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Let $Y\left(t\right)$ be a matrix function, then does $\frac{d}{dt}Y^{T}Y=O$ necessarily imply $Y^{T}\left(t\right)Y\left(t\right)\equiv I$ ? Why or why not? Here $Y^{T}$ the transpose of $Y$ , $O$ zero matrix and $I$ the identity matrix.

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What did you learn? What have you tried? –  Jack Feb 29 '12 at 21:57
    
Have you tried a 2 by 2 case to see what may happen? –  Jack Feb 29 '12 at 21:58
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Do you have any kind of initial condition? Otherwise the answer is trivially no: If $X$ is any matrix with the property that $X^tX\neq I$, then setting $Y(t) = X$ (constant matrix) gives a "no" answer. –  Jason DeVito Feb 29 '12 at 22:02
    
@Jack Let's exclude the trivial case when $Y\left(t\right)=$const. At first sight, it appears that $Y^{T}Y$ can be any constant matrix, but what I am hoping is that somebody can give me a proof that this is not the case, that the only choice is the identity matrix. I konw there is something special about $Y^{T}Y$, such as symmetric, so definitely not any constant matrix ! –  Yuanwei Feb 29 '12 at 23:30
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The correct conclusion from $\frac{d}{dt}(\text{something})=0$ is that $\text{(something)}$ does not depend on $t$. This is all one can get, because any function independent of $t$ has zero derivative with respect to $t$.

But if it is also known that $Y(t_0)$ is orthogonal for some $t_0$, then from $Y^T(t_0)Y(t_0)=I$ it indeed follows that $Y^TY\equiv I$.

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