Lower bound $\operatorname{Tr}(MB)$ where $M>0$ and $B$ is arbitrary

Question

Let $M, B$ be complex and square matrices. $M$ is positive definite (pd) and $B$ is arbitrary. If this is relevant, there is an afore-knowledge that $\operatorname{Tr}(MB)\ \ge \ 0$.

1. Please help to obtain a/the lower bound of $\operatorname{Tr}(MB)$ i.e. $\ f(.) \ \le \ \operatorname{Tr}(MB) \ $ where $f(.)$ is a function of $\operatorname{Tr}(B)$ I checked the Von Neumann's trace inequality, but couldn't find a headway. Note that $\lvert \operatorname{Tr}(MB) \rvert = \operatorname{Tr}(MB)$ in this case and $f(.)<0$ is okay.
2. If it makes a difference, let $M$ be a positive diagonal matrix of reals (such as with singular values of a pd matrix).

I have also checked some discussions such as this. However, they do not consider arbitrary $M$.

Thanks in advance.

Answer 1

If $M$ is not a multiple of the identity, there is no such lower bound: $\text{Tr}(MB)$ and $\text{Tr}(B)$ can be any two complex numbers (subject only to your requirement that $\text{Tr}(MB) \ge 0$).

WLOG we may assume $M$ is diagonal. Suppose its diagonal entries $m_{ii}$ and $m_{jj}$ are different. Take $B$ with only two (possibly) nonzero entries $b_{ii}$ and $b_{jj}$. Thus $\text{Tr}(B) = b_{ii} + b_{jj}$ while $\text{Tr}(MB) = m_{ii} b_{ii} + m_{jj} b_{jj}$. Since the matrix $\pmatrix{1 & 1\cr m_{ii} & m_{jj}\cr}$ is nonsingular, for any $y_1$ and $y_2$ we may choose $b_{ii}$ and $b_{jj}$ so that $b_{ii} + b_{jj} = y_1$ and $m_{ii} b_{ii} + m_{jj} b_{jj} = y_2$.