Finite-dimensional Golden–Thompson trace inequality

We first record two elementary trace estimates. If $A=U\mathrm{diag}(a_j)V^{\ast }$ is a singular value decomposition, cyclically expanding the finite diagonal sum gives $\mathrm{tr}(AB)={\sum }_j{a}_j⟨V^{\ast }BU{e}_j,e_j⟩$, so $\mathrm{\mid }\mathrm{tr}(AB)\mathrm{\mid }\le \mathrm{\parallel }A{\mathrm{\parallel }}_1\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$. Also $\mathrm{tr}(AB)={\sum }_{i,j}{A}_{ij}{B}_{ji}$, so scalar Cauchy–Schwarz gives $\mathrm{\mid }\mathrm{tr}(AB)\mathrm{\mid }\le \mathrm{\parallel }A{\mathrm{\parallel }}_2\mathrm{\parallel }B{\mathrm{\parallel }}_2$, since $\mathrm{\parallel }A{\mathrm{\parallel }}_2^2=\mathrm{tr}(A^{\ast }A)={\sum }_{i,j}\mathrm{\mid }{A}_{ij}{\mathrm{\mid }}^2$ by the singular value decomposition. These arguments also prove the trace cyclicity used here by interchanging two finite sums.

Set $F(z)=\exp (zH\mathrm{/}2)\exp (zK\mathrm{/}2)$ and $M=\mathrm{\parallel }\exp (H\mathrm{/}2)\exp (K\mathrm{/}2){\mathrm{\parallel }}_2>0$. The function $F$ is entrywise entire and uniformly bounded on the closed strip: for $z=x+iy$ each exponential is the product of a bounded positive real exponential with a unitary imaginary exponential of the same Hermitian matrix. On the line $x=0$, $\mathrm{\parallel }F(iy){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=1$. On $x=1$, the identity $F(1+iy)=\exp (iyH\mathrm{/}2)\exp (H\mathrm{/}2)\exp (K\mathrm{/}2)\exp (iyK\mathrm{/}2)$ and unitary invariance give $\mathrm{\parallel }F(1+iy){\mathrm{\parallel }}_2=M$.

We need a finite eigenvalue comparison only for a diagonalizable invertible matrix $A$. Order its eigenvalue moduli and its singular values decreasingly. For $1\le k\le p$, the $k$th exterior space has the orthonormal wedge basis $e_{j_1}\wedge \cdots \wedge e_{j_k}$, $j_1<\cdots <j_k$, and its induced map sends a wedge to the wedge of the images. This defines the finite matrix ${\bigwedge }^{k}A$ directly by multilinearity and alternation. The singular value decomposition makes its norm ${\prod }_{j=1}^k{s}_j(A)$: wedges of the two singular-vector orthonormal bases give diagonal coefficients consisting of all $k$-fold singular-value products. Wedges of the first $k$ independent eigenvectors are nonzero eigenvectors with eigenvalue ${\prod }_{j=1}^k{\lambda }_j(A)$; nonvanishing follows by expanding independent vectors in a basis and noting that some $k$-minor is nonzero. The norm bounds the modulus of an eigenvalue, proving ${\prod }_{j=1}^k\mathrm{\mid }{\lambda }_j(A)\mathrm{\mid }\le {\prod }_{j=1}^k{s}_j(A)$. This uses only finite bases and finite determinants.

