Upper Bound on Determinant of Matrix in terms of Trace

Question

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

Answer 1

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.

Answer 2

If your definition of a positive-definite matrix includes that it's symmetric, then it's also diagonalizable; the diagonalization leaves the trace and determinant invariant, and then you can apply your idea.

Answer 3

Your second inequality is a consequence of the Hadamard inequality, see here: http://en.wikipedia.org/wiki/Hadamard_inequality

Another way is to use the result that the eigenvalue vector of a Hermitian (hence real symmetric) matrix majorizes the vector of diagonal values, and then using that the product function $\prod_{i=1}^n x_i$ for $x_i>0$ is Schur-convex:

One source for this results is the book by Bhatia: "Matrix Analysis".