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Suppose we have two positive semi-definite matrices of dimension n, $A$ and $B$ s.t. Tr$(A)$, Tr$(B)\le1$. Can we say anything about Tr$(AB)$?

(Is Tr$(AB)\le1$ too?)

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up vote 14 down vote accepted

In the space of positive semi-definite matrices, trace is a proper inner-product (it is easy to show that), i.e. it obeys cauchy-schwarz inequality ($<x,y> <=\sqrt{ <x,x><y,y>}$). So

\begin{align} tr\{AB\}\leq \sqrt{tr\{A^2\}tr\{B^2\}} \end{align} Now since A is positive semi-definite, $tr\{A^2\}\leq tr\{A\}^2$ ( eigen values of $A^2$ are squared eigen values of A, and since they are positive, $tr\{A^2\}=\sum_{i=1}^{N}\lambda_{i}^{2}\leq (\sum_{i=1}^{N}\lambda_{i})^{2} = tr\{A\}^2 \leq 1$. A similar argument for B proves $tr\{B^2\}\leq 1$ . So $tr\{AB\}\leq 1$. Hope this answers your question.

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First, note that $(A-B)^{2}$ is positive semi-definite, so we have:



Second, for $A$ positive semi-definite, suppose that all of eigenvalues are $\lambda_{1}$, $\lambda_{2}$, $\cdots$, $\lambda_{n}$, then $\lambda_{i}\geq0$ and $\mathrm{Tr}A=\sum_{i=1}^{n}\lambda_{i}\leq1$, so $\mathrm{Tr}(A^{2})=\sum_{i=1}^{n}\lambda_{i}^{2}\leq\sum_{i=1}^{n}\lambda_{i}\leq1$.

Similarly, $\mathrm{Tr}(B^{2})\leq1$, so $\mathrm{Tr}(AB)\leq1$.

Remark. More generally, we can conclude that the range of $\mathrm{Tr}(AB)$ is $[0,1]$.

As $A$ and $B$ are positive semi-definite, so there exist $C$ and $D$ such that $A=C^{T}C$ and $B=D^{T}D$, so $\mathrm{Tr}(AB)=\mathrm{Tr}(C^{T}CD^{T}D)=\mathrm{Tr}(CD^{T}DC^{T})=\mathrm{Tr}[CD^{T}(CD^{T})^{T}]\geq0$.

Set $A=diag[1,0,0,\cdots,0]$ and $A=diag[0,1,0,\cdots,0]$, then $\mathrm{Tr}(AB)=0$.

Set $A=B=diag[1,0,0,\cdots,0]$, then $\mathrm{Tr}(AB)=1$.

Accoading to above, we can conclude that $\text{Range}(\mathrm{Tr}(AB))=[0,1]$.

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