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I‘m currently studying the fundamentals of convex analysis and the author presents an example as follows: define a function on the space of symmetric, positive definite matrices by

$$ A \mapsto f(A) = \log(\det(A^{-1})) $$

To show that this is a convex function, the author says:

... start from the inequality $$\det(t \, A + (1-t) \, B) \geq \det(A)^t \, \det(B)^{1-t}$$

where $0 \leq t \leq 1$ and $A$ and $B$ are arbitrary symmetric, positive definite matrices. I know how to proceed from there, but can someone please help me proving that this inequality holds? I don‘t see how to combine the properties of determinants and symmetric, positive definite matrices to get the result. Thanks for your help!

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    $\begingroup$ It's a good idea to add a citation for "the author" and the publication you are studying. $\endgroup$
    – hardmath
    Commented Jun 11, 2018 at 23:02
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    $\begingroup$ What an odd way to go about it. It seems to me like they're basically giving you that the determinant is log-convex, and asking you to take the logarithm. $\endgroup$ Commented Jun 12, 2018 at 0:51
  • $\begingroup$ @MichaelGrant anyway, I have to deal with this hint and get the idea how to get there. Btw, I‘m an undergraduate student in the third semester. $\endgroup$
    – user528502
    Commented Jun 12, 2018 at 5:22

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Because all the eigenvalues of $A$ and $B$ are positive, $A$ is invertible, and since $A$ is symmetric, $A^{-1}B = A^{-1/2}(A^{-1/2}BA^{-1/2})A^{1/2} $ Also, this implies that $A^{-1}B$ is similar to the symmetric matrix $A^{-1/2}BA^{-1/2}$ (and hence has the same determinant). We can thus take a factor of $\det{A}$ out of both sides and prove instead that $$ \det{(tI+(1-t)C)} \geq (\det{C})^{1-t} $$ for any symmetric positive-definite $C$. $C$'s eigenvalues are all positive; let them be $\lambda_i$. Then $$ \det{(tI+(1-t)C)} = \prod_i (t+(1-t)\lambda_i). $$ Now, $t+(1-t)\lambda_i > \lambda_i^{1-t}>0$ by the weighted AM–GM inequality (or applying Jensen's inequality to $\log$) and the result follows since $\det{C} = \prod_i \lambda_i$.

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  • $\begingroup$ Thanks, that was helpful! $\endgroup$
    – user528502
    Commented Jun 12, 2018 at 5:25

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