4
$\begingroup$

I have been trying to prove this, but I am having trouble understanding how to prove the following mapping I found is injective and surjective. Just as a side note, I am trying to show the complex ring is isomorphic to special $2\times2$ matrices in regard to matrix multiplication and addition. Showing these hold is simple enough.

$$\phi:a+bi \rightarrow \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$$

This is what I have so far:

Injective: I am also confused over the fact that there are two operations, and in turn two neutral elements (1 and 0). Showing that the kernel is trivial is usually the way I go about proving whether a mapping is injective, but I just can't grasp this.

$$ \phi(z_1) = \phi(z_2) \implies \phi(z_1)\phi(z_2)^{-1} = I = \phi(z_1)\phi(z_2^{-1}) = \phi(z_2)\phi(z_2^{-1}).$$

So if we can just show that the kernel of $\phi$ is trivial, then it also shows that $z_1 = z_2$. The only complex number that maps to the identity matrix is one where $a = 1$ and $b = 0$, $a + bi = 1 + 0i = 1$.

Using a similar argument for addition we can just say that the only complex number $z$ such that $\phi(z) = 0\text{-matrix}$, is one where $a=0$ and $b=0$, $a+bi=0+0i=0$.

Surjective:

I forgot to add this before I posted, but I honestly don't really understand how to prove this because it just seems so obvious. All possible $2\times2$ matrices of that form have a complex representation because the complex number can always be identified by its real parts and since the elements of the $2\times2$ matrix are real then the mapping is obviously onto.

I have always had trouble understanding when I can say that I have "rigorously" proved something, so any help would be appreciated!

$\endgroup$
1
$\begingroup$

You could prove injectivity simply by showing $$z\ne w \Rightarrow \phi(z) \ne \phi(w)$$ wich is next to obvious.
Regarding surjectivity: A function is always surjective on its range. All you need to show here is that any Matrix $\pmatrix{a&-b\\b&a}$ is in the range of $\phi$, explicitly $a+bi$ is the required argument (sounds like nothing to show, because it's simply the definition of $\phi$ wich guarantees this)

$\endgroup$
1
$\begingroup$

For injectivity assume that for some $(a,b), (a',b') \in \mathbb C$:

$$ \phi (a,b) = \left (\begin{array}{cc} a & -b \\ b & a \end{array}\right ) = \left (\begin{array}{cc} a' & -b' \\ b' & a' \end{array}\right ) = \phi(a',b')$$

Then since two matrices are equal if and only if each entry us equal it follows that $a=a'$ and $b=b'$ hence $\phi$ is injective.

Your argument for surjectivity is good: Let $ \left (\begin{array}{cc} a & -b \\ b & a \end{array}\right ) \in M_2(\mathbb R)$. Then $\phi$ maps $(a,b)$ to it.

$\endgroup$
1
$\begingroup$

The map is bijective by the argument you give; the trick is showing that it respects addition and (especially) multiplciation. Here's another way of doing so: Consider $\mathbb{C}$ as a $2$-dimensional real vector space with basis $\{1, i\}$. You then have a map $f:\mathbb{C} \to M_2(\mathbb{R})$ (i.e., $2$-dimensional real matrices) defined by $f(z)w = zw$ (with respect to this basis). It's clear that this map preserves addition and multiplication. If you write out $f$ explicitly, it's exactly the map $a + bi \to \begin{pmatrix} a & b\\-b & a\end{pmatrix}$ you describe (at least modulo a sign flip).

$\endgroup$
1
$\begingroup$
  • For injectivity you need to show that if $\phi(z_1)=\phi(z_2)$ then $z_1=z_2$. So assume that $$\phi(a+bi)=\begin{pmatrix}a&-b\\b&a\end{pmatrix}=\begin{pmatrix}a'&-b'\\b'&a'\end{pmatrix}=\phi(a'+b'i)$$ then $$\begin{pmatrix}a-a'&-b+b'\\b-b'&a-a'\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$ so $a=a'$, and $b=b'$ which means that $a+bi=a'+b'i$.

  • For surjectivity all the matrices are on the form $$\begin{pmatrix}a&-b\\b&a\end{pmatrix}$$ with $a,b\in\mathbb{R}$, and the element $a+bi$ maps to it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.