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I need some help proving bijections:

Suppose f is a function from $$ \mathbb R^2 \rightarrow \mathbb R^2$$

Defined by

$$f(x,y) = (ax-by,bx+ay)$$

Where a,b are numbers with $$ a^2 + b^2 \neq 0 $$

Prove that f is a bijection.

I understand that a function f is a bijection if it is both an injection and a surjection so I would need to prove both of those properties.

Could you give me a hint on how to start proving injection and surjection?

Thanks.

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First lets show that $f$ is onto $\mathbb{R}^2$. Let $(X,Y)\in \mathbb{R}^2$ then $f(x,y) = (X,Y)$ requires

$$ax-by = X$$ $$bx+ay = Y$$

which is solved by $y = \frac{aY-bX}{a^2 + b^2}$ and $x = \frac{bY+aX}{a^2+b^2}$.

Next lets show that $f$ is $1$-to-$1$. Assume $f(x_1,y_1) = f(x_2,y_2)$ then

$$ax_1-by_1 = ax_2-by_2$$ $$bx_1+ay_1 = bx_2+ay_2$$

Multiplying the equations with $b$ and $a$ respectively gives us

$$ab(x_1-x_2) = b^2(y_1-y_2)$$ $$ab(x_1-x_2) = a^2(y_2-y_1)$$

Now subtracting the equations gives $$(a^2+b^2)(y_1-y_2) = 0$$

so $y_1=y_2$ and it follows that $x_1=x_2$. Hence $f$ is a bijection.

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  • $\begingroup$ I feel kind of dumb but how'd you figure out the solutions for $$ax - by = X$$ and $$ bx + ay = Y $$ $\endgroup$ – Pandamonium Sep 23 '14 at 1:17
  • $\begingroup$ @AlexWong 1) Multiply the first eq. by $a$ and the second eq. by $b$ and add them. This gets rid of $y$ and you can solve for $x$. 2) Multiply the first eq. by $b$ and the second eq. by $a$ and subtract them. This gets rid of $x$ and you can solve for $y$ $\endgroup$ – Winther Sep 23 '14 at 2:13
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Injection:

$f(x,y)=(0,0) \iff (S)\left\{\begin{array}{l}ax-by=0\\bx+ay=0 \end{array}\right.$

Since the determinant of the linear system $(S)$ is $\Delta=a^2+b^2$ then $\Delta \neq 0$ and $(S)$ is a Cramer system, we have $x=y=0$

Now we have $f(x_1,y_1)=f(x_2,y_2) \iff f(x_1-x_2,y_1-y_2)=(0,0) \iff x_1-x_2=y_1-y_2=0$

For $X_1=(x_1,y_1)$ and $X_2=(x_2,y_2)$, we have : $$f(X_1)=f(X_2) \iff X_1=X_2$$

Surjection:

The linear system: $$(S_{u,v})\left\{\begin{array}{l}ax-by=u\\bx+ay=v \end{array}\right.$$ for $u$ and $v$ given is a Cramer system and has unique solution.

Remark: Since $S_{u,v}$ has unique solution we don't need to proof injection. It suffices to say that every $(u,v)$ has a unique antecedent by $f$

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