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I have a mapping $$\theta:S\to \mathbb C $$ where $\mathbb C$ is the field of complex numbers and $$S=\left[\begin{array}{cc} a & -b \\b & a\end{array}\right]$$ The mapping is defined by $$\theta\Bigg(\left[\begin{array}{cc} a & -b \\b & a\end{array}\right] \Bigg)=a+ib$$

where a and b belong to the real numbers.

How do I show injection and surjection?

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To show $\phi$ is injective, suppose $\phi( \left[\begin{array}{cc} a & -b \\b & a\end{array}\right])=\phi( \left[\begin{array}{cc} c & -d \\d & c\end{array}\right])$. That is, $a+ib=c+id$, which is true if and only if $a=c$ and $b=d$ Hence, $\left[\begin{array}{cc} a & -b \\b & a\end{array}\right]= \left[\begin{array}{cc} c & -d \\d & c\end{array}\right]$
To show $\phi$ is surjective, let $x+iy \in \mathbb{C}$. Now $\left[\begin{array}{cc} x & -y \\y & x\end{array}\right] \in S$ and $\phi(\left[\begin{array}{cc} x & -y \\y & x\end{array}\right])=x+iy$

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  • $\begingroup$ Is that all you need for surjection? $\endgroup$ – user204450 Mar 19 '15 at 15:00
  • $\begingroup$ A function $f: X \to Y$ is called surjective if for every $y \in Y$ there is an $x \in X$ so that $f(x)=y$. So, yes. $\endgroup$ – Sloan Mar 19 '15 at 15:23
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Observe that the map is linear, and $\text{ker}(\theta) = \{\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\}$, hence it's injective, and surjectivity is clear.

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  • $\begingroup$ why is the kernel significant? $\endgroup$ – user204450 Mar 18 '15 at 19:36
  • $\begingroup$ $\text{ker}(f) = \{x: f(x) = 0\}$ $\endgroup$ – DeepSea Mar 18 '15 at 19:46
  • $\begingroup$ Yeah I know that but why is it significant in the context of finding bijection? $\endgroup$ – user204450 Mar 18 '15 at 21:38
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    $\begingroup$ @user204450 $\ker f = 0$ if and only if $f$ is injective. $\endgroup$ – Robert Cardona Mar 19 '15 at 14:45
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Injectivity:

Suppose $$\phi\left(\begin{pmatrix} a & -b \\ b & a \end{pmatrix}\right) = 0 + 0i$$ What can you say about $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$? Is this sufficient (i.e. why do I only need to look at the pre-image of $0 + 0i$?)

Surjectivity:

Choose any $z = x + iy \in \Bbb C$. Can you find $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ so that

$$ \phi\left(\begin{pmatrix} a & -b \\ b & a \end{pmatrix}\right) = x + yi? $$

(In other words, how can you choose $a,b$ so this happens?)

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  • $\begingroup$ I do not know why $\endgroup$ – user204450 Mar 18 '15 at 19:36
  • $\begingroup$ for surjecticity, how can you choose those a and b? $\endgroup$ – user204450 Mar 19 '15 at 12:06

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