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let $T$ be a continuously differentiable transformation from $3$-space to $3$-space.

then clearly, $J(p)$ which is the jacobian at point $p$ is continuous because each component in $dT$is continuous as given in the question. But what about the rank of $dT$? Seems like it is not , but can someone give a proof?

  • Related: http://math.stackexchange.com/questions/1913394/how-to-prove-that-the-rank-of-a-matrix-is-a-lower-semi-continuous-function – Aloizio Macedo Apr 09 '17 at 21:02

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Consider $T(x,y,z)=(x^2,y^2,z^2)$. The rank of its differential is discontinuous at $0$.

Tsemo Aristide
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