Show $\log(\det(A))\le \operatorname{tr}(A)-n$ 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.

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 $x>0$
