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Let $$ [x,y] = T[u,v]=[u]=[u^2-v^2,2uv] \\$$

  1. Show that T is injective on R.
  2. Sketch the square image by T.
  3. Sketch thequare image by $$ \left.dT\right|_{(1,1)}(u,v)$$ This image is a square?
  4. Show that translation of the image found by vector $$ T(1,1)- \left.dT\right|_{(1,1)}(1,1) $$ is the square image by linear approximation $$A(uv)$$ to near $(1,1)$.

I tried to do and I did not succeed. If you can help me with any items, I'll be grateful.

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  • $\begingroup$ Welcome Felipe. Try in your future post to ask just one question. $\endgroup$
    – Piquito
    Nov 24, 2016 at 13:05
  • $\begingroup$ Thank you. I might have done it, but it seems there is a dependency from one issue to another. In this way it is more consistent to leave the issues and expect someone to "help with any items". $\endgroup$ Nov 24, 2016 at 17:45

1 Answer 1

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I understand that the question is about the map $$T:\quad(u,v)\mapsto (x,y)=T(u,v):=(u^2-v^2,2uv)\ ,$$ restricted to the given square $Q$.

${\bf 1\ }$ Assume we have two points $(u_i, v_i)\in Q$ with $T(u_1,v_1)=T(u_2,v_2)=:(x,y)$. Then $$x^2+y^2=(u_i^2-v_i^2)^2+4u_i^2v_i^2=(u_i^2+v_i^2)^2\qquad(i=1, 2)\ .$$ This allows to conclude that $u_1^2+v_1^2=u_2^2+v_2^2$, and together with $u_1^2-v_1^2=x=u_2^2-v_2^2$ it follows that $u_1^2=u_2^2$, $v_1^2=v_2^2$. Since $Q$ is in the first quadrant we may take square roots here, and obtain $(u_1,v_1)=(u_2,v_2)$.

${\bf 2\ }$ The following picture incorporates also part ${\bf 4}$:

enter image description here

${\bf 3\ }$ The Jacobian is $$dT(u,v)=\left[\matrix{2u&-2v\cr 2v&2u\cr}\right]\ ,$$ hence $$dT(1,1)=\left[\matrix{2&-2\cr 2&2\cr}\right]=\sqrt{8}\left[\matrix{{1\over\sqrt{2}}&{-1\over\sqrt{2}}\cr {1\over\sqrt{2}}&{1\over\sqrt{2}}\cr}\right]\ .$$ The last matrix describes a counterclockwise rotation by $45^\circ$, followed by a scaling. Therefore it transforms squares into squares.

${\bf 4\ }$ The map $$(u,v)\mapsto T(1,1)+dT(1,1).(u-1,v-1)$$ is an approximation to $T$ in the neighborhood of $(1,1)$ and maps $Q$ to some euclidean square $Q'$ which "approximates" the actual image $T(Q)$.

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  • $\begingroup$ Hi, Mr. Christian. Good job! Fantastic your idea in tem 1. I had tried it another way, trying to get different images for u and u', v and v'. I could not come toa conclusion with 100% certainty.In the third item would derive dT in relation to variable u? $\endgroup$ Nov 24, 2016 at 17:56

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