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Let $\bf P$ be a $n \times n$ symmetric positive definite matrix. It is known that

$$ \operatorname{tr} ({\bf P}) \geq n + \log \det({\bf P})$$

Since ${\bf P} \succ {\bf O}_n$, it is also known that $\bf P$ is invertible. Hence,

$$ \operatorname{tr} \left( {\bf P}^{-1} \right) \geq n + \log \det \left( {\bf P}^{-1} \right) = n - \log \det({\bf P}) $$

Adding these two inequalities,

$$ \color{blue}{\operatorname{tr} ({\bf P}) + \operatorname{tr} \left( {\bf P}^{-1} \right) \geq 2 n} $$

Has this trace inequality earned a name?


Alternatively, this trace inequality can be derived from the fact that $x + \frac{1}{x} \geq 2$ for all $x > 0$.

adding x > 0 and its reciprocal


Related

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    We also have $P+P^{-1}\ge 2 I$, see here, but I didn't find a special name. – Dietrich Burde Feb 16 '25 at 11:30
  • @DietrichBurde Good find! I appended a link to it. – Rodrigo de Azevedo Feb 16 '25 at 11:35
  • Given that $x+\frac1x\geq 2,$ and the trace is the sum of the (positive real) eigenvalues, and the eigenvalues of $P^{-1}$ are the inverses of the eigenvalues of $P,$ you don't need the inequality with $\log.$ It doesn't seem like a deep enough result result to require a seperate name. – Thomas Andrews Feb 16 '25 at 11:40
  • @ThomasAndrews Sure, but it was the "inequality with log" that led me to this trace inequality, rather than the inequality $x + \frac1x \geq 2$. I assume that this trace inequality is indeed kind of useless but it's nice to have a question to link to in case I ever find a use for it – Rodrigo de Azevedo Feb 16 '25 at 11:42
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    We have several hundred inequalities in use, so it would be rather difficult to name every one of them. Still, it is a nice inequality. – Dietrich Burde Feb 16 '25 at 11:59
  • I didn't say the question was not good, I was saying there probably isn't a name, because, in fact, it is not very deep. That you got this trivial result from a deeper result was deceptive, making it look deeper than it is. – Thomas Andrews Feb 16 '25 at 12:31
  • @ThomasAndrews If it does not have a name, now it has a page that can be cited along the lines of "[...] using the trivial inequality in this question" – Rodrigo de Azevedo Feb 16 '25 at 12:35
  • Again, I am not criticizing the question. You are writing to me like I am. @RodrigodeAzevedo – Thomas Andrews Feb 16 '25 at 15:36
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    Back in my high school days, I posted questions on StackMath about reducing the number of matrix multiplications, and you helped me a lot. When I was in university, there was a period when I came back and saw that your account had been suspended for a while. Now I'm a graduate student, and I'm still happy to see you're still here with interesting matrix problems. – Dang Dang Apr 21 '25 at 13:24
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    @DangDang Thank you for the kind words. I am happy to hear that my activity on StackMath has helped at least one person – Rodrigo de Azevedo Apr 21 '25 at 14:13

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