An upper bound for $trace(XYXZ)$ in terms of $trace(X^2YZ)$

Question

I want to find an upper bound for $trace(XYXZ)$ in terms of $trace(X^2YZ)$. $X, Y, Z$ are Hermitian positive definite matrices.

Any idea/help is appreciated.

Update: Assume that $Y$ and $Z$ commute: $YZ=ZY$.

Answer 1

It is possible for $\text{trace}(X^2 Y Z) = 0$ with $X,Y,Z$ positive definite Hermitian. Try $$ X^2 = \pmatrix{2 & -1\cr -1 & 1\cr},\ Y = \pmatrix{1/10 & 0\cr 0 & 1\cr},\ Z = \pmatrix{2 & 1\cr 1 & 7/10}$$ However, this will not work with the added condition that $Y$ and $Z$ commute. If $Y$ and $Z$ commute, so do $Y$ and $Z^{1/2}$, and $YZ = Z^{1/2} Y Z^{1/2}$ is positive definite. Then $$\text{trace}(X^2 Y Z) = \text{trace}((YZ)^{1/2} X^2 (YZ)^{1/2}) > 0$$

Answer 2

Short answer: None of the two has a reason to be larger than the other. Some bounds in terms of the condition numbers of $Y,Z$ are $$trace(XYXZ) \leq \min(\kappa(Y), \kappa(Z)) \cdot trace(X^2YZ)$$ $$trace(X^2YZ) \leq \min(\kappa(Y), \kappa(Z)) \cdot trace(XYXZ)$$

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First, we can simultaneously diagonalize $Y$ and $Z$, say $Y = PY'P^{-1}$, $Z = PZ'P^{-1}$ with $P$ unitary, then we get $$trace(X'Y'X'Z') \quad, \quad trace(X'^2Y'Z')$$ with $X' = P^{-1}XP$ still hermitian. Thus we may suppose $Y,Z$ are diagonal. Then $$\begin{align*} trace(XYXZ) &= \sum_{i,j}|X_{ij}|^2Y_{jj}Z_{ii} \\ trace(X^2YZ) &= \sum_{i,j}|X_{ij}|^2Y_{ii}Z_{ii} \end{align*}$$ The $Y_{ii},Z_{ii}>0$ are arbitrary, so all we can say is that one is bounded by $$\kappa(Y) = \max_{i \neq j}|Y_{ii}/Y_{jj}|$$ (or $\kappa(Z) = \max_{i \neq j}|Z_{ii}/Z_{jj}|$) times the other.