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Let $T:\mathbb R^n \to \mathbb R^n $ be an isometry and $T(0)=0$ , then $T$ is linear and $T(B[0,1])\subseteq B[0,1]$ so $T:B[0,1]\to B[0,1]$ is an isometry and since $B[0,1]$ is compact so $T|_{B[0,1]}$ is surjective ( It is well known that if $K$ is a compact metric space and $f:K \to K$ is isometry then $f$ is surjective ) i.e. $T(B[0,1])=B[0,1]$ , so then $T$ is surjective . My question is , suppose we drop the assumption $T(0)=0$ , Is $T$ still surjective ?

Please help . Thanks in advance

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    What is the difficulty in taking the compact $K$ any closed ball $\overline{B_R(0)}$ rather than just only $\overline{B_1(0)} $ as you did? (Assuming that the theorem you cite is correct) – guestDiego Jun 17 '16 at 13:16
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    http://people.math.gatech.edu/~ghomi/LectureNotes/LectureNotes2U.pdf Might be of your interest. – Ningxin Jun 17 '16 at 13:19
  • In the first sentence, you have "and $T(0)$". Do you mean "and $T(0)=0$"? – robjohn Jun 17 '16 at 14:09
  • @robjohn : yes , thank you for pointing out :) –  Jun 17 '16 at 14:37
  • Related (for the claim "then $T$ is linear"): http://math.stackexchange.com/questions/194538 – Watson Sep 01 '16 at 10:43

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You already know that $T$ is linear if $T(0) = 0$. However, consider the maps $$ S(x) = T(x) - T(0)\\ R(x) = x + T(0) $$ Then clearly both $R$ and $S$ are isometries with $T = R \circ S$. However, since $S(0) = 0$, $S$ is linear, and by the rank-nullity theorem any injective linear map is surjective. Moreover, the translation $R$ is clearly surjective. Thus, the composition $R \circ S = T$ must also be surjective.

So, $T$ is indeed necessarily surjective.

Ben Grossmann
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  • Not every linear mao is surjective , but any linear injective map is surjective and indeed , any isometry is injective –  Jun 17 '16 at 13:30
  • @user228169 yes, that's what I meant. Hence "rank nullity theorem" – Ben Grossmann Jun 17 '16 at 13:31
  • very clever solution , it doesn't even need to use the fact about compact metric spaces –  Jun 17 '16 at 13:32