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What are the conditions on $\text{tr}(AB) \leq \text{tr(A)} \text{tr(B)}$ to be true?

I started the service today, so I don't have any reputation points and can't make any comments. So, let me ask a question here.

I don't understand the part of this question that says tr(^1/2^1/2)≤tr(^1/2()^1/2).

if A<=aI is true, why is the inequality holds?

numpy
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1 Answers1

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The trace of a PSD matrix is non-negative. This means that the trace function is increasing with respect to the PSD partial order $\preceq$ (the Löwner order). If $A \preceq B$ then $B - A$ is PSD so

$$ \operatorname{tr}(B - A) = \operatorname{tr} B - \operatorname{tr} A \ge 0. $$

(The trace is also linear.) Therefore $\operatorname{tr} A \le \operatorname{tr} B$.

Sera Gunn
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