# Upper Bound on Determinant of Matrix in terms of Trace

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

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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.

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Which of the following always holds(if any)? Without Positive Definite Assumption. 1) Det(A) is equal to the product of its eigen values. 2) Trace(A) is equal to the sum of its eigen values. –  damned Sep 26 '12 at 5:15
Both of them are true always, i.e.: for any square matrix over any field. This follows from the fact that any square matrix is similar to its Jordan Canonical Form. –  DonAntonio Sep 26 '12 at 8:33

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.

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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".

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