Suppose that $A$ is a real, symmetric, positive definite $n\times n$ matrix. Show that $$\log(\det(A))\le \operatorname{tr}(A)-n \quad \text{and} \quad \log(\det(I_n+A))\le \operatorname{tr}(A).$$

Since $A=CDC^{-1}$ we can say the following: $$\det(A)=\det(C)\det(D)\det(C^{-1})=\det(D)=\Pi \lambda_i$$

But I'm not sure how to proceed from there. I need to somehow show that the trace is greater than the eigenvalues multiplied.

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    $\begingroup$ you should state what $X$ is and what $A$ is and how they relate. As currently written the post cannot be true in any meaningful sense $\endgroup$ – user8675309 Jul 10 '20 at 23:19
  • $\begingroup$ We also know that $Tr(A) = \sum \lambda_i$. But what's up with $x$? What is it? $\endgroup$ – ir7 Jul 10 '20 at 23:55
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    $\begingroup$ my B I meant to say A Jesus Christ my typo $\endgroup$ – John Rawls Jul 11 '20 at 0:23
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    $\begingroup$ You need that the eigenvalues are positive real. This doesn't hold for the $2\times 2$ matrix $A=-I$, where $\log(\det(-I))=\log(1)=0$ but $\tr(-I)-n=-2-2=-4$. This means there is some condition that you omitted from your problem. You need to be careful to state ALL conditions when asking a question or you could very well ask something false (as was the case here). $\endgroup$ – Aaron Jul 11 '20 at 2:08
  • $\begingroup$ @Aaron Can I edit it like this: Let $A$ be a $n\times n$ real matrix whose eigenvalues all have positive real parts. Prove that $\log \det A \le \mathrm{tr}(A) - n$? $\endgroup$ – River Li Jul 11 '20 at 2:34

We have:

$$\mathrm{Tr}(A) = \sum \lambda_i $$

$$\det(A) = \prod \lambda_i $$

So, if eigenvalues are positive reals, we have to show

$$ \sum \ln \lambda_i \leq \sum \lambda_i -n $$

which is true as $$\ln x \leq x -1 $$ for all $x>0$


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