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I am looking for a source for the following "well known" inequality that I found here. The corresponding matrix properties/assumtions are also mentioned there in more detail.

$$ \mbox{Tr} \left( A^{-1} \right) \ge n^2 \, \mbox{Tr}(A)^{-1} $$

Can someone point me to a paper or book to which I can cite this inequality in my thesis? I looked up several known linear algebra books, but couldn't find it anywhere.

LissaB
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    It seems that your inequality is a special case of Theorem 2.1 of this, so you may cite this one. – Seewoo Lee Aug 19 '22 at 21:46
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    VanBaffo suggested a linked question, which had been linked by the OP herself. It contains enough information to construct the proof. However, the present question requests a reference to use in a thesis, not a proof. – paperskilltrees Aug 19 '22 at 23:48
  • @SeewooLee Thm 2.1 and further theorems in this paper are only for positive exponents, right? In addition, I don't think that the inverse of a matrix can be dealt with as a "simple exponeniation". – LissaB Aug 20 '22 at 09:03
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    This is just a rearrangement of $|A^{-1/2}|_F^2|A^{1/2}|_F^2\ge\langle A^{-1/2},A^{1/2}\rangle_F^2$, which is essentially Cauchy-Schwarz inequality. I don't think you need any citation at all. – user1551 Aug 24 '22 at 11:12
  • Related – Rodrigo de Azevedo Feb 16 '25 at 09:52

2 Answers2

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Just so you're aware, you need more assumptions for this inequality to be true. Let $A$ be a diagonal matrix with values $-1, -1/2$ on the diagonal.

Then $$ \mathrm{Tr}(A^{-1}) = -3 < -8/3 = 2^2 \mathrm{Tr}(A)^{-1}. $$

For $A$ with positive eigenvalues, the inequality is true and can be seen as a simple consequence of Cauchy-Schwarz.

Drew Brady
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  • The assumptions can be found in more detail in the link that I provided to the original post to which I was referring to. But you are right, I probably should mention the matrix properties again in my own post/question. @DrewBrady And thanks for the hint regarding Cauchy-Schwarz, I didn't see that connection and I will look further into it. – LissaB Aug 24 '22 at 08:37
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I found a reference to a general version of this inequality (inverse log-convex property). It can be found here. This specific case uses $\lambda=1$ and $\mu=0$ which leads to this simplified form of the inequality.

LissaB
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