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Here's a little problem I ran across and could use a hint or two on:

Let $U \subset \mathfrak{R}^{n}$ be open, $f:U \rightarrow \mathfrak{R}^{m}$ differential on $U$ and satisfying $\|f(X)\|=1$ on $U$. Then $Df(X)^{T}f(X)=0$ on $U$, where $Df(X)^{T}$ is the transpose of the Jacobian matrix of $f$ at $X$.


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up vote 2 down vote accepted

Your hypothesis is equivalent to $\sum_i f_i^2(x)=1$ for all $x\in U$. Taking the partial derivative with respect to $x_j$ we get $\sum_i 2f_i(x)\frac{\partial f_i}{\partial x_j}(x)=0$. Now, $Df(x)^t f(x)$ is a vector whose $j$-th coordinate is $\sum_i \frac{\partial f_i}{\partial x_j}(x) f_i(x)$. The previous computations imply that it is zero.

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