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Suppose I

$$\begin{array}{ll} \underset{\Omega \in \Bbb R^{n \times n}}{\text{maximize}} & \text{Tr}(\Omega A)\\ \text{subject to} & \Omega^\top \Omega = I_n\end{array}$$

What am I looking at here? My instinct tells me to use the SVD decomposition $U \Sigma V^T$ of $A$ and simply argue that I must pick $V U^T$ so that then I am left with $\text{Tr} (\Sigma)$ which is then the sum of the square roots of the eigenvalues of $A A^T$, which is positive semidefinite. But why is this the highest the trace can get?

Rodrigo de Azevedo
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Karl
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1 Answers1

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The trace of a matrix product is usually better interpreted as the natural inner product of matrix space, cf. Frobenius inner product and Tensor product of Hilbert spaces.

$$\operatorname{tr}(A^⊤ B) = ⟨A ∣ B⟩_F = ⟨A ∣ B⟩_{ℝ^m⊗ℝ^n} = \sum_{ij} A_{ij}B_{ij}$$

Next, note that:

$$ f(X) = ‖A-X‖_F^2 = ‖A‖_F^2 - 2⟨A ∣ X⟩_F + ‖X‖_F^2$$

Since $‖A‖_F^2$ is independent of $X$, and $‖X‖_F^2 = \operatorname{tr}(X^⊤ X) = \operatorname{tr}(_n) = n$ is constant for $X$ orthogonal, minimizing $f$ is equivalent to the original maximization problem:

$$ \operatorname*{arg\,max}_{X∈O_n}\operatorname{tr}(A^⊤ X) = \operatorname*{arg\,min}_{X∈O_n} ‖A-X‖_F^2 $$

which is the closest orthogonal matrix problem, cf. Showing that matrix $Q=UV^T$ is the nearest orthogonal matrix to $A$.

Hyperplane
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