I need to check if if the following application is surjective or injective and justify why it is in each case

$f (u,v) = (au + bv, -\overline{b}u + \overline{a}v)$

Where $f$ is a complex function

I've tried to start from the definition of injective applications :

$f$ is injective if for for every $x_1,y_1$ we have $f(x_1) = f(y_1) \rightarrow x_1 = y_1 $

So using this definition i got here :

$$ \left\{ \begin{array}{c} au_1 + bv_1 = au_2 + bv_2 \\ -\overline{b}u_1 + \overline{a}v_1 = -\overline{b}u_2 + \overline{a}v_2 \\ f(u_1,v_1)=f(u_2, v_2) \end{array} \right. $$

But I don't know how to go further and check if $u_1 = u_2$ , also How can I check if $f$ is injective or not in this case ? what about surjectivity ?

  • $\begingroup$ $f: ? \rightarrow ?$ ? $\endgroup$ – user261263 Dec 9 '17 at 11:02
  • $\begingroup$ takes from C x C to C x C (I couldn't find how to write this properly in mathjax) $\endgroup$ – Anis Souames Dec 9 '17 at 11:03

Your map is a linear map, whise matrix with respect to the standard basis is $\left(\begin{smallmatrix}a&b\\-\overline b&\overline a\end{smallmatrix}\right)$. The determinant of this matrix is $|a|^2+|b|^2$, which is $0$ if and only if $a=b=0$. Therefore, your function is injective if and only if $a\neq0$ or $b\neq0$.

Here's a more elementary apporach. Of course, if $a=b=0$, then $f$ is neither injective nor surjective. So, suppose that $a\neq0$ or that $b\neq0$. Let us try to solve the system$$\tag{1}\left\{\begin{array}{l}au+bv=\alpha\\-\overline bu+\overline av=\beta.\end{array}\right.$$If $a=0$ (in which case we are supposing that $b\neq0$) this is just$$\left\{\begin{array}{l}bv=\alpha\\-\overline bu=\beta\end{array}\right.$$which has one and only one solution: $(u,v)=\left(\frac\alpha b,-\frac\beta{\overline b}\right)$. If $a\neq0$, then, multiplying the first equation of $(1)$ by $\overline a$, the second one by $-b$ and adding them, one gets $u=\frac{\overline a\alpha-\overline b\beta}{|\alpha|^2+|\beta|^2}$. So, there is only one possible $u$ and now it is easy to get from $(1)$ that there is only one possible $v$. Therefore $f$ is bijective.

  • $\begingroup$ is there another way to determine this without using matrix ? Also how do i check if it's surjective ? $\endgroup$ – Anis Souames Dec 9 '17 at 11:05
  • $\begingroup$ @AnisSouames I've added a more elementary anwser. $\endgroup$ – José Carlos Santos Dec 9 '17 at 11:15
  • $\begingroup$ Thank's but I cant understand $\alpha$ and $\beta$ are equals to what here ? Also what's the logic you used to prove that it's injective and surjective at the sametime ? $\endgroup$ – Anis Souames Dec 9 '17 at 11:32
  • $\begingroup$ @AnisSouames Let $f$ be a function from $A$ into $B$. Then $f$ is injective if and only if no equation $f(x)=b$ has more than one solution and $f$ is surjective if and only if every equation $f(x)=b$ has a solution. Therefore, $f$ is both injective and surjective if and only if each equation $f(x)=b$ has one and only one solution. $\endgroup$ – José Carlos Santos Dec 9 '17 at 12:01

Use the notations $p=u_1-u_2, q=v_1-v_2$ then solve the resulting two equations system. You'll find that the only solution is $(0,0)$ iff $a \ne 0$ or $b \ne 0$. For surjectivity you have a similar equations system.


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