Let $B$ is a Hermitian matrix such that $\operatorname{trace}(B)>0$ (but $B$ is "not" positive definite), and $C$ is a Hermitian positive definite matrix. Can we conclude $\operatorname{trace}(BC)\geq 0$? I tried two approaches, but neither worked.
1) We can write
$$\operatorname{trace}(BC)=\operatorname{trace}(C^{1/2}BC^{1/2})$$
where $C^{1/2}$ is positive definite, and thus $C^{1/2}BC^{1/2}$ is congruent with $BC$. But congruence just preserves the number of positive/negative/zero eigenvalues, and doesn't say anything about the magnitude, so apparently this doesn't say anything about the sign of $\operatorname{trace}(BC)$.
2) We can also write eigenvalue decomposition of $B$ and $C$:
$$B=\sum_i \beta_ib_ib_i^T \quad\text{and}\quad C=\sum_i \gamma_ic_ic_i^T$$
Hence,
$$\operatorname{trace}(BC)=\sum_i\sum_j\beta_i\gamma_j|b_i^Ta_j|^2$$
We know $\gamma_j>0$ (from positive-definiteness of $C$) and $|b_i^Ta_j|^2\geq 0$, but for $\beta_i$, we only know $\sum_i\beta_i>0$. So I don't know how we can conclude $\sum_i\sum_j\beta_i\gamma_j|b_i^Ta_j|^2\geq 0$?
Best Answer
Not necessarily. For instance, consider $$ B = \pmatrix{2&0\\0&-1}, \quad C = \pmatrix{1&0\\0&3} $$