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Suppose that $A, B \in \mathbf R^{m \times n}$ with $m \leq n$. Let $\|\,\cdot\,\|$ denote the Frobenius norm and let $\langle \,\cdot\,, \,\cdot\,\rangle$ denote the Frobenius inner product. Note that $$ \|A^T A\|^2 = \sum_{j=1}^m \sigma_j(A)^4 \geq \frac{1}{m} \left( \sum_{j=1}^m \sigma_j(A)^2 \right)^2 = \frac{1}{m} \|A\|^4, $$ where $\sigma_j(A)$ are the singular values of $A$. Is it true that, similarly, $$ \|A^T B\|^2 \geq \frac{1}{m} \langle A, B\rangle^2 \, ? $$

Roberto Rastapopoulos
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1 Answers1

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By Cauchy-Schwarz inequality, we obtain \begin{align*} \langle A,B\rangle^2 =\left(\operatorname{tr}(A^TB)\right)^2 =\left(\operatorname{tr}(AB^T)\right)^2 =\langle I_m,AB^T\rangle^2 \le\|I_m\|_F^2\|AB^T\|_F^2 =m\|\color{red}{AB^T}\|_F^2, \end{align*} which is not exactly your inequality. However, we may modify the proof above to prove your inequality. Since the rank of $A^TB$ is at most $m$, it admits a singular value decomposition $U\Sigma V^T$ where $\Sigma=S\oplus0$ for some nonnegative $m\times m$ diagonal matrix $S$ and $U,V$ are some orthogonal $n\times n$ matrices. It follows that \begin{align*} \langle A,B\rangle^2 &=\left(\operatorname{tr}(A^TB)\right)^2\\ &=\left(\operatorname{tr}(U\Sigma V^T)\right)^2\\ &=\left(\operatorname{tr}(V^TU\Sigma)\right)^2\\ &\le\left(\operatorname{tr}(\Sigma)\right)^2\\ &=\left(\operatorname{tr}(S)\right)^2\\ &=\langle I_m,S\rangle^2\\ &\le\|I_m\|_F^2\|S\|_F^2\\ &=m\|\Sigma\|_F^2\\ &=m\|U\Sigma V^T\|_F^2\\ &=m\| \color{red}{A^TB}\|_F^2.\\ \end{align*} Equality holds if and only if $A^TB\in M_n(\mathbb R)$ is a scalar multiple of a rank-$m$ orthogonal projector.

user1551
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