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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 $)$

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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? –  Martin 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
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2 Answers

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 $\sqrt{A^T A}$.)

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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).

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