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Let $f:\mathbb{R}^2\to \mathbb{R}^2$ be defined by the equation \begin{equation} f(x,y)=(x^2-y^2,2xy)\end{equation}

Parts (a) and (b) of the problem asked me to show that $f$ is one-to-one on the set $A$ consisting of all $(x,y)$ with $x>0$, and to find set $B=f(A)$, which was easy enough to do. Part (c) asks

If $g$ is the inverse function, find $Dg(0,1)$.

What I thought to do was to find the matrix $Df$ and find the inverse of it, but I am finding that very challenging to do (here I thought I had my linear algebra skills in check!) Because what I get is that

$Df= \begin{bmatrix} 2x &2y\\ -2y & 2x \end{bmatrix}$

And when I try to find the inverse, i get to the step

$ \begin{bmatrix} x & y & 1/2 & 0\\ 0 & \frac{x^2+y^2}{x} & \frac{y}{2x} & 1/2 \end{bmatrix}$

And i feel like maybe this is not the correct method to finding the inverse...?

Thank you!

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  • $\begingroup$ Do you know the formula for the inverse of a $2 \times 2$ matrix? $\endgroup$ Commented Jun 25, 2017 at 7:10
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    $\begingroup$ Hint : in complex coordinate $f(z) = z^2$ $\endgroup$
    – user171326
    Commented Jun 25, 2017 at 7:10
  • $\begingroup$ @N.H. Thanks for the hint. I got the correct answer using your method, but it's weird that when I use the other method being suggested that I do not get the right answer. $\endgroup$ Commented Jun 25, 2017 at 7:28

2 Answers 2

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First, fix the differential ($f(x,y)=(u(x,y),v(x,y))\;;u=x^2-y^2\;;v=2xy$):

$$Df= \begin{pmatrix} \partial u/\partial x &\partial u/\partial y\\ \partial v/\partial x & \partial v/\partial y \end{pmatrix}= \begin{pmatrix} 2x &-2y\\ 2y & 2x \end{pmatrix}$$

Now, with the invaluable aid of the commentators, the inverse is

$$Dg(f(x,y))=\dfrac{1}{\vert Df\vert}\begin{pmatrix}2x &2y\\ -2y & 2x\end{pmatrix}=\dfrac{1}{4(x^2+y^2)}\begin{pmatrix}2x &2y\\ -2y & 2x\end{pmatrix}=$$

$$=\dfrac{1}{2(x^2+y^2)}\begin{pmatrix}x &y\\ -y & x\end{pmatrix}$$

We have to find the values for wich $f(x,y)=(0,1)$, so is, $x^2-y^2=0$ and $2xy=1\implies (x,y)=(1/\sqrt{2},1/\sqrt{2})$ or $(x,y)=(-1/\sqrt{2},-1/\sqrt{2})$. But $x\gt0$, so we drop the solution with the negatives. Then,

$$Dg(0,1)=Dg(f(1/\sqrt{2},1/\sqrt{2}))=\dfrac{1}{2}\begin{pmatrix}\sqrt{2}/2 &\sqrt{2}/2\\-\sqrt{2}/2 &\sqrt{2}/2\end{pmatrix}=$$

$$Dg(0,1)=\dfrac{1}{4}\begin{pmatrix}\sqrt{2} &\sqrt{2}\\-\sqrt{2} & \sqrt{2}\end{pmatrix}$$

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The inverse of a $2$x$2$ matrix is as follows $$\begin{bmatrix} a & b\\c & d\end{bmatrix}^{-1} = \frac{1}{ad-bc} \begin{bmatrix}d &-b\\ -c & a\end{bmatrix}$$

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  • $\begingroup$ Ok I guess the reason I didnt think it was that is because my professor gave me the answer to be $ 1/4 \begin{bmatrix} \sqrt 2 & \sqrt 2 \\ -\sqrt 2 & \sqrt 2 \end{bmatrix}$ which doesnt match... $\endgroup$ Commented Jun 25, 2017 at 7:19
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    $\begingroup$ You're forgetting to divide by the determinant! That's just the adjugate. $\endgroup$ Commented Jun 25, 2017 at 7:41
  • $\begingroup$ Yea, I still don't get the correct answer guys. I appreciate your help, but I don't understand what I'm doing wrong. $\endgroup$ Commented Jun 25, 2017 at 7:45
  • $\begingroup$ When I do it this way, I get the inverse of $Df$ to be $\frac{1}{2x^2+2y^2} \begin{bmatrix} 2x& 2y\\ -2y & 2x\end{bmatrix}$, and when evaluated at $(0,1)$, it is incorrect. $\endgroup$ Commented Jun 25, 2017 at 7:47
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    $\begingroup$ @jgcello I guess you shall evaluate that at $(1/\sqrt{2}, 1/\sqrt{2})$? The inverse function thm says if $g$ is the inverse, then $Dg(f(a))=[Df(a)]^{-1}$. $\endgroup$ Commented Jun 25, 2017 at 8:36

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