Let $T_1 \in \mathcal{L}(V)$ and $T_2 \in \mathcal{L}(V)$ be positive operators. Prove that the trace of their product is non-negative i.e., tr($T_1 T_2) \geq 0$
Attempt 1: Obviously, a positive operator has a positive trace because all of its eigenvalues are positive. However, the product of two positive operators is not necessarily positive.
Attempt 2: Finding a pattern between the positivity of eigenvalues during multiplication
Both of my attempts fizzled out and I am pretty stuck. How would one solve this?
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$\begingroup$Let $e_1, \ldots, e_n$ be an orthonormal basis of eigenvectors of $T_2$, corresponding respectively to $\lambda_i$. Then,$$\operatorname{tr}(T_1T_2) = \sum_{i=1}^n \langle T_1T_2v_i, v_i\rangle = \sum_{i=1}^n \lambda_i \langle T_1v_i, v_i\rangle,$$which is non-negative, as $T_1$ is positive and $\lambda_i \ge 0$, hence all the terms are non-negative.
$\endgroup$ 1 $\begingroup$The product of two positive matrices $A,B \ge 0$ always has nonnegative eigenvalues. Indeed, since $B$ is positive it has a unique positive square root $B^{1/2}$ so
$$\sigma(AB) = \sigma(AB^{1/2}B^{1/2}) = \sigma(B^{1/2}AB^{1/2}).$$Notice that $B^{1/2}AB^{1/2}$ is a positive matrix since it is self-adjoint and for all $x \in V$ holds$$\langle B^{1/2}AB^{1/2}x, x\rangle = \langle AB^{1/2}x, B^{1/2}x\rangle \ge 0$$because $A \ge 0$. In particular, $B^{1/2}AB^{1/2}$ has nonnegative spectrum and hence the same holds for $AB$.
Since the trace is the sum of eigenvalues, it follows that $\operatorname{Tr}(AB) \ge 0$.
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