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 $)$
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 $)$ 


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


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). 

