For an $n\times n$ positive definite matrix $A$, I wish to prove that
$$\det(A) \leq \bigg(\frac{Trace(A)}{n}\bigg)^n$$
To me this seems some form of AM-GM Inequality (Arithmatic Mean-Geometric Mean Inequality). Therefore If I can show the following, above inequality follows :
$$\det(A) \leq \prod_{i=1}^{i=n} A_{ii}$$
Any idea how to prove the above.
Thanks
Best Answer
Let $\,\lambda_1,...,\lambda_n\,$ be the matrix's eigenvalues (perhaps in some field extension of the original one), which are all positive (of course, it is customary to consider only Hermitian, or symmetric, matrices when defining positive definite), then
$$\det A=\prod_{k=1}^n \lambda_k\,\,\,,\,\,\,\operatorname{tr.}A=\sum_{k=1}^n\lambda_k$$
Thus we're required to prove
$$\prod_{k=1}^n\lambda_k\leq\left(\frac{\sum_{k=1}^n\lambda_k}{n}\right)^n\Longleftrightarrow \sqrt[n]{\prod_{k=1}^n\lambda_k}\leq \,\frac{1}{n}\sum_{k=1}^n\lambda_k$$
which is precisely the AM-GM inequality, as you mentioned.