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!
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 $\det{C} = \prod_i \lambda_i$.