Basic trace inequality Suppose $A$ and $B$ are self-adjoint matrices. Why is it true that 
$$Tr(A^2) \le Tr(Ae^{-tB}Ae^{tB})$$
for $t\in\mathbb R$, where $e^x$ denotes the matrix exponential? 
 A: Since trace is invariant under conjugation, take $C$ such that $CBC^{-1}$ is diagonal, say $(\lambda_1,\cdots, \lambda_n)$. By replacing $A$ with $CAC^{-1}$, it suffices to show the inequality with $e^{tB} = diag(e^{\lambda_1},\cdots,e^{\lambda_n})$. It then suffices to check a few things


*

*$Tr(XY) = \sum_{i,j} X_{ij}Y_{ji}$, where $X,Y$ are two matrices that can be multiplied.

*$e^{-tB}Ae^{tB}$ has $(i,j)$ entry being $e^{\lambda_i - \lambda_j} a_{ij}$.

*Using 1 and 2, for $(X,Y) = (A,A)$ and $(A, e^{-tB}Ae^{tB})$, we need to show that 
$\sum_{i \neq j} 2a_{ij}^2 \leq \sum_{i \neq j} (e^{\lambda_i - \lambda_j} + e^{\lambda_j - \lambda_i}) a_{ij}^2$.
(Here, we used the self adjointness of $A$ (in creating the squares), and that of $B$ (to make sure that the eigenvalues are real). This is clear by AMGM.

A: Let $X,Y,Z$ be square matrices of the same size. Denote the commutator $[X,Y]=XY-YX$. Note that $\mathrm{Tr}[X,Y]=0$ and $\mathrm{Tr}(X[Y,Z])=-\mathrm{Tr}([X,Z]Y)$. In particular, $\mathrm{Tr}(X[X,Y])=0$.
Let $A_0=A$, $A_n=[A_{n-1},B]$ for each $n\ge 1$. Let $f(t)=\mathrm{Tr}(Ae^{-tB}Ae^{tB})$, which is an analytic function of $t$ with inifite convergence radius. By induction, it is easy to see that $f^{(n)}(t)=\mathrm{Tr}(Ae^{-tB}A_ne^{tB})$. In particular, $f^{(n)}(0)=\mathrm{Tr}(AA_n)$, so 
$$f(t)=\sum_{n=0}^\infty \frac{\mathrm{Tr}(AA_n)}{n!}t^n.$$
Let $a_{ij}=\mathrm{Tr}(A_iA_j)$ for $i,j\ge 0$. Then
$$a_{i,j+1}=\mathrm{Tr}(A_i[A_j,B])=-\mathrm{Tr}([A_i,B]A_j)=-a_{i+1,j}.$$ 
In particular, 
$$a_{0,2n}=(-1)^na_{n,n}=(-1)^n\mathrm{Tr}(A_n^2),$$ 
and
$$a_{0,2n+1}=(-1)^na_{n,n+1}=(-1)^n\mathrm{Tr}(A_n[A_n,B])=0.$$
Since both $A$ and $B$ are self-adjoint, by induction, it is easy to see that $A_n$ is self-adjoint when $n$ is even, and it is anti-self-adjoint when $n$ is odd. Therefore, The eigenvalues of $A_n$ are real when $n$ is even and are $0$ or pure imaginary when $n$ is odd. It follows that $a_{0,2n}\ge 0$ for every $n$.
Therefore, in the Taylor expansion of $f(t)$, the coefficients of the odd terms are zero, but the coefficients of the even terms are nonnegative. As a result, the conclusion follows from $f(t)\ge f(0)$. 
