An inequality on trace of product of two matrices

Question

Suppose we have two $n \times n$ positive semidefinite matrices, $A$ and $B$, such that $\mbox{tr}(A), \mbox{tr}(B) \le 1$.

Can we say anything about $\mbox{tr}(AB)$? Is $\mbox{tr}(AB) \le 1 $ too?

Answer 1

In the space of positive semi-definite matrices, trace is a proper inner-product (it is easy to show that), i.e. it obeys the Cauchy-Schwarz inequality: $\langle x,y \rangle \leq \sqrt{ \langle x,x \rangle \langle y,y\rangle}$. So

$$\mbox{tr}\{AB\}\leq \sqrt{\mbox{tr}\{A^2\} \mbox{tr}\{B^2\}}$$

Now, since $A$ is positive semidefinite, $\mbox{tr}\{A^2\} \leq \mbox{tr}\{A\}^2$, i.e., the eigenvalues of $A^2$ are squared eigenvalues of $A$, and since they are positive

$$\mbox{tr} \{A^2\} = \sum_{i=1}^{N}\lambda_{i}^{2}\leq \left( \sum_{i=1}^{N}\lambda_{i} \right)^{2} = \mbox{tr}\{A\}^2 \leq 1$$

A similar argument for B proves $\mbox{tr}\{B^2\}\leq 1$ . So $\mbox{tr}\{AB\}\leq 1$. Hope this answers your question.

Answer 2

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

$$0\leq\mathrm{Tr}(A-B)^{2}=\mathrm{Tr}(A^{2})+\mathrm{Tr}(B^{2})-2\mathrm{Tr}(AB)$$

$$\mathrm{Tr}(AB)\leq\frac{1}{2}(\mathrm{Tr}(A^{2})+\mathrm{Tr}(B^{2}))$$

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 $\text{Range}(\mathrm{Tr}(AB))=[0,1]$.

Answer 3

An extension to @dineshdileep 's answer can also be found here, in which it shows that: $$tr(A^TB)\le \sqrt{tr(A^TA)tr(B^TB)}$$

Answer 4

Let's write $A = \sum_{i=1}^{n}\lambda_i u_iu_i^T$ and $B = \sum_{i=1}^{n}\mu_i v_iv_i^T$.

$$AB = \sum_{i,j} \lambda_i \mu_j u_iu_i^Tv_jv_j^T$$

and so $$\text{Tr}(AB) = \sum_{i=1}^n \sum_{j=1}^n \lambda_i \mu_j \left|u_i^Tv_j\right|^2 \le \sum_{i=1}^n \sum_{j=1}^n \lambda_i \mu_j \left\|u_i\right\|^2\left\|v_j\right\|^2 = \sum_{i=1}^n \lambda_i \sum_{j=1}^n\mu_j = \text{Tr}(A)\text{Tr}(B)$$

Answer 5

When $A$ and $B$ are positive semidefinite, so are $\operatorname{tr}(A)I-A$ (by considering its eigenvalues) and $B^{1/2}\left(\operatorname{tr}(A)I-A\right)B^{1/2}$. Hence $$ \operatorname{tr}(A)\operatorname{tr}(B)-\operatorname{tr}(AB) =\operatorname{tr}\big(\left(\operatorname{tr}(A)I-A\right)B\big) =\operatorname{tr}\left(B^{1/2}\left(\operatorname{tr}(A)I-A\right)B^{1/2}\right) \ge0 $$ and the result follows.

Using the same idea, one can actually obtain a tighter inequality $\operatorname{tr}(AB)\le\rho(A)\operatorname{tr}(B)$.