Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your second inequality is a consequence of the Hadamard inequality, see here:

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".

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.