If $\operatorname{Tr}(H)\ge 0$, is $\operatorname{Tr}(PH)\ge 0?\ $ $H$ is Hermitian, $P$ is positive definite.

Question

Old: Given $\operatorname{Tr}(DH)\ge 0$ for positive real diagonal $D$ and Hermitian $H$, is $\operatorname{Tr}(H)\ge 0$?

Let $D$ be positive real diagonal and $H$ be Hermitian such that $H$ may have negative eigenvalues. Both matrices are of course of the same size $n\times n$.

Given that $\operatorname{Tr}(DH)\ge 0. \ $ Is $\operatorname{Tr}(H)\ge 0$?

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Post-edit:

Thanks @Aweygan for providing a counter-example to the initial question above.
Let's look at it the other way round by letting $\operatorname{Tr}(H)\ge 0$. $H$ is $n\times n$ Hermitian, which may have negative eigenvalue(s).

1. $D$ be positive real diagonal. Is $\operatorname{Tr}(DH)\ge 0$?
2. Let us generalize the question for when $n\times n$ complex matrix $P$ is positive definite. Is $\operatorname{Tr}(PH)\ge 0$?

I have edited the title to accommodate the changes made.

Answer 1

$P$ is positive definite, and $H$ is Hermitian. Take a look at this thread, we have $$\operatorname{Tr}(PH) \ge \lambda_n(\bar{P})\operatorname{Tr}(H) - \lambda_n(H)\Big(n \lambda_n(\bar{P}) - \operatorname{Tr}(P) \Big) \qquad (1) $$ where $\lambda_n$ denotes the minimum eigenvalue, and $\bar{P}=(P+P^*)/2 = P. \ $ The bound can be rewritten as $$\operatorname{Tr}(PH) \ge \lambda_n(P)\operatorname{Tr}(H) + \lambda_n(H)\Big( \operatorname{Tr}(P) - n \lambda_n(P) \Big)$$ Let's seek the condition on $\lambda_n(H)$ for \begin{align}\lambda_n(P)\operatorname{Tr}(H) + &\lambda_n(H)\Big( \operatorname{Tr}(P) - n \lambda_n(P) \Big) \ge 0 \\ &\lambda_n(H) \ge -\frac{\lambda_n(P)\operatorname{Tr}(H)}{\operatorname{Tr}(P) - n \lambda_n(P)} \end{align} is required for $\operatorname{Tr}(PH) \ge 0$ based on the bound stated in (1).

Meanwhile, we know that $\operatorname{Tr}(H)\ge0$ from the assumption, $ \ \lambda_n(P)>0, \ $ and $\operatorname{Tr}(P) - n \lambda_n(P)\ge 0$.

Answer 2

Not necessarily. Consider the following example:

$$D=\begin{pmatrix}2&0\\0&1\end{pmatrix},\qquad H=\begin{pmatrix}2&0\\0&-3\end{pmatrix}.$$