# Is there a trace inequality for the product of a sequence of hermitian postive definite matrices?

Let $A$ and $B$ be two Hermitian matrices with positive eigenvalues. Let $k>0$ be a integer. Let $P=(P_1,P_2,\dots,P_{2k})$ be a sequence of $k$ $A$s and $k$ $B$s in any given order.

Do we have ${\rm tr}\,\prod^{2k}_{i=1} P_i \leq {\rm tr}\,A^k B^k$ ?

Although the general inequality is false (counterexamples can be found easily), for words where $A$ and $B$ occur in pairs, a stronger inequality is known. Indeed, a classic theorem of Ando and Hiai (1994) yields that for $A, B \ge 0$ and arbitrary reals $p_1,\ldots,p_k \ge 0$ we have \begin{equation*} \text{tr}|A^{p_1}B^{p_1}\cdots A^{p_k}B^{p_k}| \le \text{tr}|A^{p_1+\cdots p_k}B^{p_1+\cdots+p_k}|, \end{equation*} where $|X|=(X^*X)^{1/2}$ denotes the matrix absolute value.

Remark. An easy way to obtain counterexamples for $3\times 3$ matrices similar to the one mentioned in Francois's answer is to use matrices of the form (the values are somewhat arbitrary, other choices also work): \begin{equation*} X = \begin{bmatrix}a & b& 1\\ 0 & c & -2\\ 0 &0 &0\end{bmatrix},\quad Y = \begin{bmatrix}p & q& 1\\ 0 & r & -1\\ 0 &0 &-1\end{bmatrix}, \end{equation*} then to define $A=XX^T$ and $B=YY^T$. Then, try to numerically minimize the quantity $g(a,b,c,p,q,r) := \text{tr}(A^5B^5)-\text{tr}(A^4BAB^4)$. Doing so, for instance in Mathematica, easily yields $g<0$, and thus, a desired counterexample. By playing around with different $X$ and $Y$, one can easily generate additional counterexamples.

• Just curious, which theorem is it in the Ando-Hiai paper? Mar 4 '17 at 13:24
• Theorem 4.1 (actually, it's corollary lists a stronger log majorization from which the above trace inequality follows). Mar 4 '17 at 15:05
• (the above comment should read: "actually its corollary....") Mar 4 '17 at 15:58

A counterexample was apparently first found by Plevnik (2016) (pdf): $$A = \begin{bmatrix}76&0&0\\0&0&0\\0&0&1\end{bmatrix} \qquad\text{and}\qquad B = \begin{bmatrix}20&-14&13\\-14&2880&3100\\13&3100&3380\end{bmatrix}$$ give $$\operatorname{tr} A^4BAB^4 = 7608677695167720100 > 7566365725138281700 = \operatorname{tr} A^5B^5.$$ To get one with $A$ positive definite as requested, replace entry $A_{22}$ by $0.01$: the two traces become $$7584680876077508226.18611992\ \ > \ \ 7566365725573314229.03610008.$$

• @Suvrit A 2 x 2 example could be interesting; before posting this I tried and failed generating one with numpy.random. Mar 5 '17 at 2:23
• Actually, it seems that my $2\times 2$ counterexamples may have been due to "numerical error" --- I tried to obtain them by solving a numerical minimization problem. The $3\times 3$ case seems easier! I take back my previous comment. Mar 5 '17 at 12:00