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Let $Z \in \mathbb{R}^{m \times n}$ be a tall matrix ($m > n$). Solve the following optimization problem in $Q \in \mathbb{R}^{m \times n}$

$$\begin{array}{ll} \text{maximize} & \mbox{Tr} \left(Q^T Z \right)\\ \text{subject to} & Q^T Q = I_{n \times n}\end{array}$$

I'm thinking about the QR factorization of $Z$: let $QR=Z$, then $Trace(Q^TQR)=Trace(R)$. But this does not achieve the upper bound, since $\langle Q, Z\rangle \leq \|Q\|_F \|Z\|_F=\sqrt{n}\|R\|_F$. Am I missing something obvious here?

Thanks in advance!

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kkcocoqq
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1 Answers1

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What you want to do is the polar decomposition on $Z$, so $Z= QP$. Note that all singular values of $Q$ are necessarily one. And let $\sigma_k$ be singular values of $Z$.

So you have
$\text{trace}\big(Q^T Z\big) = \text{trace}\big(Q^T QP\big)= \text{trace}\big(P\big) = \sigma_1 + \sigma_2 + ... + \sigma_n \leq 1\cdot \sigma_1 + 1\cdot \sigma_2 + ... + 1\cdot \sigma_n$

where the upper bound is the von Neumann trace inequality, and it is met with equality, so you cannot improve on this.

addendum:
for a lighter weight solution, at least in the special case when $Q^T$ and $Z$ are square, you can reduce this to a standard inequality that you may prove e.g. with Cauchy Schwarz or triangle inequality

prove:
$\text{trace}\big(UB\big) \leq \text{trace}\big(B\big)$
for any orthogonal $U$ and real symmetric positive semidefinite $B$.

(The reduction occurs by polar decomposition on Z and the fact that that the product of orthogonal matrices gives an orthogonal matrix.)

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