[Math] Upper and lower bounds log determinant

Question

I found an inequality in Wikipedia that i want to know how to prove it.

For a positive definite matrix A, the trace operator gives the following tight lower and upper bounds on the log determinant.

$tr(I- A^{-1}) ≤ \log\det(A) ≤ tr(A – I)$.

Answer 1

Assuming that $\lambda_1,\ldots,\lambda_n\in\mathbb{R}^+$ are the eigenvalues, we are just claiming:

$$\sum_{k=1}^{n}\left(1-\frac{1}{\lambda_k}\right) \leq \sum_{k=1}^{n}\log(\lambda_k)\leq \sum_{k=1}^{n}\left(\lambda_k-1\right)\tag{1}$$ or, by setting $\lambda_k=e^{u_k}$: $$\sum_{k=1}^{n}\left(1-e^{-u_k}\right) \leq \sum_{k=1}^{n} u_k \leq \sum_{k=1}^{n}\left(e^{u_k}-1\right)\tag{2}$$ that trivially follows from the convexity of the exponential function.