For any two matrices $\mathbf{A},\mathbf{B} \in \mathbb{C}^{n \times n}$, we know that the following majorization inequality holds

$$ \tag{1} \label{grz} \sigma^{\downarrow}(\mathbf{A}\mathbf{B}) \prec_w \sigma^{\downarrow}(\mathbf{A})\sigma^{\downarrow}(\mathbf{B}), $$ where $\sigma^{\downarrow}(\cdot)$ denotes the vector of singular values, ordered in the decreasing order. This is equivalent to the following system of inequalities $$ \tag{2} \label{sysineq} \sum_{i=1}^k\sigma_i^{\downarrow}(\mathbf{A}\mathbf{B}) \leq \sum_{i=1}^k \sigma_i^{\downarrow}(\mathbf{A})\sigma_i^{\downarrow}(\mathbf{B}), $$ for $k=1,\dots,n$.

## Proof:

In all the textbooks or papers that I have seen, the proof of this majorization inequality is as follows. By the sub-multiplicativity of the spectral norm, one has $$ \sigma_1^{\downarrow}(\mathbf{A}\mathbf{B}) \leq \sigma_1^{\downarrow}(\mathbf{A})\sigma_1^{\downarrow}(\mathbf{B}). $$ By employing this inequality to the anti-symmetric tensor powers (i.e. the compound matrices) $\wedge^k(\mathbf{A})$ and $\wedge^k(\mathbf{B})$, we have $$ \sigma_1^{\downarrow}\big((\wedge^k \mathbf{A})(\wedge^k \mathbf{B})\big) \leq \sigma_1^{\downarrow}\big(\wedge^k \mathbf{A}\big)\sigma_1^{\downarrow}\big(\wedge^k \mathbf{B}\big), $$ for $k=1,\dots,n$. Then using the facts that $\wedge^k(\mathbf{A}\mathbf{B}) = (\wedge^k \mathbf{A})(\wedge^k \mathbf{B})$ and $\sigma_1^{\downarrow}\big(\wedge^k \mathbf{A}\big) = \prod_{i=1}^k \sigma_i^{\downarrow}(\mathbf{A})$, it follows that

$$ \tag{3} \label{lwm} \prod_{i=1}^k\sigma_i^{\downarrow}(\mathbf{A}\mathbf{B}) \leq \prod_{i=1}^k \sigma_i^{\downarrow}(\mathbf{A})\sigma_i^{\downarrow}(\mathbf{B}), $$ for $k=1,\dots,n$. Finally, inequality \eqref{grz} follows using the fact that log-weak majorization inequality \eqref{lwm} implies weak majorization inequality \eqref{grz} [Bhatia, Matrix analysis, Example II.3.5 (vi)].

## Question:

Can we prove the majorization inequality \eqref{grz} without resorting to the tensor products and employing no facts about them?

Thanks in advance!

## My attempt:

By the maximal characteristic of the singular values, we know that \begin{equation} \sigma_i(\mathbf{A}) = \max_{\substack{\|\bf{x}_i\|=\|\bf{y}_i\|=1 \\ \bf{x}_i \bot \text{span}\{\bf{x}_1,\dots, \bf{x}_{i-1}\} \\ \bf{y}_i \bot \text{span}\{\bf{y}_1,\dots, \bf{y}_{i-1}\}}}\big|\langle \mathbf{A}\bf{x}_i,\bf{y}_i \rangle\big|, \end{equation} for $i=1,\dots,n$. Using this formula, we can demonstrate that the inequalities \eqref{sysineq} are equivalent to the following system of inequalities: \begin{equation} \max_{\substack{\|\bf{x}_i\|=\|\bf{y}_i\|=1, \;i \in [k] \\ \bf{x}_1 \bot \dots \bot \bf{x}_k \\ \bf{y}_1 \bot \dots \bot \bf{y}_k}} \sum_{i=1}^k \big|\langle \mathbf{A}\mathbf{B} \bf{x}_i,\bf{y}_i \rangle\big| \leq \max_{\substack{\|\bf{x}_i\|=\|\hat{\bf{x}}_i\|=1, \;i \in [k] \\ \bf{x}_1 \bot \dots \bot \bf{x}_k \\ \hat{\bf{x}}_1 \bot \dots \bot \hat{\bf{x}}_k}} \max_{\substack{\|\bf{y}_i\|=\|\hat{\bf{y}}_i\|=1, \;i \in [k] \\ \bf{y}_1 \bot \dots \bot \bf{y}_k \\ \hat{\bf{y}}_1 \bot \dots \bot \hat{\bf{y}}_k}} \sum_{i=1}^k\big| \langle \mathbf{B}\bf{x}_i,\hat{\bf{x}}_i \rangle \langle \mathbf{A}\bf{y}_i,\hat{\bf{y}}_i \rangle\big|, \end{equation} for $k=1,\dots,n$. All I can show is that for each $i=1,\dots,k$, we have \begin{equation} \begin{split} \big|\langle \mathbf{A}\mathbf{B} \bf{x}_i,\bf{y}_i \rangle\big| &= \big|\langle \mathbf{B} \bf{x}_i, \mathbf{A}^\mathsf{H}\bf{y}_i \rangle\big| \\ & \leq \|\mathbf{B}\bf{x}_i\| \|\mathbf{A}^\mathsf{H}\bf{y}_i\| \\ & = \max_{\|\hat{\bf{x}}_i\|=1} \big|\langle \mathbf{B}\bf{x}_i,\hat{\bf{x}}_i \rangle\big| \max_{\|\hat{\bf{y}}_i\|=1} \big|\langle \mathbf{A}^\mathsf{H}\bf{y}_i,\hat{\bf{y}}_i \rangle\big|, \end{split} \end{equation} where $\mathbf{A}^\mathsf{H}$ is the conjugate transpose of $\mathbf{A}$. The inequality and the last equality follow by the Cauchy-Schwarz inequality. Therefore \begin{equation} \max_{\substack{\|\bf{x}_i\|=\|\bf{y}_i\|=1 \\ \bf{x}_1 \bot \dots \bot \bf{x}_k \\ \bf{y}_1 \bot \dots \bot \bf{y}_k}} \sum_{i=1}^k \big|\langle \mathbf{A}\mathbf{B} \bf{x}_i,\bf{y}_i \rangle\big| \leq \max_{\substack{\|\bf{x}_i\|=\|\hat{\bf{x}}_i\|=1 \\ \bf{x}_1 \bot \dots \bot \bf{x}_k}} \max_{\substack{\|\bf{y}_i\|=\|\hat{\mathbf{y}}_i\|=1 \\ \mathbf{y}_1 \bot \dots \bot \mathbf{y}_k}} \sum_{i=1}^k\big| \langle B\mathbf{x}_i,\hat{\mathbf{x}}_i \rangle \langle A\hat{\mathbf{y}}_i,\bf{y}_i \rangle\big|. \end{equation} However, these inequalities are weaker than what we want.

*Bhatia, Rajendra*, Matrix analysis, Graduate Texts in Mathematics. 169. New York, NY: Springer. xi, 347 p. (1996).