trace inequality for trace(ABA*C)

Question

I have three $n \times n$ matrices $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ and I calculate the trace \begin{gather} \mathrm{trace}\left(\mathbf{A} \mathbf{B} \mathbf{A}^* \mathbf{C}\right), \end{gather} where $\mathbf{A}^*$ denotes the conjugate transpose (Hermitian) of $\mathbf{A}$.

Matrix $\mathbf{A}$ is positive definite with $\left\|\mathbf{A}\right\|^2 \leq n$.

Matrices $\mathbf{B}$, $\mathbf{C}$ are Hermitian positive semidefinite, with $\mathbf{B} \mathbf{B}^* = \mathbf{B}$ and $\mathbf{C} \mathbf{C}^* = \mathbf{C}$, and $\left\|\mathbf{B}\right\|^2 = \left\|\mathbf{C}\right\|^2 = m < n$.

I would like to prove/disprove the following inequality: \begin{gather} \mathrm{trace}\left(\mathbf{A}\mathbf{B}\mathbf{A}^*\mathbf{C}\right) \leq \mathrm{trace}\left(\mathbf{A}\mathbf{A}^*\right) \mathrm{trace}\left(\mathbf{B}\mathbf{C}\right) \end{gather}

Any hint would be welcome.

Answer 1

The trace $\mathop{\rm trace}(U^\ast V)$ of the product of two matrices $U,V\in \mathbb{C}^{n\times n}$ behaves as an inner product

$$ \langle U, V \rangle= \mathop{\rm trace}(U^\ast V) =\sum_{i=1}^{n}\sum_{j=1}^{n} U_{ij}\cdot \overline{V_{ij}} $$ So worth the inequality of CAUCHY-SCHWARZ $$ \langle U, V \rangle \leq \| U \|\cdot \|V\| $$ Then $$ \mathop{\rm trace }\left(U V W Z \right) \leq \|UV\|\|WZ\| $$ Since for matrices norm $\|\;\;\|$ is worth the following inequality $\|T\cdot S\| \leq \|T\|\cdot \|S\|$ we have $$ \mathop{\rm trace }\left(UVWZ \right) \leq \|U\|\|V\|\|W\|\|Z\| $$ If $\|U\|\leq 1 $, $\|V\|\leq 1$, $\|W\|= 1$ and $\|Z\|= 1$ we have $$ \mathop{\rm trace }\left(UVWZ \right)\leq 1 \qquad \mathop{\rm trace }\left(UV\right)\leq 1 \qquad \mathop{\rm trace }\left(WZ \right)= 1 $$ Now choose $U$, $V$, $W$ and $Z$ properly in function of $A$,$A^\ast$, $B$, $C$, $\sqrt{n}\;$ and $\;\sqrt{m}$ to obtain desired inequality.

UPDATE 09/08/2021

To make sure you correctly use the hypotheses:

1. $\mathbf{A} $ is Hermitian positive definite ( $v^TAv>0 \quad \forall v\in \mathbb{C}^{n\times 1}$);
2. $\mathbf{B}$ is Hermitian positive semidefinite ( $v^TBv\geq 0 \quad \forall v\in \mathbb{C}^{n\times 1}$);
3. $\mathbf{C}$ is Hermitian positive semidefinite ( $v^TCv\geq 0 \quad \forall v\in \mathbb{C}^{n\times 1}$);;
4. $\left\|\mathbf{A}\right\|^2 \leq n$ ( and by 1. we have $\left\|\mathbf{A}\right\|^2>0$) ;
5. $\mathbf{B} \mathbf{B}^* = \mathbf{B}$;
6. $\mathbf{C} \mathbf{C}^* = \mathbf{C}$;
7. $\left\|\mathbf{B}\right\|^2 = m < n$.
8. $\left\|\mathbf{C}\right\|^2 = m < n$.

Answer 2

I have found now a counterexample that shows that the considered inequality does generally not hold. Specifically, if B and C are generated as $B = Q_B Q_B^*$ and $C = Q_C Q_C^*$, where $Q_B$ and $Q_C$ are bases for orthogonal $m$-dimensional subspaces, then $\mathrm{trace}(B C)$ is zero, whereas $\mathrm{trace}(A B A^* C)$ is generally not zero.