# A cute trace inequality

Theorem. If $A$ and $B$ are positive semi-definite matrices then $\mathrm{Tr} (AB)^2 \leq \mathrm{Tr} (A^2 B^2)$

Note that both sides contain the same matrices the same number of times, just in a different order ( $ABAB$ versus $AABB$). Here is a cute proof due to Bellman.

Proof. First, note that $(AB-BA)^2 = (AB)^2 + (BA)^2 - A B^2 A - B A^2 B.$

Next, note that the commutator $AB-BA$ is skew-Hermitian, that is, $(AB-BA)^\dagger = -(AB-BA)$. In other words, $AB-BA = iH$ for some Hermitian matrix $H$. This means that all eigenvalues of $AB-BA$ are purely imaginary. When we square the matrix, they become negative, i.e., $(AB-BA)^2$ is negative semi-definite. From this we get $0 \geq \mathrm{Tr} (AB-BA)^2 = \mathrm{Tr} \Bigl[ (AB)^2 + (BA)^2 - A B^2 A - B A^2 B \Bigr].$

Using the cyclic property of the trace, this simplifies to $0 \geq 2 \mathrm{Tr}(AB)^2 - 2 \mathrm{Tr} (A^2 B^2)$

and we are done.

Here is an interesting open question.

Christmas Conjecture. Let $A$ and $B$ be $n \times n$ density matrices with fixed spectrum: the eigenvalues of $A$ are $\alpha_1 \geq \alpha_2 \geq \dots \geq \alpha_n$ while the eigenvalues of $B$ are $\beta_1 \geq \beta_2 \geq \dots \geq \beta_n$. Then $\max_{A,B} \mathrm{Tr} \Bigl[ AB + A^2 B^2 - (AB)^2 \Bigr] = \sum_{i=1}^n \alpha_i \beta_i.$

where the maximum is over all $A,B$ with the given spectrum.

If the last two terms $A^2 B^2 - (AB)^2$ would not be there, then showing this would be quite straightforward. Indeed, the left-hand side becomes simply $\max_P \sum_{i,j=1}^n \alpha_i P_{ij} \beta_j$

where the maximum is over doubly-stochastic matrices $P$. By the Birkhoff–von Neumann theorem (see Theorem 13.1 in John Watrous’ lecture notes), the maximal value is achieved when $P$ is a permutation matrix. In fact, by the rearrangement theorem, $P$ must be the identity matrix, in which case the two lists of eigenvalues are sorted in the same order.

However, we are dealing with a slightly more complicated expression in the conjecture. As we just saw above, $\mathrm{Tr} \bigl[ A^2 B^2 - (AB)^2 \bigr] \geq 0$, so the last two terms can only increase the left-hand side and thus make the inequality harder to prove. Nevertheless, I conjecture that this stronger inequality still holds, and I can prove it for $n = 2$.

Update: I recently received an e-mail from Jianxin Chen containing a proof of the Christmas Conjecture!