Fix $0<s<1$, put $q=2\mathrm{/}s$ and $q^{\mathrm{\prime }}=2\mathrm{/}(2-s)$, and choose a singular value decomposition $F(s)=U\mathrm{diag}(a_j)V^{\ast }$. All $a_j$ are positive because $F(s)$ is invertible. Set $N=({\sum }_j{a}_j^q{)}^{1\mathrm{/}q}>0$ and $b_j=(a_j\mathrm{/}N{)}^{q-1}$, so ${\sum }_j{b}_j^{q^{\mathrm{\prime }}}=1$. Define $G(z)=V\mathrm{diag}(\exp (q^{\mathrm{\prime }}(1-z\mathrm{/}2)\log b_j))U^{\ast }$ and $g(z)=\mathrm{tr}(G(z)F(z))$. These functions are continuous and bounded on the closed strip and holomorphic inside: their entries are finite sums of entire exponentials and their moduli are bounded uniformly in the imaginary coordinate. On its left boundary $\mathrm{\parallel }G(iy){\mathrm{\parallel }}_1={\sum }_j{b}_j^{q^{\mathrm{\prime }}}=1$, and on its right boundary $\mathrm{\parallel }G(1+iy){\mathrm{\parallel }}_2=({\sum }_j{b}_j^{q^{\mathrm{\prime }}}{)}^{1\mathrm{/}2}=1$. The two trace estimates of step 1.1 therefore bound $\mathrm{\mid }g\mathrm{\mid }$ by $1$ and $M$ on the respective boundaries. Since $q^{\mathrm{\prime }}(1-s\mathrm{/}2)=1$, one has $g(s)={\sum }_j{a}_j{b}_j=N$. Applying [F3] yields $\mathrm{\parallel }F(s){\mathrm{\parallel }}_{2\mathrm{/}s}\le M^s$.

Put $x_j=\log \mathrm{\mid }{\lambda }_j(A)\mathrm{\mid }$ and $y_j=\log s_j(A)$ in decreasing order. Step 1.3 gives ${\sum }_{j=1}^k{x}_j\le {\sum }_{j=1}^k{y}_j$ for every $k$. For any real $u$, the positive terms form an initial segment, so ${\sum }_j(x_j-u{)}_{+}={\max }_{0\le k\le p}({\sum }_{j=1}^k{x}_j-ku)\le {\sum }_j(y_j-u{)}_{+}$. For every $c>0$, multiply by $c^2{e}^{cu}$ and integrate: $c^2{\int }_{-\mathrm{\infty }}^x(x-u)e^{cu}du=e^{cx}$, as follows by evaluating the elementary exponential primitive at finite lower limits and sending that limit to $-\mathrm{\infty }$. The finite sums can be integrated termwise and all integrands are nonnegative, giving ${\sum }_j\mathrm{\mid }{\lambda }_j(A){\mathrm{\mid }}^c\le {\sum }_j{s}_j(A{)}^c$.

For each integer $n\ge 2$, let $A_n=F(1\mathrm{/}n)$. It is similar to the positive definite matrix $\exp (K\mathrm{/}4n)\exp (H\mathrm{/}2n)\exp (K\mathrm{/}4n)$, since conjugating $A_n$ by $\exp (K\mathrm{/}4n)$ gives that matrix. Thus [F2] makes $A_n$ diagonalizable with positive eigenvalues. Apply step 2.2 with $c=2n$ and step 2.1 with $s=1\mathrm{/}n$ to obtain $\mathrm{tr}(A_n^{2n})\le {\sum }_j{s}_j(A_n{)}^{2n}=\mathrm{\parallel }{A}_n{\mathrm{\parallel }}_{2n}^{2n}\le M^2$.

The exponential series gives $A_n=I+(H+K)\mathrm{/}(2n)+O(n^{-2})$ and $B_n:=\exp ((H+K)\mathrm{/}(2n))=I+(H+K)\mathrm{/}(2n)+O(n^{-2})$ in operator norm, where the constants depend only on the fixed matrices. Also $\mathrm{\parallel }{A}_n\mathrm{\parallel },\mathrm{\parallel }{B}_n\mathrm{\parallel }\le e^{c\mathrm{/}n}$ for one fixed $c$. The telescoping identity $A^N-B^N={\sum }_{j=0}^{N-1}{A}^j(A-B)B^{N-1-j}$, valid by cancellation without commutativity, gives $\mathrm{\parallel }{A}_n^{2n}-B_n^{2n}\mathrm{\parallel }=O(n^{-1})$. The addition law for exponential series of scalar multiples of one matrix gives $B_n^{2n}=\exp (H+K)$. Therefore $\mathrm{tr}\exp (H+K)={\lim }_n\mathrm{tr}(A_n^{2n})\le M^2=\mathrm{tr}(\exp (H)\exp (K))$, using $\mathrm{\mid }\mathrm{tr}C\mathrm{\mid }\le p\mathrm{\parallel }C{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ and cyclicity to evaluate the Hilbert–Schmidt square $M^2$. This is the claimed inequality.