# 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?

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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

1. $Tr(XY) = \sum_{i,j} X_{ij}Y_{ji}$, where $X,Y$ are two matrices that can be multiplied.
2. $e^{-tB}Ae^{tB}$ has $(i,j)$ entry being $e^{\lambda_i - \lambda_j} a_{ij}$.
3. 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.
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+1 Nice answer. I tried to use the fact that the trace of the product of three symmetric matrices is invariant under any permutation (but failed :-(). – copper.hat Nov 5 '12 at 18:49
Did you mean replace $B$ with $CBC^{-1}$? I don't see how replacing $A$ in that way helps. – Potato Nov 5 '12 at 18:50
Both. I'm conjugating the matrices on both sides by $C$. – user27126 Nov 5 '12 at 18:50
Number $3$ is wrong. Try it with $A=\pmatrix{0&1\\1&0}$ and $\mathrm e^{tB}=\pmatrix{a\\&b}$. – joriki Nov 5 '12 at 18:51
Thanks. It's edited. – user27126 Nov 5 '12 at 18:55

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 $f(t)\ge f(0)$.

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+1, very nice! :-) – joriki Nov 5 '12 at 20:15
@joriki: Thank you for editing and appreciation!:-) – 23rd Nov 5 '12 at 20:29