have $$ \operatorname{tr}(A^HX) \leq \sigma_\max(X) \cdot \operatorname{tr}(\sqrt{A^HA}) = \operatorname{tr}(\sqrt{A^HA}). $$ Thus, it suffices to find an $X$ that satisfies

  1. $\operatorname{tr}(A^HX) = \operatorname{tr}(\sqrt{A^HA})$,
  2. $X^HX = I$.
Ben Grossmann
  • 225,327
  • Thank you for the hint. I first thought to solve it by KKT condition, as I treat it as an optimization problem at the first sight (it is actually an exercise question of the Matrix analysis course). But after I relaxed the equality constraint to $\le$ and applied the KKT condition, it is still hard to proceed. Then I thought it may be solved by SVD as the constraint is equivalent to specify that $\mathbf{X}$ is a semi-unitary matrix. – bitvt Jun 15 '21 at 08:27