Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

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.

share|cite|improve this question
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
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

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 $Y^TY\equiv I$.

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.