MathOverflow is a question and answer site for professional mathematicians. 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

Suppose you have an nxn parametric square matrix A(t). I am wondering if I can prove this:

$(lim\_{t\rightarrow\infty}det(A(t)) \ne 0) \Rightarrow (\int^{\infty}A^T(t)A(t)dt$ Does not Converge $)$

share|cite|improve this question
If the integral converges, the $t\mapsto A(t)$ is square integrable, and therefore $\int^{+\infty}|\det A(t)|^{2/n}dt<\infty$. If the determinant has a limit, this limit has to be zero. – Denis Serre Mar 1 '12 at 12:20
I think you need either a decay condition for $t \rightarrow -\infty$, or start the integral at, say, $0$, don't you? – shuhalo Mar 1 '12 at 14:22
@Martin: Looking at the first version of the question I think the OP means an integral over $[0,\infty)$ or so, but in any case the convergence of the integral here is the assumption (writing my answer I did not notice Denis' one, which is better and more concise). – Pietro Majer Mar 2 '12 at 7:31

Yes, because for any square matrix $A$ there holds $|\operatorname{det}A|^{2/n}\le \frac{1}{n}\operatorname{tr}(A^T A)$. (The inequality is just an instance of the inequality of geometric and arithmetic means in the particular case of a positive diagonal matrix; for a general matrix $A$ you may reduce to the particular case diagonalizing $\sqrt{A^T A}$.)

share|cite|improve this answer

Just a comment. The questions in the post and in the title are different, and the answers also.

The answer to the question in the tile is NO, since $\det( A(t)) $ does not converge to 0 does not imply that $\det(A(t))$ converges (and a square integrable function does not have to tend to zero at infinity if the limit does not exist).

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.