0

Show that the function $f: \mathbb{R}\times\mathbb{R}\to \mathbb{R}\times\mathbb{R}$ be defined by $f(x, y) = (x+y, x-y)$ is a bijection.

I tries this by matrix inverse method but i am not able to solve the problem.

Please someone guide me to the solution.

Asaf Karagila
  • 393,674
Prasoon Shukla
  • 1
  • 3
    There are many ways to do that, probably one of the easiest is to show that $f(x,y)=(0,0)$ has the unique solution $(x,y)=(0,0)$. – leoli1 Jan 07 '21 at 11:18
  • 1
    Let $a=x+y$ and $b=x-y$. If you add them, you can solve for $x$ in terms of $a$ and $b$. Similarly, try to find $b$ in terms of $x$ and $y$. What does this tell you about the existence of an inverse function? What can you now deduce about bijectivity of $f$? – peek-a-boo Jan 07 '21 at 11:19
  • @leoli1 yes, that's an elegant way. The kernel of the transform is {0} iff it is 1-1. – Hank Igoe Jan 07 '21 at 11:24
  • 1
    You can check this - similar problem https://math.stackexchange.com/questions/1104870/prove-that-fx-y-2xy-x4y-is-bijective – Math Lover Jan 07 '21 at 11:26
  • Interesting aside. $f(x, y) = (x+y, x-y)$ on $\mathbb Z \times \mathbb Z$ is not bijective. – GEdgar Jan 07 '21 at 12:09

2 Answers2

1

  1. Injectivity: $f(x_1,y_1)=f(x_2,y_2)\implies(x_1+y_1,x_1-y_1)=(x_2+y_2,x_2-y_2)$. So $(x_1+y_1)+(x_1-y_1)=(x_2+y_2)+(x_2-y_2)$ and $(x_1+y_1)-(x_1-y_1)=(x_2+y_2)-(x_2-y_2)$.

  2. Surjectivity: For $(k,m)\in\Bbb R^2,f\left(\frac{k+m}2,\frac{k-m}2\right)=(k,m)$.

Alternatively note that $f$ is a linear transformation with the matrix \begin{bmatrix}1&1\\1&-1\end{bmatrix} which is invertible.

Shubham Johri
  • 17,659
1

Solution using linear algebra: I suggest the notation $\mathbb{R}^{2}$ instead of $\mathbb{R}\times \mathbb{R}$, but as Arthur pointed out, there is no need to change the notation. Now returning to the problem in your question, note that the $f$ is a linear map and the $\color{blue}{\text{matrix representation}}$ is $$[f]=\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}$$ and we can see that $$\det([f])=\det \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}=(-1\cdot 1)-(1\cdot 1)=-1-1=-2\not=0.$$ So, $[f]$ is an $\color{blue}{\text{invertible matrix}}$ and therefore $f$ is a $\color{blue}{\text{bijective function}}$.

David C. Ullrich
  • 89,985
  • 1
    I see no reason to require $\Bbb R^2$ over $\Bbb R\times \Bbb R$. They mean the exact same thing (at least for linear algebra purposes). – Arthur Jan 07 '21 at 11:26
  • Dear @Arthur of course $\Bbb R^{n}\times \Bbb R^{m}$ is the same that $\Bbb R^{n+m}$, but in linear algebra the notation of $\mathbb{R}^{2}$ is more common than (at least for me) $\mathbb{R}\times \mathbb{R}$. However, thank you very much for your observation. –  Jan 07 '21 at 11:35